The fraction is called correct if the highest power of the numerator is less than the highest power of the denominator. The integral of a proper rational fraction has the form:

$$ \int \frac(mx+n)(ax^2+bx+c)dx $$

The formula for integrating rational fractions depends on the roots of the polynomial in the denominator. If the polynomial $ ax^2+bx+c $ has:

1. Only complex roots, then it is necessary to select a full square from it: $$ \int \frac(mx+n)(ax^2+bx+c) dx = \int \frac(mx+n)(x^2 \pm a ^2) $$
2. Different real roots $ x_1 $ and $ x_2 $, then you need to expand the integral and find the indefinite coefficients $ A $ and $ B $: $$ \int \frac(mx+n)(ax^2+bx+c) dx = \int \frac(A)(x-x_1) dx + \int \frac(B)(x-x_2) dx $$
3. One multiple root $ x_1 $, then we expand the integral and find the indefinite coefficients $ A $ and $ B $ for this formula: $$ \int \frac(mx+n)(ax^2+bx+c) dx = \int \frac(A)((x-x_1)^2)dx + \int \frac(B)(x-x_1) dx $$

If the fraction is wrong, that is, the highest degree in the numerator is greater than or equal to the highest degree of the denominator, then first it must be reduced to correct mind by dividing the polynomial from the numerator by the polynomial from the denominator. In this case, the formula for integrating a rational fraction is:

$$ \int \frac(P(x))(ax^2+bx+c)dx = \int Q(x) dx + \int \frac(mx+n)(ax^2+bx+c)dx $$

Solution examples

Example 1

Find the integral of a rational fraction: $$ \int \frac(dx)(x^2-10x+16) $$

Decision

The fraction is regular and the polynomial has only complex roots. Therefore, we select a full square: $$ \int \frac(dx)(x^2-10x+16) = \int \frac(dx)(x^2-2\cdot 5 x+ 5^2 - 9) = $$ We collapse the full square and sum under the differential sign $ x-5 $: $$ = \int \frac(dx)((x-5)^2 - 9) = \int \frac(d(x-5))((x-5)^2-9) = $$ Using the table of integrals, we get: $$ = \frac(1)(2 \cdot 3) \ln \bigg | \frac(x-5 - 3)(x-5 + 3) \bigg | + C = \frac(1)(6) \ln \bigg |\frac(x-8)(x-2) \bigg | +C$$ If you cannot solve your problem, then send it to us. We will provide a detailed solution. You will be able to familiarize yourself with the progress of the calculation and gather information. This will help you get a credit from the teacher in a timely manner!

Answer

$$ \int \frac(dx)(x^2-10x+16) = \frac(1)(6) \ln \bigg |\frac(x-8)(x-2) \bigg | +C$$

Example 2

Integrate rational fractions: $$ \int \frac(x+2)(x^2+5x-6) dx $$

Decision

Solve the quadratic equation: $$ x^2+5x-6 = 0 $$ $$ x_(12) = \frac(-5\pm \sqrt(25-4\cdot 1 \cdot (-6)))(2) = \frac(-5 \pm 7)(2) $$ Let's write down the roots: $$ x_1 = \frac(-5-7)(2) = -6; x_2 = \frac(-5+7)(2) = 1 $$ Taking into account the obtained roots, we transform the integral: $$ \int \frac(x+2)(x^2+5x-6) dx = \int \frac(x+2)((x-1)(x+6)) dx = $$ We perform the expansion of a rational fraction: $$ \frac(x+2)((x-1)(x+6)) = \frac(A)(x-1) + \frac(B)(x+6) = \frac(A(x -6)+B(x-1))((x-1)(x+6)) $$ Equate the numerators and find the coefficients $ A $ and $ B $: $$ A(x+6)+B(x-1)=x+2 $$ $$ Ax + 6A + Bx - B = x + 2 $$ $$ \begin(cases) A ​​+ B = 1 \\ 6A - B = 2 \end(cases) $$ $$ \begin(cases) A ​​= \frac(3)(7) \\ B = \frac(4)(7) \end(cases) $$ We substitute the found coefficients into the integral and solve it: $$ \int \frac(x+2)((x-1)(x+6))dx = \int \frac(\frac(3)(7))(x-1) dx + \int \frac (\frac(4)(7))(x+6) dx = $$ $$ = \frac(3)(7) \int \frac(dx)(x-1) + \frac(4)(7) \int \frac(dx)(x+6) = \frac(3) (7) \ln |x-1| + \frac(4)(7) \ln |x+6| +C$$

Answer

$$ \int \frac(x+2)(x^2+5x-6) dx = \frac(3)(7) \ln |x-1| + \frac(4)(7) \ln |x+6| +C$$

As I already noted, in integral calculus there is no convenient formula for integrating a fraction. And therefore, there is a sad trend: the more “fancy” the fraction, the more difficult it is to find the integral from it. In this regard, one has to resort to various tricks, which I will now discuss. Prepared readers can immediately use table of contents:

• The method of subsuming under the sign of the differential for simple fractions

Numerator Artificial Transformation Method

Example 1

By the way, the considered integral can also be solved by the change of variable method, denoting , but the solution will be much longer.

Example 2

Find the indefinite integral. Run a check.

This is a do-it-yourself example. It should be noted that here the variable replacement method will no longer work.

Attention important! Examples No. 1, 2 are typical and are common. In particular, such integrals often arise in the course of solving other integrals, in particular, when integrating irrational functions (roots).

The above method also works in the case if the highest power of the numerator is greater than the highest power of the denominator.

Example 3

Find the indefinite integral. Run a check.

Let's start with the numerator.

The numerator selection algorithm is something like this:

1) In the numerator I need to organize , but there . What to do? I enclose in brackets and multiply by: .

2) Now I try to open these brackets, what happens? . Hmm ... already better, but there is no deuce with initially in the numerator. What to do? You need to multiply by:

3) Opening the brackets again: . And here is the first success! Needed turned out! But the problem is that an extra term has appeared. What to do? In order for the expression not to change, I must add the same to my construction:
. Life has become easier. Is it possible to organize again in the numerator?

4) You can. We try: . Expand the brackets of the second term:
. Sorry, but I actually had in the previous step, and not . What to do? We need to multiply the second term by:

5) Again, for verification, I open the brackets in the second term:
. Now it's normal: obtained from the final construction of paragraph 3! But again there is a small “but”, an extra term has appeared, which means that I must add to my expression:

If everything is done correctly, then when opening all the brackets, we should get the original numerator of the integrand. We check:
Good.

Thus:

Ready. In the last term, I applied the method of bringing the function under the differential.

If we find the derivative of the answer and bring the expression to a common denominator, then we get exactly the original integrand. The considered method of expansion into a sum is nothing more than the reverse action to bring the expression to a common denominator.

