Peevid and its properties

An antiderivative function f(x) on the interval (a; b) is such a function F(x) that equality holds for any x from a given interval.

If we take into account the fact that the derivative of the constant C is equal to zero, then the equality . Thus, the function f(x) has a set of antiderivatives F(x)+C, for an arbitrary constant C, and these antiderivatives differ from each other by an arbitrary constant value.

properties of the primitive.

If the function F(x) is an antiderivative for the function f(x) on an interval X, then the function f(x) + C, where C is an arbitrary constant, will also be an antiderivative for f(x) on this interval.

If the function F(x) is some antiderivative for the function f(x) on the interval X=(a,b), then any other antiderivative F1(x) can be represented as F1(x) = F(x) + C, where C is a constant function on X.

2 Definition of the indefinite integral.

The whole set of antiderivatives of the function f(x) is called the indefinite integral of this function and is denoted .

The expression is called the integrand, and f(x) is called the integrand. The integrand is the differential of the function f(x).

The action of finding an unknown function by its given differential is called indefinite integration, because the result of integration is not one function F(x), but the set of its antiderivatives F(x)+C.

properties of the indefinite integral (properties of the antiderivative).

The derivative of the integration result is equal to the integrand.

The indefinite integral of the differential of a function is equal to the sum of the function itself and an arbitrary constant.

where k is an arbitrary constant. The coefficient can be taken out of the sign of the indefinite integral.

The indefinite integral of the sum/difference of functions is equal to the sum/difference of the indefinite integrals of functions.

Change of variable in indefinite integral

Variable substitution in the indefinite integral is performed using substitutions of two types:

a) where is a monotone, continuously differentiable function of the new variable t. Variable substitution formula in this case:

Where U is a new variable. Variable substitution formula for this substitution:

Integration by parts

Finding the integral by the formula is called integration by parts. Here U=U(x), υ=υ(x) are continuously differentiable functions of x. With the help of this formula, finding the integral is reduced to finding another integral; its use is expedient in cases where the last integral is either simpler than the original one or similar to it.

In this case, υ is taken to be such a function that simplifies upon differentiation, and dU is that part of the integrand, the integral of which is known or can be found.

Newton–Leibniz formula

Continuity of a definite integral as a function of the upper limit

If the function y = f (x) is integrable on the segment , then, obviously, it is also integrable on an arbitrary segment [a, x] embedded in . Function ,

where x О , is called an integral with a variable upper limit. The value of the function Ф (x) at the point x is equal to the area S (x) under the curve y \u003d f (x) on the segment [a, x]. This is the geometric meaning of an integral with a variable upper limit.

Theorem. If the function f(x) is continuous on the interval, then the function Φ(x) is also continuous on [a, b].

Let Δх be such that x + Δ x О . We have

By the mean value theorem, there is such a value with н [ x, x + Δ x] that Since with н , and the function f (x) is bounded, passing to the limit as Δ x → 0, we obtain

ODR 1st order

What is the difference between homogeneous differential equations and other types of DE? This is easiest to explain right away with a specific example.

Solve differential equation

What should be analyzed first of all when solving any first-order differential equation? First of all, it is necessary to check whether it is possible to immediately separate the variables using "school" actions? Usually such an analysis is carried out mentally or trying to separate the variables in a draft.

In this example, the variables cannot be divided (you can try to flip the terms from part to part, take factors out of brackets, etc.). By the way, in this example, the fact that the variables cannot be divided is quite obvious due to the presence of the multiplier

The question arises - how to solve this diffur?

Need to check if this equation is homogeneous? The verification is simple, and the verification algorithm itself can be formulated as follows:

To the original equation:

Instead of x, substitute instead of y, substitute the derivative, do not touch: The letter lambda is some abstract numerical parameter, the point is not in the lambdas themselves, and not in their values, but the point is this:

If, as a result of transformations, it is possible to reduce ALL "lambdas" (i.e., obtain the original equation), then this differential equation is homogeneous.

Obviously, the lambdas immediately cancel out in the exponent: Now, on the right side, we take the lambda out of brackets: Both parts of the equation can be reduced to this same lambda: As a result, all lambdas disappeared like a dream, like a morning mist, and we got the original equation.