The numerator selection algorithm in such examples is best performed on a draft. With some skills, it will also work mentally. I remember a record time when I did a selection for the 11th power, and the expansion of the numerator took almost two lines of Werd.

Example 4

Find the indefinite integral. Run a check.

This is a do-it-yourself example.

The method of subsuming under the sign of the differential for simple fractions

Let's move on to the next type of fractions.
, , , (the coefficients and are not equal to zero).

In fact, a couple of cases with arcsine and arctangent have already slipped in the lesson Variable change method in indefinite integral. Such examples are solved by bringing the function under the sign of the differential and then integrating using the table. Here are some more typical examples with a long and high logarithm:

Example 5

Example 6

Here it is advisable to pick up a table of integrals and follow what formulas and as transformation takes place. Note, how and why squares are highlighted in these examples. In particular, in Example 6, we first need to represent the denominator as , then bring under the sign of the differential. And you need to do all this in order to use the standard tabular formula .

But what to look at, try to solve examples No. 7,8 on your own, especially since they are quite short:

Example 7

Example 8

Find the indefinite integral:

If you can also check these examples, then great respect is your differentiation skills at their best.

Full square selection method

Integrals of the form , (coefficients and are not equal to zero) are solved full square selection method, which has already appeared in the lesson Geometric Plot Transformations.

In fact, such integrals reduce to one of the four table integrals that we have just considered. And this is achieved using the familiar abbreviated multiplication formulas:

Formulas are applied in this direction, that is, the idea of ​​the method is to artificially organize the expressions either in the denominator, and then convert them, respectively, to or .

Example 9

Find the indefinite integral

This is the simplest example where with the term - unit coefficient(and not some number or minus).

We look at the denominator, here the whole thing is clearly reduced to the case. Let's start converting the denominator:

Obviously, you need to add 4. And so that the expression does not change - the same four and subtract:

Now you can apply the formula:

After the conversion is finished ALWAYS it is desirable to perform a reverse move: everything is fine, there are no errors.

The clean design of the example in question should look something like this:

Ready. Bringing a "free" complex function under the differential sign: , in principle, could be neglected

Example 10

Find the indefinite integral:

This is an example for self-solving, the answer is at the end of the lesson.

Example 11

Find the indefinite integral:

What to do when there is a minus in front? In this case, you need to take the minus out of brackets and arrange the terms in the order we need:. Constant("double" in this case) do not touch!

Now we add one in parentheses. Analyzing the expression, we come to the conclusion that we need one behind the bracket - add:

Here is the formula, apply:

ALWAYS we perform a check on the draft:
, which was to be verified.

The clean design of the example looks something like this:

We complicate the task

Example 12

Find the indefinite integral:

Here, with the term, it is no longer a single coefficient, but a “five”.

(1) If a constant is found at, then we immediately take it out of brackets.

(2) In general, it is always better to take this constant out of the integral, so that it does not get in the way.

(3) It is obvious that everything will be reduced to the formula . It is necessary to understand the term, namely, to get a "two"

(4) Yep, . So, we add to the expression, and subtract the same fraction.

(5) Now select a full square. In the general case, it is also necessary to calculate , but here we have a long logarithm formula , and the action does not make sense to perform, why - it will become clear a little lower.

(6) Actually, we can apply the formula , only instead of "x" we have, which does not negate the validity of the tabular integral. Strictly speaking, one step is missing - before integration, the function should have been brought under the differential sign: , but, as I have repeatedly noted, this is often neglected.

(7) In the answer under the root, it is desirable to open all the brackets back:

Complicated? This is not the most difficult in integral calculus. Although, the examples under consideration are not so much complicated as they require good calculation technique.

Example 13

Find the indefinite integral:

This is a do-it-yourself example. Answer at the end of the lesson.

There are integrals with roots in the denominator, which, with the help of a replacement, are reduced to integrals of the considered type, you can read about them in the article Complex integrals, but it is designed for highly prepared students.

Bringing the numerator under the sign of the differential

This is the final part of the lesson, however, integrals of this type are quite common! If fatigue has accumulated, maybe it is better to read tomorrow? ;)

The integrals that we will consider are similar to the integrals of the previous paragraph, they have the form: or (the coefficients , and are not equal to zero).

That is, we have a linear function in the numerator. How to solve such integrals?

The problem of finding the indefinite integral of a fractionally rational function is reduced to integrating simple fractions. Therefore, we recommend that you first familiarize yourself with the section on the theory of decomposition of fractions into simple ones.

Example.

Find the indefinite integral.

Decision.

Since the degree of the numerator of the integrand is equal to the degree of the denominator, first we select the integer part by dividing the polynomial by the polynomial by the column:

So, .

The decomposition of the obtained proper rational fraction into simple fractions has the form . Hence,

The resulting integral is an integral of the simplest fraction of the third type. Looking ahead a little, we note that it can be taken by bringing it under the differential sign.

As , then . So

Hence,

Now let's move on to describing the methods for integrating the simplest fractions of each of the four types.

Integration of the simplest fractions of the first type

The method of direct integration is ideal for solving this problem:

Example.

Find the set of antiderivatives of a function

Decision.

Let's find the indefinite integral using the properties of the antiderivative, the table of antiderivatives and the integration rule.

Top of page

Integration of the simplest fractions of the second type

The method of direct integration is also suitable for solving this problem:

Example.

Decision.

Top of page

Integration of the simplest fractions of the third type

First, we present the indefinite integral as a sum:

We take the first integral by the method of subsuming under the sign of the differential:

So,

We transform the denominator of the resulting integral:

Hence,

The formula for integrating the simplest fractions of the third type takes the form:

Example.

Find the indefinite integral .

Decision.

We use the resulting formula:

If we didn't have this formula, what would we do:

Top of page

Integration of the simplest fractions of the fourth type

The first step is to sum it up under the differential sign:

The second step is to find an integral of the form . Integrals of this type are found using recurrent formulas. (See section integrating using recursive formulas). For our case, the following recursive formula is suitable:

Example.

Find the indefinite integral

Decision.

For this type of integrand, we use the substitution method. Let's introduce a new variable (see the section on integrating irrational functions):

After substitution we have:

We came to finding the integral of a fraction of the fourth type. In our case, we have the coefficients M=0, p=0, q=1, N=1 and n=3. We apply the recursive formula:

After the reverse substitution, we get the result:

Integration of trigonometric functions

1. Integrals of the form are calculated by converting the product of trigonometric functions into a sum according to the formulas: For example, 2. Integrals of the form , where m or n- an odd positive number, are calculated by subsuming under the sign of the differential. For example, 3. Integrals of the form , where m and n- even positive numbers, are calculated using the reduction formulas: For example, 4. Integrals where are calculated by changing the variable: or For example, 5. Integrals of the form are reduced to integrals of rational fractions using a universal trigonometric substitution then (because =[after dividing the numerator and denominator by ]= ; For example, It should be noted that the use of a universal substitution often leads to cumbersome calculations.

§5. Integration of the simplest irrationalities

Consider methods for integrating the simplest types of irrationality. one. Functions of this type are integrated in the same way as the simplest rational fractions of the 3rd type: in the denominator, a full square is extracted from the square trinomial and a new variable is introduced. Example. 2. (under the sign of the integral - the rational function of the arguments). Integrals of this kind are calculated using the substitution . In particular, in integrals of the form we denote . If the integrand contains roots of different degrees: , then denote , where n is the least common multiple of the numbers m,k. Example 1 Example 2 is an improper rational fraction, select the integer part: – – – 3. Integrals of the form are calculated using trigonometric substitutions:

44

45 Definite Integral

Definite integral is an additive monotone normalized functional defined on a set of pairs, the first component of which is an integrable function or functional, and the second is an area in the set of this function (functional).

Definition

Let it be defined on . Let's break it into parts with several arbitrary points. Then we say that the segment has been partitioned Next, we choose an arbitrary point , ,

The definite integral of a function on a segment is the limit of integral sums as the partition rank tends to zero, if it exists regardless of the partition and choice of points, that is

If this limit exists, then the function is said to be Riemann integrable on.

Notation

· - lower limit.

· - upper limit.

· - integrand function.

· - length of a partial segment.

· is the integral sum of the function on the corresponding partition .

· - maximum length of a partial segment.

Properties

If a function is Riemann-integrable on , then it is bounded on it.

geometric sense

The definite integral as the area of ​​a figure

The definite integral is numerically equal to the area of ​​the figure bounded by the x-axis, straight lines and and the function graph.

Newton-Leibniz theorem

[edit]

(redirected from "Newton-Leibniz formula")

Newton - Leibniz formula or fundamental theorem of analysis gives the relation between two operations: taking a definite integral and calculating an antiderivative.

Proof

Let an integrable function be given on the segment. Let's start by noting that

that is, it does not matter which letter ( or ) is under the sign in a definite integral over the interval .

Set an arbitrary value and define a new function . It is defined for all values ​​of , because we know that if there is an integral of on , then there is also an integral of on , where . Recall that we consider by definition

(1)

notice, that

Let us show that it is continuous on the segment . Indeed, let ; then

and if , then

Thus, is continuous on regardless of whether it has discontinuities or not; it is important that it is integrable on .

The figure shows a graph. The area of ​​the variable figure is . Its increment is equal to the area of ​​the figure , which, due to the boundedness of , obviously tends to zero at regardless of whether it is a point of continuity or discontinuity, for example, a point .

Now let the function not only be integrable on , but be continuous at the point . Let us prove that then has a derivative at this point equal to

(2)

Indeed, for the given point

(1) , (3)

We put , and since the constant is relative to ,TO . Further, due to the continuity at the point, for any one can specify such that for .

which proves that the left side of this inequality is o(1) for .

Passing to the limit in (3) at shows the existence of the derivative of at the point and the validity of equality (2). Here we are talking about the right and left derivatives, respectively.

If a function is continuous on , then, based on what was proved above, the corresponding function

(4)

has a derivative equal to . Therefore, the function is antiderivative for on .

This conclusion is sometimes called the Variable Upper Limit Integral Theorem or Barrow's Theorem.

We have proved that an arbitrary continuous function on an interval has an antiderivative on this interval, defined by equality (4). This proves the existence of an antiderivative for any function continuous on an interval.

Let now be an arbitrary antiderivative of a function on . We know that , where is some constant. Assuming in this equality and taking into account that , we obtain .

Thus, . But

Improper integral

[edit]

From Wikipedia, the free encyclopedia

Definite integral called improper if at least one of the following conditions is true:

· The limit a or b (or both limits) is infinite;

· The function f(x) has one or more breakpoints inside the segment .

[edit] Improper integrals of the first kind

. Then:

1. If and the integral is called . In this case is called convergent.

, or simply divergent.

Let be defined and continuous on the set from and . Then:

1. If , then the notation and the integral is called improper Riemann integral of the first kind. In this case is called convergent.

2. If there is no final ( or ), then the integral is said to be divergent to , or simply divergent.

If the function is defined and continuous on the entire real line, then there may be an improper integral of this function with two infinite limits of integration, which is determined by the formula:

, where c is an arbitrary number.

[edit] The geometric meaning of the improper integral of the first kind

The improper integral expresses the area of ​​an infinitely long curvilinear trapezoid.

[edit] Examples

[edit] Improper integrals of the second kind

Let it be defined on , suffer an infinite discontinuity at the point x=a and . Then:

1. If , then the notation and the integral is called

is called divergent to , or simply divergent.

Let it be defined on , suffer an infinite discontinuity at x=b and . Then:

1. If , then the notation and the integral is called improper Riemann integral of the second kind. In this case, the integral is called convergent.

2. If or , then the designation is preserved, and is called divergent to , or simply divergent.

If the function suffers a discontinuity at an internal point of the segment , then the improper integral of the second kind is determined by the formula:

[edit] Geometric meaning of improper integrals of the second kind

The improper integral expresses the area of ​​an infinitely high curvilinear trapezoid

[edit] Example

[edit] Special case

Let the function be defined on the entire real axis and have a discontinuity at points .

Then we can find the improper integral

[edit] Cauchy criterion

1. Let be defined on the set from and .

Then converges

2. Let is defined on and .

Then converges

[edit] Absolute convergence

Integral called absolutely convergent, if converges.
If an integral converges absolutely, then it converges.

[edit] Conditional convergence

The integral is called conditionally convergent if converges and diverges.

48 12. Improper integrals.

When considering definite integrals, we assumed that the region of integration is bounded (more specifically, it is the segment [ a ,b ]); for the existence of a definite integral, the boundedness of the integrand on [ a ,b ]. We will call definite integrals for which both of these conditions are satisfied (boundedness of both the integration domain and the integrand) own; integrals for which these requirements are violated (i.e., either the integrand, or the domain of integration, or both, is unbounded) non-own. In this section, we will study improper integrals.