Conclusion: This equation is homogeneous

LOU.General properties of solutions

that is, is linear with respect to the unknown function y and its derivatives and . The coefficients and and the right side of this equation are continuous.

If the right side of the equation , then the equation is called linear inhomogeneous. If , then the equation has the form

(9)

and is called linear homogeneous.

Let and be any particular solutions of equation (9), that is, they do not contain arbitrary constants.

Theorem 1. If and are two partial solutions of a linear homogeneous equation of the second order, then it is also a solution of this equation.

Since and are solutions of equation (9), they turn this equation into an identity, that is,

And

(10)

Substitute into equation (9). Then we have:

Due to (10). Hence, is a solution to the equation.

Theorem 2. If is a solution of a linear homogeneous equation of the second order, and C is a constant, then it is also a solution to this equation.

Proof. Substitute into equation (9). We get: that is, the solution of the equation.

Consequence. If and are solutions of equation (9), then so is its solution by virtue of theorems (1) and (2).

Definition. Two solutions and equations (9) are called linearly dependent (on the segment ) if it is possible to choose such numbers and , which are not simultaneously equal to zero, such that the linear combination of these solutions is identically equal to zero on , that is, if .

If such numbers cannot be chosen, then the solutions and are called linearly independent (on the interval ).

Obviously, solutions and will be linearly dependent if and only if their ratio is constant, that is (or vice versa).

Indeed, if and are linearly dependent, then , where at least one constant or is different from zero. Let, for example, . Then , , Denoting we get , that is, the ratio is constant.

Conversely, if . Here, the coefficient at , i.e., is different from zero, which by definition means that and are linearly dependent.

Comment. From the definition of linearly independent solutions and the reasoning above, we can conclude that if and are linearly independent, then their ratio cannot be constant.

For example, the functions and for are linearly independent, since , because . And here are 5 features x And x are linearly dependent, since their ratio is .

Theorem. If and are linearly independent particular solutions of a second-order linear homogeneous equation, then their linear combination , where and are arbitrary constants, is a general solution to this equation.

Proof. By virtue of Theorems 1 and 2 (and corollaries to them), is a solution of Eq. (9) for any choice of constants and .

If the solutions and are linearly independent, then is a general solution, since this solution contains two arbitrary constants that cannot be reduced to one.

At the same time, even if they were linearly dependent solutions, it would no longer be a general solution. In this case where α -constant. Then , where is a constant. cannot be a general solution to a second-order differential equation, since it depends on only one constant.

So, the general solution of equation (9):

19. The concept of a linearly independent system of functions. Vronsky's determinant. sufficient condition for linear independence. the concept of a fundamental system of a function. Examples. Necessary and sufficient condition for the difference from zero of the Wronsky determinant on the interval [a, c]

The concept of a linearly independent system of functions

Functions are called linearly dependent on if one of them is a linear combination of the others. In other words, the functions are called linearly dependent on if there exist numbers , of which at least one is not equal to zero, such that

If identity (4) holds only if all , then the functions are called linearly independent on .

System of solutions linearly independent on an interval

homogeneous differential equation of the th order (3) with continuous coefficients is called the fundamental system of solutions of this equation.

To solve a linear homogeneous differential equation of the th order (3) with continuous coefficients , we need to find its fundamental system of solutions.

According to Theorem 1, an arbitrary linear combination of solutions , i.e., the sum

, (5)

where are arbitrary numbers, is in turn the solution of equation (3) on . But it turns out that vice versa, any solution of the differential equation (3) on an interval is a certain linear combination of the indicated (interdependent) particular solutions of it (see Theorem 4 below), which form a fundamental system of solutions.

Thus, the general solution of the homogeneous differential equation (3) has the form (5), where are arbitrary constants, and are particular solutions (3), which form the fundamental system of solutions of the homogeneous equation.

Note that the general solution of the inhomogeneous equation (1) is the sum of any of its particular solutions and the general solution of the homogeneous equation

. (6)

Indeed,

.