• 12.1. Improper integrals over an unbounded interval (improper integrals of the first kind).
• 12.1.1. Definition of an improper integral over an infinite interval. Examples.
• 12.1.2. The Newton-Leibniz formula for the improper integral.
• 12.1.3. Comparison criteria for non-negative functions.
• 12.1.3.1. Sign of comparison.
• 12.1.3.2. A sign of comparison in the limiting form.
• 12.1.4. Absolute convergence of improper integrals over an infinite interval.
• 12.1.5. Convergence criteria for Abel and Dirichlet.
• 12.2. Improper integrals of unbounded functions (improper integrals of the second kind).
• 12.2.1. Definition of an improper integral of an unbounded function.
• 12.2.1.1. Singularity at the left end of the interval of integration.
• 12.2.1.2. Application of the Newton-Leibniz formula.
• 12.2.1.3. Singularity at the right end of the interval of integration.
• 12.2.1.4. Singularity at an interior point of the integration interval.
• 12.2.1.5. Several singularities on the interval of integration.
• 12.2.2. Comparison criteria for non-negative functions.
• 12.2.2.1. Sign of comparison.
• 12.2.2.2. A sign of comparison in the limiting form.
• 12.2.3. Absolute and conditional convergence of improper integrals of discontinuous functions.
• 12.2.4. Convergence criteria for Abel and Dirichlet.

12.1. Improper integrals over an unbounded interval

(improper integrals of the first kind).

12.1.1. Definition of an improper integral over an infinite interval. Let the function f (x ) is defined on the half-line and is integrable over any interval [ from, implying in each of these cases the existence and finiteness of the corresponding limits. Now the solutions of the examples look more simple: .

12.1.3. Comparison criteria for non-negative functions. In this section, we will assume that all integrands are non-negative over the entire domain of definition. Until now, we have determined the convergence of the integral by calculating it: if there is a finite limit of the antiderivative with the corresponding aspiration ( or ), then the integral converges, otherwise it diverges. When solving practical problems, however, it is important first of all to establish the very fact of convergence, and only then calculate the integral (besides, the antiderivative is often not expressed in terms of elementary functions). We formulate and prove a number of theorems that allow us to establish the convergence and divergence of improper integrals of nonnegative functions without calculating them.
12.1.3.1. Comparison sign. Let the functions f (x ) and g (x ) integr

The material presented in this topic is based on the information presented in the topic "Rational fractions. Decomposition of rational fractions into elementary (simple) fractions". I strongly advise you to at least skim through this topic before proceeding to reading this material. In addition, we will need a table of indefinite integrals.

Let me remind you of a couple of terms. They were discussed in the relevant topic, so here I will limit myself to a brief formulation.

The ratio of two polynomials $\frac(P_n(x))(Q_m(x))$ is called a rational function or a rational fraction. The rational fraction is called correct if $n< m$, т.е. если степень многочлена, стоящего в числителе, меньше степени многочлена, стоящего в знаменателе. В противном случае (если $n ≥ m$) дробь называется wrong.

Elementary (simplest) rational fractions are rational fractions of four types:

1. $\frac(A)(x-a)$;
2. $\frac(A)((x-a)^n)$ ($n=2,3,4, \ldots$);
3. $\frac(Mx+N)(x^2+px+q)$ ($p^2-4q< 0$);
4. $\frac(Mx+N)((x^2+px+q)^n)$ ($p^2-4q< 0$; $n=2,3,4,\ldots$).

Note (desirable for a better understanding of the text): show\hide

Why is the $p^2-4q condition necessary?< 0$ в дробях третьего и четвертого типов? Рассмотрим квадратное уравнение $x^2+px+q=0$. Дискриминант этого уравнения $D=p^2-4q$. По сути, условие $p^2-4q < 0$ означает, что $D < 0$. Если $D < 0$, то уравнение $x^2+px+q=0$ не имеет действительных корней. Т.е. выражение $x^2+px+q$ неразложимо на множители. Именно эта неразложимость нас и интересует.

For example, for the expression $x^2+5x+10$ we get: $p^2-4q=5^2-4\cdot 10=-15$. Since $p^2-4q=-15< 0$, то выражение $x^2+5x+10$ нельзя разложить на множители.

By the way, for this check it is not necessary that the coefficient in front of $x^2$ equals 1. For example, for $5x^2+7x-3=0$ we get: $D=7^2-4\cdot 5 \cdot (-3)=109$. Since $D > 0$, the expression $5x^2+7x-3$ is factorizable.

Examples of rational fractions (regular and improper), as well as examples of the expansion of a rational fraction into elementary ones, can be found. Here we are only interested in questions of their integration. Let's start with the integration of elementary fractions. So, each of the four types of the above elementary fractions is easy to integrate using the formulas below. Let me remind you that when integrating fractions of type (2) and (4) $n=2,3,4,\ldots$ is assumed. Formulas (3) and (4) require the condition $p^2-4q< 0$.

\begin(equation) \int \frac(A)(x-a) dx=A\cdot \ln |x-a|+C \end(equation) \begin(equation) \int\frac(A)((x-a)^n )dx=-\frac(A)((n-1)(x-a)^(n-1))+C \end(equation) \begin(equation) \int \frac(Mx+N)(x^2 +px+q) dx= \frac(M)(2)\cdot \ln (x^2+px+q)+\frac(2N-Mp)(\sqrt(4q-p^2))\arctg\ frac(2x+p)(\sqrt(4q-p^2))+C \end(equation)

For $\int\frac(Mx+N)((x^2+px+q)^n)dx$ the replacement $t=x+\frac(p)(2)$ is made, after which the resulting integral is split into two. The first one will be calculated by inserting it under the differential sign, and the second one will look like $I_n=\int\frac(dt)((t^2+a^2)^n)$. This integral is taken using the recurrence relation

\begin(equation) I_(n+1)=\frac(1)(2na^2)\frac(t)((t^2+a^2)^n)+\frac(2n-1)(2na ^2)I_n, \; n\in N \end(equation)

The calculation of such an integral is analyzed in example No. 7 (see the third part).

Scheme for calculating integrals from rational functions (rational fractions):

1. If the integrand is elementary, then apply formulas (1)-(4).
2. If the integrand is not elementary, then represent it as a sum of elementary fractions, and then integrate using formulas (1)-(4).

The above algorithm for integrating rational fractions has an undeniable advantage - it is universal. Those. Using this algorithm, one can integrate any rational fraction. That is why almost all replacements of variables in the indefinite integral (Euler, Chebyshev substitutions, universal trigonometric substitution) are done in such a way that after this replacement we get a rational fraction under the interval. And apply the algorithm to it. We will analyze the direct application of this algorithm using examples, after making a small note.

$$ \int\frac(7dx)(x+9)=7\ln|x+9|+C. $$

In principle, this integral is easy to obtain without mechanical application of the formula. If we take the constant $7$ out of the integral sign and take into account that $dx=d(x+9)$, then we get:

$$ \int\frac(7dx)(x+9)=7\cdot \int\frac(dx)(x+9)=7\cdot \int\frac(d(x+9))(x+9 )=|u=x+9|=7\cdot\int\frac(du)(u)=7\ln|u|+C=7\ln|x+9|+C. $$

For detailed information I recommend to look at the topic. It explains in detail how such integrals are solved. By the way, the formula is proved by the same transformations that were applied in this paragraph when solving "manually".