On the other hand, if there is an arbitrary solution to equation (1), then

and, therefore, is a solution of the homogeneous equation; but then there are numbers such that

,

i.e., equality (6) holds for these numbers.

Vronsky's determinant.

Theorem 2. If the functions are linearly dependent on and have derivatives up to the th order, then the determinant

. (7)

I

The determinant (7) is called the Wronsky determinant or the Wronskian and is denoted by the symbol .

Proof. Since the functions are linearly dependent on , then there are non-zero numbers such that identity (4) holds on . Differentiating it times, we obtain the system of equations

This homogeneous system, by assumption, has a nontrivial solution (i.e., at least one ) for . The latter is possible when the determinant of the system, which is the Wronsky determinant, is identically equal to zero. The theorem has been proven.

Comment. Theorem 2 implies that if at least at one point , then the functions are linearly independent on .

Example 2. The functions are linearly independent on any , because

.

Example 3. Functions are linearly independent on any , if - different numbers (real or complex).

Indeed.

,

since the last determinant is a Vandermonde determinant, which is not equal to zero for various.

Example 4 Functions are linearly independent on any .

Since and

then the linear independence of these functions follows from the second example.

Theorem 3. In order for the solutions linear differential homogeneous equation with continuous coefficients are linearly independent on , it is necessary and sufficient that for all .

Proof. 1) If on , then the functions are linearly independent, regardless of whether they are solutions to the equation or not (see Remark).

2) Let are linearly independent functions on and be solutions of the equation .

Let us prove that everywhere on . Assume to the contrary that there exists a point at which . We choose numbers that are not equal to zero at the same time so that they are solutions to the system

(8)

This can be done, since the determinant of system (8) is . Then, by virtue of Theorem 1, the function will be a solution to the equation with zero initial conditions (by (8))

But the trivial solution also satisfies the same conditions. By virtue of the existence and uniqueness theorem, there can be only one solution that satisfies these initial conditions; therefore, on , i.e., the functions are linearly dependent on , which was not assumed. The theorem has been proven.

If are discontinuous functions in the interval where we are looking for a solution, then the equation may have more than one solution that satisfies the initial conditions , and then it is possible that on .

Example 5. It is easy to check that the functions

are linearly independent on and for them on .

This is because the function is a general solution to the equation

,

Where discontinuous at the point . For this equation, the existence and uniqueness theorem does not hold (in a neighborhood of the point ). Not only the function , but also the function is a solution of a differential equation that satisfies the conditions and for .

The structure of the general solution.

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Theorem 4. If - linearly independent on solutions of a linear homogeneous differential equation of the th order with continuous coefficients , then the function

, (9)

where are arbitrary constants, is a general solution of the equation , i.e. the sum (9) for any , is a solution to this equation and, conversely, any solution to this equation can be represented as a sum (9) for the corresponding values ​​of .

Proof. We already know that the sum (9) for any is a solution to the equation . Let, conversely, be an arbitrary solution of this equation. Let's put

For the received numbers we will compose a linear system of equations for unknown numbers : , it is enough to find some - real constants. To find the general solution of Eq. (8), we proceed as follows. We compose a characteristic equation for equation (8): . Using the initial conditions, we define

Consider the linear differential equation n-th order

y (n) + a n -1 (x)y (n- 1) + ... + a 1 (x)y" + a 0 (x)y = f(x).

with continuous coefficients a n -1 (x), a n -2 (x), ..., a 1 (x), a 0 (x) and continuous right hand side f(x).

Superposition principle based on the following properties of solutions of linear differential equations.

1. If y 1 (x) And y 2 (x) are two solutions of the linear homogeneous differential equation

y (n) + a n -1 (x)y (n- 1) + ... + a 1 (x)y" + a 0 (x)y = 0

then any linear combination of them y(x) = C 1 y 1 (x) + C 2 y 2 (x) is a solution to this homogeneous equation.

2. If y 1 (x) And y 2 (x) are two solutions of the linear inhomogeneous equation L(y) = f(x) , then their difference y(x) = y 1 (x) − y 2 (x) is a solution to the homogeneous equation L(y) = 0 .