2) Again, there are two ways: to apply a ready-made formula or to do without it. If you apply the formula, then you should take into account that the coefficient in front of $x$ (the number 4) will have to be removed. To do this, we simply take out the four of them in brackets:

$$ \int\frac(11dx)((4x+19)^8)=\int\frac(11dx)(\left(4\left(x+\frac(19)(4)\right)\right)^ 8)= \int\frac(11dx)(4^8\left(x+\frac(19)(4)\right)^8)=\int\frac(\frac(11)(4^8)dx) (\left(x+\frac(19)(4)\right)^8). $$

Now it's time to apply the formula:

$$ \int\frac(\frac(11)(4^8)dx)(\left(x+\frac(19)(4)\right)^8)=-\frac(\frac(11)(4 ^8))((8-1)\left(x+\frac(19)(4) \right)^(8-1))+C= -\frac(\frac(11)(4^8)) (7\left(x+\frac(19)(4) \right)^7)+C=-\frac(11)(7\cdot 4^8 \left(x+\frac(19)(4) \right )^7)+C. $$

You can do without using the formula. And even without putting the constant $4$ out of the brackets. If we take into account that $dx=\frac(1)(4)d(4x+19)$, then we get:

$$ \int\frac(11dx)((4x+19)^8)=11\int\frac(dx)((4x+19)^8)=\frac(11)(4)\int\frac( d(4x+19))((4x+19)^8)=|u=4x+19|=\\ =\frac(11)(4)\int\frac(du)(u^8)=\ frac(11)(4)\int u^(-8)\;du=\frac(11)(4)\cdot\frac(u^(-8+1))(-8+1)+C= \\ =\frac(11)(4)\cdot\frac(u^(-7))(-7)+C=-\frac(11)(28)\cdot\frac(1)(u^7 )+C=-\frac(11)(28(4x+19)^7)+C. $$

Detailed explanations on finding such integrals are given in the topic "Integration by substitution (introduction under the differential sign)" .

3) We need to integrate the fraction $\frac(4x+7)(x^2+10x+34)$. This fraction has the structure $\frac(Mx+N)(x^2+px+q)$, where $M=4$, $N=7$, $p=10$, $q=34$. However, to make sure that this is indeed an elementary fraction of the third type, you need to check the condition $p^2-4q< 0$. Так как $p^2-4q=10^2-4\cdot 34=-16 < 0$, то мы действительно имеем дело с интегрированием элементарной дроби третьего типа. Как и в предыдущих пунктах есть два пути для нахождения $\int\frac{4x+7}{x^2+10x+34}dx$. Первый путь - банально использовать формулу . Подставив в неё $M=4$, $N=7$, $p=10$, $q=34$ получим:

$$ \int\frac(4x+7)(x^2+10x+34)dx = \frac(4)(2)\cdot \ln (x^2+10x+34)+\frac(2\cdot 7-4\cdot 10)(\sqrt(4\cdot 34-10^2)) \arctg\frac(2x+10)(\sqrt(4\cdot 34-10^2))+C=\\ = 2\cdot \ln (x^2+10x+34)+\frac(-26)(\sqrt(36)) \arctg\frac(2x+10)(\sqrt(36))+C =2\cdot \ln (x^2+10x+34)+\frac(-26)(6) \arctg\frac(2x+10)(6)+C=\\ =2\cdot \ln (x^2+10x +34)-\frac(13)(3) \arctg\frac(x+5)(3)+C. $$

Let's solve the same example, but without using the ready-made formula. Let's try to isolate the derivative of the denominator in the numerator. What does this mean? We know that $(x^2+10x+34)"=2x+10$. It is the expression $2x+10$ that we have to isolate in the numerator. So far, the numerator contains only $4x+7$, but this is not for long. Apply the following transformation to the numerator:

$$ 4x+7=2\cdot 2x+7=2\cdot (2x+10-10)+7=2\cdot(2x+10)-2\cdot 10+7=2\cdot(2x+10) -thirteen. $$

Now the required expression $2x+10$ has appeared in the numerator. And our integral can be rewritten as follows:

$$ \int\frac(4x+7)(x^2+10x+34) dx= \int\frac(2\cdot(2x+10)-13)(x^2+10x+34)dx. $$

Let's break the integrand into two. Well, and, accordingly, the integral itself is also "split":

$$ \int\frac(2\cdot(2x+10)-13)(x^2+10x+34)dx=\int \left(\frac(2\cdot(2x+10))(x^2 +10x+34)-\frac(13)(x^2+10x+34) \right)\; dx=\\ =\int \frac(2\cdot(2x+10))(x^2+10x+34)dx-\int\frac(13dx)(x^2+10x+34)=2\cdot \int \frac((2x+10)dx)(x^2+10x+34)-13\cdot\int\frac(dx)(x^2+10x+34). $$

Let's talk about the first integral first, i.e. about $\int \frac((2x+10)dx)(x^2+10x+34)$. Since $d(x^2+10x+34)=(x^2+10x+34)"dx=(2x+10)dx$, then the denominator differential is located in the numerator of the integrand. In short, instead of the expression $( 2x+10)dx$ we write $d(x^2+10x+34)$.

Now let's say a few words about the second integral. Let's single out the full square in the denominator: $x^2+10x+34=(x+5)^2+9$. In addition, we take into account $dx=d(x+5)$. Now the sum of integrals obtained by us earlier can be rewritten in a slightly different form:

$$ 2\cdot\int \frac((2x+10)dx)(x^2+10x+34)-13\cdot\int\frac(dx)(x^2+10x+34) =2\cdot \int \frac(d(x^2+10x+34))(x^2+10x+34)-13\cdot\int\frac(d(x+5))((x+5)^2+ nine). $$

If we make the change $u=x^2+10x+34$ in the first integral, then it will take the form $\int\frac(du)(u)$ and is taken by simply applying the second formula from . As for the second integral, the replacement $u=x+5$ is feasible for it, after which it takes the form $\int\frac(du)(u^2+9)$. This is the purest water, the eleventh formula from the table of indefinite integrals. So, returning to the sum of integrals, we will have:

$$ 2\cdot\int \frac(d(x^2+10x+34))(x^2+10x+34)-13\cdot\int\frac(d(x+5))((x+ 5)^2+9) =2\cdot\ln(x^2+10x+34)-\frac(13)(3)\arctg\frac(x+5)(3)+C. $$

We got the same answer as when applying the formula , which, in fact, is not surprising. In general, the formula is proved by the same methods that we used to find this integral. I believe that an attentive reader may have one question here, therefore I will formulate it:

Question #1

If we apply the second formula from the table of indefinite integrals to the integral $\int \frac(d(x^2+10x+34))(x^2+10x+34)$, then we get the following:

$$ \int \frac(d(x^2+10x+34))(x^2+10x+34)=|u=x^2+10x+34|=\int\frac(du)(u) =\ln|u|+C=\ln|x^2+10x+34|+C. $$

Why was the module missing from the solution?