3. Any solution of an inhomogeneous linear equation L(y) = f(x) is the sum of any fixed (particular) solution of a non-homogeneous equation and some solution of a homogeneous equation.

4. If y 1 (x) And y 2 (x) - solutions of linear inhomogeneous equations L(y) = f 1 (x) And L(y) = f 2 (x) respectively, then their sum y(x) =y 1 (x) + y 2 (x) is a solution to the inhomogeneous equation L(y) = f 1 (x) + f 2 (x).

This last statement is usually called superposition principle.

Constant Variation Method

Consider the inhomogeneous equation of the -th order

where the coefficients and the right side are given continuous functions on the interval .

Let us assume that we know the fundamental system of solutions corresponding homogeneous equation

As we showed in § 1.15 (formula (6)), the general solution of equation (1) is equal to the sum of the general solution of equation (2) and any solution of equation (1).

The solution of the inhomogeneous equation (1) can be

The change of variable in the indefinite integral is used to find integrals in which one of the functions is the derivative of another function. Let there be an integral $ \int f(x) dx $, let's make the replacement $ x=\phi(t) $. Note that the function $ \phi(t) $ is differentiable, so $ dx = \phi"(t) dt $ can be found.

Now we substitute $ \begin(vmatrix) x = \phi(t) \\ dx = \phi"(t) dt \end(vmatrix) $ into the integral and get that:

$$ \int f(x) dx = \int f(\phi(t)) \cdot \phi"(t) dt $$

This one is variable change formula in indefinite integral.

Variable replacement method algorithm

Thus, if an integral of the form is given in the problem: $$ \int f(\phi(x)) \cdot \phi"(x) dx $$ It is advisable to replace the variable with a new one: $$ t = \phi(x) $ $ $$ dt = \phi"(t) dt $$

After that, the integral will be presented in a form that is easy to take with the main integration methods: $$ \int f(\phi(x)) \cdot \phi"(x) dx = \int f(t)dt $$

Don't forget to also set the replaced variable back to $x$.

Solution examples

Example 1

Find the indefinite integral by the change of variable method: $$ \int e^(3x) dx $$

Solution

We change the variable in the integral to $ t = 3x, dt = 3dx $: $$ \int e^(3x) dx = \int e^t \frac(dt)(3) = \frac(1)(3) \int e^t dt = $$ The exponential integral is still the same according to the integration table, although $ t $ is written instead of $ x $: $$ = \frac(1)(3) e^t + C = \frac(1)(3) e^(3x) + C $$ 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

$$ \int e^(3x) dx = \frac(1)(3) e^(3x) + C $$

Integration by substitution (change of variable). Let it be required to calculate the integral, which is not tabular. The essence of the substitution method is that in the integral the variable x is replaced by the variable t according to the formula x \u003d q (t), whence dx \u003d q "(t) dt.

Theorem. Let the function x=u(t) be defined and differentiable on some set T and let X be the set of values ​​of this function on which the function f(x) is defined. Then if on the set X the function f(x) has an antiderivative, then on the set T the formula is true:

Formula (1) is called the change of variable formula in the indefinite integral.

Integration by parts. The method of integration by parts follows from the formula for the differential of the product of two functions. Let u(x) and v(x) be two differentiable functions of x. Then:

d(uv)=udv+vdu. - (3)

Integrating both parts of equality (3), we obtain:

But since then:

Relation (4) is called the integration-by-parts formula. Using this formula, find the integral. It is advisable to use it when the integral on the right side of formula (4) is easier to calculate than the original one.

In formula (4) there is no arbitrary constant C, since the right side of this formula contains an indefinite integral containing an arbitrary constant.

Let us present some common types of integrals calculated by the method of integration by parts.

I. Integrals of the form, (P n (x) is a polynomial of degree n, k is some number). To find these integrals, it is enough to put u=P n (x) and apply formula (4) n times.

II. Integrals of the form, (Pn(x) is a polynomial of degree n with respect to x). They can be found by frequent ones, taking for u the function that is a factor at P n (x).

With the help of a change of variable, you can calculate simple integrals and, in some cases, simplify the calculation of more complex ones.

The variable replacement method is that we go from the original integration variable, let it be x , to another variable, which we denote as t . At the same time, we assume that the variables x and t are related by some relation x = x (t), or t = t (x). For example x = log t, x = sin t, t = 2 x + 1, and so on. Our task is to choose such a relationship between x and t so that the original integral either reduces to a tabular one or becomes simpler.

Basic Variable Change Formula

Consider the expression that is under the integral sign. It consists of the product of the integrand, which we will denote as f (x) and differential dx : . Let us pass to a new variable t by choosing some relation x = x (t). Then we have to express the function f (x) and the differential dx in terms of the variable t .

To express the integrand f (x) through the variable t , you just need to substitute the chosen ratio x = x instead of the variable x (t).

The differential transformation is done like this:
.
That is, the differential dx is equal to the product of the derivative of x with respect to t and the differential dt.

Then
.

In practice, the most common case is when we perform a replacement by choosing a new variable as a function of the old one: t = t (x). If we guessed that the integrand can be represented as
,
where t′ (x) is the derivative of t with respect to x, then
.

So, the basic variable change formula can be represented in two forms.
(1) ,
where x is a function of t .
(2) ,
where t is a function of x .

Important note

In tables of integrals, the integration variable is most often denoted as x . However, it is worth considering that the integration variable can be denoted by any letter. Moreover, any expression can be used as the integration variable.

As an example, consider the table integral
.

Here x can be replaced by any other variable or a function of a variable. Here are examples of possible options:
;
;
.

In the last example, you need to take into account that when passing to the integration variable x , the differential is transformed as follows:
.
Then
.

This example is the essence of substitution integration. That is, we must guess that
.
After that, the integral is reduced to a tabular one.
.

You can evaluate this integral using a change of variable, applying the formula (2) . Let t = x 2+x. Then
;
;

.

Examples of integration by change of variable

1) We calculate the integral
.
We notice that (sin x)′ = cos x. Then

.
Here we have applied the substitution t = sin x.

2) We calculate the integral
.
We notice that . Then

.
Here we have performed the integration by changing the variable t = arctg x.

3) Let's integrate
.
We notice that . Then

. Here, during integration, the change of variable t = x 2 + 1 .

Linear substitutions

Perhaps the most common are linear substitutions. This is a substitution of the form variable
t = ax + b
where a and b are constants. Under such a change, the differentials are related by the relation
.

Examples of integration by linear substitutions

A) Calculate Integral
.
Solution.
.

b) Find the integral
.
Solution.
Let's use the properties of the exponential function.
.
ln 2- is a constant. We calculate the integral.

.

c) Calculate Integral
.
Solution.
We bring the square polynomial in the denominator of a fraction to the sum of squares.
.
We calculate the integral.

.

D) Find the integral
.
Solution.
We transform the polynomial under the root.

.
We integrate using the change of variable method.

.
We have previously obtained the formula
.
From here
.
Substituting this expression, we get the final answer.

e) Calculate Integral
.
Solution.
Apply the formula for the product of sine and cosine.
;
.
We integrate and make substitutions.


.

References:
N.M. Gunther, R.O. Kuzmin, Collection of problems in higher mathematics, Lan, 2003.

In this lesson, we will get acquainted with one of the most important and most common tricks that is used in the course of solving indefinite integrals - the change of variable method. For successful mastering of the material, initial knowledge and integration skills are required. If there is a feeling of an empty full teapot in integral calculus, then you should first read the material, where I explained in an accessible form what an integral is and analyzed in detail the basic examples for beginners.

Technically, the method of changing a variable in an indefinite integral is implemented in two ways:

– Bringing the function under the sign of the differential;
– The actual change of variable.

In fact, it's the same thing, but the design of the solution looks different.

Let's start with a simpler case.

Bringing a function under the differential sign

At the lesson Indefinite integral. Solution examples we learned how to open the differential, I recall the example that I gave:

That is, to open the differential is formally almost the same as to find the derivative.

Example 1

Run a check.

We look at the table of integrals and find a similar formula: . But the problem is that we have under the sine not just the letter "x", but a complex expression. What to do?