Answer to question #1

The question is completely legitimate. The modulus was absent only because the expression $x^2+10x+34$ for any $x\in R$ is greater than zero. This is quite easy to show in several ways. For example, since $x^2+10x+34=(x+5)^2+9$ and $(x+5)^2 ≥ 0$, then $(x+5)^2+9 > 0$ . It is possible to judge in a different way, without involving the selection of a full square. Since $10^2-4\cdot 34=-16< 0$, то $x^2+10x+34 >0$ for any $x\in R$ (if this logical chain is surprising, I advise you to look at the graphical method for solving square inequalities). In any case, since $x^2+10x+34 > 0$, then $|x^2+10x+34|=x^2+10x+34$, i.e. you can use normal brackets instead of a module.

All points of example No. 1 are solved, it remains only to write down the answer.

Answer:

1. $\int\frac(7dx)(x+9)=7\ln|x+9|+C$;
2. $\int\frac(11dx)((4x+19)^8)=-\frac(11)(28(4x+19)^7)+C$;
3. $\int\frac(4x+7)(x^2+10x+34)dx=2\cdot\ln(x^2+10x+34)-\frac(13)(3)\arctg\frac(x +5)(3)+C$.

Example #2

Find the integral $\int\frac(7x+12)(3x^2-5x-2)dx$.

At first glance, the integrand $\frac(7x+12)(3x^2-5x-2)$ is very similar to an elementary fraction of the third type, i.e. to $\frac(Mx+N)(x^2+px+q)$. It seems that the only difference is the coefficient $3$ in front of $x^2$, but it won't take long to remove the coefficient (out of brackets). However, this similarity is apparent. For the fraction $\frac(Mx+N)(x^2+px+q)$ the condition $p^2-4q< 0$, которое гарантирует, что знаменатель $x^2+px+q$ нельзя разложить на множители. Проверим, как обстоит дело с разложением на множители у знаменателя нашей дроби, т.е. у многочлена $3x^2-5x-2$.

Our coefficient in front of $x^2$ is not equal to one, so check the condition $p^2-4q< 0$ напрямую мы не можем. Однако тут нужно вспомнить, откуда взялось выражение $p^2-4q$. Это всего лишь дискриминант квадратного уравнения $x^2+px+q=0$. Если дискриминант меньше нуля, то выражение $x^2+px+q$ на множители не разложишь. Вычислим дискриминант многочлена $3x^2-5x-2$, расположенного в знаменателе нашей дроби: $D=(-5)^2-4\cdot 3\cdot(-2)=49$. Итак, $D >0$, so the expression $3x^2-5x-2$ can be factorized. And this means that the fraction $\frac(7x+12)(3x^2-5x-2)$ is not an elementary fraction of the third type, and apply to the integral $\int\frac(7x+12)(3x^2- 5x-2)dx$ formula is not allowed.

Well, if the given rational fraction is not elementary, then it must be represented as a sum of elementary fractions, and then integrated. In short, trail take advantage of . How to decompose a rational fraction into elementary ones is written in detail. Let's start by factoring the denominator:

$$ 3x^2-5x-2=0;\\ \begin(aligned) & D=(-5)^2-4\cdot 3\cdot(-2)=49;\\ & x_1=\frac( -(-5)-\sqrt(49))(2\cdot 3)=\frac(5-7)(6)=\frac(-2)(6)=-\frac(1)(3); \\ & x_2=\frac(-(-5)+\sqrt(49))(2\cdot 3)=\frac(5+7)(6)=\frac(12)(6)=2.\ \ \end(aligned)\\ 3x^2-5x-2=3\cdot\left(x-\left(-\frac(1)(3)\right)\right)\cdot (x-2)= 3\cdot\left(x+\frac(1)(3)\right)(x-2). $$

We represent the subinternal fraction in the following form:

$$ \frac(7x+12)(3x^2-5x-2)=\frac(7x+12)(3\cdot\left(x+\frac(1)(3)\right)(x-2) )=\frac(\frac(7)(3)x+4)(\left(x+\frac(1)(3)\right)(x-2)). $$

Now let's expand the fraction $\frac(\frac(7)(3)x+4)(\left(x+\frac(1)(3)\right)(x-2))$ into elementary ones:

$$ \frac(\frac(7)(3)x+4)(\left(x+\frac(1)(3)\right)(x-2)) =\frac(A)(x+\frac( 1)(3))+\frac(B)(x-2)=\frac(A(x-2)+B\left(x+\frac(1)(3)\right))(\left(x+ \frac(1)(3)\right)(x-2));\\ \frac(7)(3)x+4=A(x-2)+B\left(x+\frac(1)( 3)\right). $$

To find the coefficients $A$ and $B$ there are two standard ways: the method of indeterminate coefficients and the method of substitution of partial values. Let's apply the partial value substitution method by substituting $x=2$ and then $x=-\frac(1)(3)$:

$$ \frac(7)(3)x+4=A(x-2)+B\left(x+\frac(1)(3)\right).\\ x=2;\; \frac(7)(3)\cdot 2+4=A(2-2)+B\left(2+\frac(1)(3)\right); \; \frac(26)(3)=\frac(7)(3)B;\; B=\frac(26)(7).\\ x=-\frac(1)(3);\; \frac(7)(3)\cdot \left(-\frac(1)(3) \right)+4=A\left(-\frac(1)(3)-2\right)+B\left (-\frac(1)(3)+\frac(1)(3)\right); \; \frac(29)(9)=-\frac(7)(3)A;\; A=-\frac(29\cdot 3)(9\cdot 7)=-\frac(29)(21).\\ $$

Since the coefficients have been found, it remains only to write down the finished expansion:

$$ \frac(\frac(7)(3)x+4)(\left(x+\frac(1)(3)\right)(x-2))=\frac(-\frac(29)( 21))(x+\frac(1)(3))+\frac(\frac(26)(7))(x-2). $$

In principle, you can leave this entry, but I like a more accurate version:

$$ \frac(\frac(7)(3)x+4)(\left(x+\frac(1)(3)\right)(x-2))=-\frac(29)(21)\ cdot\frac(1)(x+\frac(1)(3))+\frac(26)(7)\cdot\frac(1)(x-2). $$

Returning to the original integral, we substitute the resulting expansion into it. Then we divide the integral into two, and apply the formula to each. I prefer to immediately take out the constants outside the integral sign:

$$ \int\frac(7x+12)(3x^2-5x-2)dx =\int\left(-\frac(29)(21)\cdot\frac(1)(x+\frac(1) (3))+\frac(26)(7)\cdot\frac(1)(x-2)\right)dx=\\ =\int\left(-\frac(29)(21)\cdot\ frac(1)(x+\frac(1)(3))\right)dx+\int\left(\frac(26)(7)\cdot\frac(1)(x-2)\right)dx =- \frac(29)(21)\cdot\int\frac(dx)(x+\frac(1)(3))+\frac(26)(7)\cdot\int\frac(dx)(x-2 )dx=\\ =-\frac(29)(21)\cdot\ln\left|x+\frac(1)(3)\right|+\frac(26)(7)\cdot\ln|x- 2|+C. $$

Answer: $\int\frac(7x+12)(3x^2-5x-2)dx=-\frac(29)(21)\cdot\ln\left|x+\frac(1)(3)\right| +\frac(26)(7)\cdot\ln|x-2|+C$.

Example #3

Find the integral $\int\frac(x^2-38x+157)((x-1)(x+4)(x-9))dx$.

We need to integrate the fraction $\frac(x^2-38x+157)((x-1)(x+4)(x-9))$. The numerator is a polynomial of the second degree, and the denominator is a polynomial of the third degree. Since the degree of the polynomial in the numerator is less than the degree of the polynomial in the denominator, i.e. $2< 3$, то подынтегральная дробь является правильной. Разложение этой дроби на элементарные (простейшие) было получено в примере №3 на странице, посвящённой разложению рациональных дробей на элементарные. Полученное разложение таково:

$$ \frac(x^2-38x+157)((x-1)(x+4)(x-9))=-\frac(3)(x-1)+\frac(5)(x +4)-\frac(1)(x-9). $$

We just have to break the given integral into three, and apply the formula to each. I prefer to immediately take out the constants outside the integral sign:

$$ \int\frac(x^2-38x+157)((x-1)(x+4)(x-9))dx=\int\left(-\frac(3)(x-1) +\frac(5)(x+4)-\frac(1)(x-9) \right)dx=\\=-3\cdot\int\frac(dx)(x-1)+ 5\cdot \int\frac(dx)(x+4)-\int\frac(dx)(x-9)=-3\ln|x-1|+5\ln|x+4|-\ln|x- 9|+C. $$

Answer: $\int\frac(x^2-38x+157)((x-1)(x+4)(x-9))dx=-3\ln|x-1|+5\ln|x+ 4|-\ln|x-9|+C$.

A continuation of the analysis of examples of this topic is located in the second part.

Before proceeding with the integration of the simplest fractions to find the indefinite integral of a fractionally rational function, it is recommended to refresh the memory of the section “Decomposition of a fraction into simplest”.

Example 1

Find the indefinite integral ∫ 2 x 3 + 3 x 3 + x d x .

Decision

We select the integer part by dividing the column of the polynomial by the polynomial, taking into account the fact that the degree of the numerator of the integrand is equal to the degree of the denominator:

So 2 x 3 + 3 x 3 + x = 2 + - 2 x + 3 x 3 + x . We got a proper rational fraction - 2 x + 3 x 3 + x, which we now expand into simple fractions - 2 x + 3 x 3 + x \u003d 3 x - 3 x + 2 x 2 + 1. Hence,

∫ 2 x 3 + 3 x 3 + x d x = ∫ 2 + 3 x - 3 x + 2 x 2 + 1 d x = ∫ 2 d x + ∫ 3 x d x - ∫ 3 x + 2 x 2 + 1 d x = 2 x + 3 log x - ∫ 3 x + 2 x 2 + 1 d x

We have obtained an integral of the simplest fraction of the third type. You can take it by bringing it under the sign of the differential.

Since d x 2 + 1 = 2 x d x , then 3 x d x = 3 2 d x 2 + 1 . So
∫ 3 x + 2 x 2 + 1 d x = ∫ 3 x x 2 + 1 d x + ∫ 2 x 2 + 1 = 3 2 ∫ d x 2 + 1 x 2 + 1 + 2 ∫ d x x 2 + 1 = 3 2 ln x 2 + 1 + 2 a r c t g x + C 1

Hence,
∫ 2 x 3 + 3 x 3 + x d x = 2 x + 3 ln x - ∫ 3 x + 2 x 2 + 1 d x = 2 x + 3 ln x - 3 2 ln x 2 + 1 - 2 a r c tan x + C , where C \u003d - C 1

Let us describe methods for integrating the simplest fractions of each of the four types.

Integration of the simplest fractions of the first type A x - a

We use the direct integration method to solve this problem:

∫ A x - a d x = A ∫ d x x - a = A ln x - a + C

Example 2

Find the set of antiderivatives of the function y = 3 2 x - 1 .

Decision

Using the integration rule, the properties of the antiderivative and the table of antiderivatives, we find the indefinite integral ∫ 3 d x 2 x - 1: ∫ f k x + b d x = 1 k F k x + b + C

∫ 3 d x 2 x - 1 = 3 ∫ d x 2 x - 1 2 = 3 2 ∫ d x x - 1 2 = 3 2 ln x - 1 2 + C

Answer: ∫ 3 d x 2 x - 1 = 3 2 ln x - 1 2 + C

Integration of simple fractions of the second type A x - a n

Here we also apply the method of direct integration: ∫ A x - a n d x = A ∫ x - a - n d x = A - n + 1 x - a - n + 1 + C = A 1 - n x - a n - 1 + C

Example 3

It is necessary to find the indefinite integral ∫ d x 2 x - 3 7 .

Decision

∫ d x 2 x - 3 7 = ∫ d x 2 x - 3 2 7 = 1 2 7 ∫ x - 3 2 - 7 d x = = 1 2 7 1 - 7 + 1 x - 3 2 - 7 + 1 + C = 1 2 7 - 6 x - 3 2 6 + C = = 1 2 - 6 2 6 x - 3 2 6 + C = - 1 12 1 2 x - 3 6 + C

Answer:∫ d x 2 x - 3 7 = - 1 12 1 2 x - 3 6 + C

Integration of simple fractions of the third type M x + N x 2 + p x + q , D = p 2 - 4 q< 0

As a first step, we represent the indefinite integral ∫ M x + N x 2 + p x + q as a sum:

∫ M x + N x 2 + p x + q d x = ∫ M x x 2 + p x + q d x + N ∫ d x x 2 + p x + q

In order to take the first integral, we use the method of subsuming under the differential sign:

∫ M x x 2 + p x + q d x = d x 2 + p x + q = 2 x + p d x = 2 x d x + p d x ⇒ 2 x d x = d x 2 + p x + q - p d x ⇒ M x d x = M 2 d x 2 + p x + q - p M 2 d x = = ∫ M 2 d x 2 + p x + q - p M 2 d x x 2 + p x + q = = M 2 ∫ d x 2 + p x + q x 2 + p x + q - p M 2 ∫ d x x 2 + p x + q = = M 2 log x 2 + p x + q - p M 2 ∫ d x x 2 + p x + q

So,
∫ M x + N x 2 + p x + q d x = ∫ M x x 2 + p x + q d x + N ∫ d x x 2 + p x + q = = M 2 ln x 2 + p x + q - p M 2 ∫ d x x 2 + p x + q + N ∫ d x x 2 + p x + q = = M 2 ln x 2 + p x + q + 2 N - p M 2 ∫ d x x 2 + p x + q

We have obtained the integral ∫ d x x 2 + p x + q . Let's transform its denominator:

∫ d x x 2 + p x + q = ∫ d x x 2 + p x + p 2 2 - p 2 2 + q = = ∫ d x x + p 2 2 - p 2 4 + q = ∫ d x x + p 2 2 - p 2 4 + q = = ∫ d x x + p 2 2 + 4 q - p 2 4 = 2 4 q - p 2 a r c t g 2 x + p 2 4 q - p 2 + C 1

Hence,

∫ M x + N x 2 + p x + q d x = M 2 ln x 2 + p x + q + 2 N - p M 2 ∫ d x x 2 + p x + q = = M 2 ln x 2 + p x + q + 2 N - p M 2 2 4 q - p 2 a r c t g 2 x + p 2 4 q - p 2 + C 1

The formula for integrating the simplest fractions of the third type takes the form:
∫ M x + N x 2 + p x + q d x = M 2 ln x 2 + p x + q + 2 N - p M 4 q - p 2 a r c t g 2 x + p 2 4 q - p 2 + C

Example 4

It is necessary to find the indefinite integral ∫ 2 x + 1 3 x 2 + 6 x + 30 d x .

Decision

Let's apply the formula:

∫ 2 x + 1 3 x 2 + 6 x + 30 d x = 1 3 ∫ 2 x + 1 x 2 + 2 x + 10 d x = M = 2 , N = 1 , p = 2 , q = 10 = = 1 3 2 2 ln x 2 + 2 x + 10 + 2 1 - 2 2 4 10 - 2 2 a r c t g 2 x + 2 2 4 10 - 2 2 + C = = 1 3 ln x 2 + 2 x + 10 - 1 9 a r c t g x + 1 3 + C

The second solution looks like this:

∫ 2 x + 1 3 x 2 + 6 x + 30 d x = 1 3 ∫ 2 x + 1 x 2 + 2 x + 10 d x = d (x 2 + 2 x + 10 = (2 x + 2) d x = = 1 3 ∫ 2 x + 2 - 1 x 2 + 2 x + 10 d x = 1 3 ∫ d (x 2 + 2 x + 10) x 2 + 2 x + 10 = 1 3 ∫ d x x 2 + 2 x + 10 = \u003d r a n d e n t a n d a n t = 1 3 ln x 2 + 2 x + 10 - 1 3 ∫ d (x) x + 1 2 + 9 = = 1 3 ln x 2 + 2 x + 10 - 1 9 a r c t g x + 1 3 + C

Answer: ∫ 2 x + 1 3 x 2 + 6 x + 30 d x = 1 3 ln x 2 + 2 x + 10 - 1 9 a r c t g x + 1 3 + C

Integration of the simplest fractions of the fourth type M x + N (x 2 + p x + q) n , D = p 2 - 4 q< 0

First of all, we perform summing under the sign of the differential:

∫ M x + N x 2 + p x + q d x = d (x 2 + p x + q) = (2 x + p) d x = = M 2 ∫ d (x 2 + p x + q) (x 2 + p x + q ) n + N - p M 2 ∫ d x (x 2 + p x + q) n = = M 2 (- n + 1) 1 (x 2 + p x + q) n - 1 + N - p M 2 ∫ d x (x 2 + p x + q) n

Then we find an integral of the form J n = ∫ d x (x 2 + p x + q) n using recurrent formulas. Information about recurrent formulas can be found in the topic "Integration using recurrent formulas".

To solve our problem, a recurrent formula of the form J n \u003d 2 x + p (n - 1) (4 q - p 2) (x 2 + p x + q) n - 1 + 2 n - 3 n - 1 2 4 q - p 2 · J n - 1 .

Example 5

It is necessary to find the indefinite integral ∫ d x x 5 x 2 - 1 .

Decision

∫ d x x 5 x 2 - 1 = ∫ x - 5 (x 2 - 1) - 1 2 d x

We will use the substitution method for this type of integrand. Let's introduce a new variable x 2 - 1 = z 2 x = (z 2 + 1) 1 2 d x = z (z 2 + 1) - 1 2 d x

We get:

∫ d x x 5 x 2 - 1 = ∫ x - 5 (x 2 - 1) - 1 2 d x = = ∫ (z 2 + 1) - 5 2 z - 1 z (z 2 + 1) - 1 2 d z = ∫ d z (z 2 + 1) 3

We came to finding the integral of a fraction of the fourth type. In our case, we have the coefficients M=0, p=0, q=1, N=1 and n=3. We apply the recursive formula:

J 3 \u003d ∫ d z (z 2 + 1) 3 \u003d 2 z + 0 (3 - 1) (4 1 - 0) z 2 + 1 3 - 1 + 2 3 - 3 3 - 1 2 4 1 - 0 ∫ d z (z 2 + 1) 2 = = z 4 (z 2 + 1) 2 + 3 4 2 z (2 - 1) (4 1 - 0) (z 2 + 1) 2 - 1 + 2 2 - 3 2 - 11 2 4 1 - 0 ∫ d z z 2 + 1 = = z 4 (z 2 + 1) 2 + 3 8 z z 2 + 1 + 3 8 a r c t g (z) + C

After the reverse substitution z = x 2 - 1 we get the result:
∫ d x x 5 x 2 - 1 = x 2 - 1 4 x 4 + 3 8 x 2 - 1 x 2 + 3 8 a r c t g x 2 - 1 + C

Answer:∫ d x x 5 x 2 - 1 = x 2 - 1 4 x 4 + 3 8 x 2 - 1 x 2 + 3 8 a r c t g x 2 - 1 + C

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