We bring the function under the sign of the differential:

Expanding the differential, it is easy to check that:

In fact and is a record of the same.

But, nevertheless, the question remains, how did we come to the idea that at the first step we need to write our integral exactly like this: ? Why so, and not otherwise?

Formula (and all other tabular formulas) are valid and applicable NOT ONLY for a variable, but also for any complex expression ONLY THE FUNCTION ARGUMENT(- in our example) AND THE EXPRESSION UNDER THE DIFFERENTIAL SIGN WERE THE SAME .

Therefore, mental reasoning when solving should be something like this: “I need to solve the integral. I looked at the table and found a similar formula . But I have a complex argument and I can’t immediately use the formula. However, if I manage to get under the sign of the differential, then everything will be fine. If I write , then . But there is no triple factor in the original integral, therefore, in order for the integrand to not change, I need to multiply it by ". In the course of approximately such mental reasoning, a record is born:

Now you can use the spreadsheet :

Ready

The only difference is that we do not have the letter "x", but a complex expression.

Let's do a check. Open the table of derivatives and differentiate the answer:

The original integrand was obtained, which means that the integral was found correctly.

Please note that during the verification we used the rule of differentiation of a complex function . In fact, bringing the function under the sign of the differential and are two mutually inverse rules.

Example 2

We analyze the integrand function. Here we have a fraction, and the denominator is a linear function (with "x" in the first degree). We look in the table of integrals and find the most similar thing: .

We bring the function under the sign of the differential:

Those who find it difficult to immediately figure out which fraction to multiply by can quickly reveal the differential on a draft:. Yeah, it turns out, so that nothing changes, I need to multiply the integral by .
Next, we use the spreadsheet formula :

Examination:


The original integrand was obtained, which means that the integral was found correctly.

Example 3

Find the indefinite integral. Run a check.

Example 4

Find the indefinite integral. Run a check.

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

With some experience in solving integrals, such examples will seem easy, and will crack like nuts:

At the end of this paragraph, I would also like to dwell on the “free” case when a variable enters a linear function with a unit coefficient, for example:

Strictly speaking, the solution should look like this:

As you can see, bringing the function under the sign of the differential went “painlessly”, without any multiplications. Therefore, in practice, such a long solution is often neglected and immediately written down as . But be prepared, if necessary, to explain to the teacher how you decided! Since there is no integral in the table at all.

Variable change method in indefinite integral

We turn to the consideration of the general case - the method of changing variables in the indefinite integral.

Example 5

Find the indefinite integral.

As an example, I took the integral, which we considered at the very beginning of the lesson. As we have already said, to solve the integral, we liked the tabular formula , and I would like to reduce the whole thing to her.

The idea behind the replacement method is to replace a complex expression (or some function) with one letter.
In this case it asks:
The second most popular replacement letter is the letter .
In principle, you can use other letters, but we still stick to the traditions.

So:
But when replacing, we have left! Probably, many have guessed that if a transition is made to a new variable, then in the new integral everything must be expressed through the letter, and there is no place for the differential at all.
It follows a logical conclusion that it is necessary turn into some expression that depends only on.

The action is the following. After we have chosen a replacement, in this example, , we need to find the differential . With differentials, I think, friendship has already been established for everyone.

Since then

After the showdown with the differential, I recommend rewriting the final result as briefly as possible:
Now, according to the rules of proportion, we express the one we need:

Eventually:
Thus:

And this is the most tabular integral (the table of integrals, of course, is also valid for the variable).

In conclusion, it remains to carry out the reverse replacement. We remember that .

Ready.

The final design of this example should look something like this:

“

Let's replace:

“

The icon does not carry any mathematical meaning, it means that we have interrupted the solution for intermediate explanations.

When making an example in a notebook, it is better to superscript the reverse substitution with a simple pencil.

Attention! In the following examples, finding the differential will not be described in detail.

And now it's time to remember the first solution:

What is the difference? There is no fundamental difference. It's actually the same thing. But from the point of view of the design of the task, the method of bringing the function under the sign of the differential is much shorter.

The question arises. If the first way is shorter, then why use the replace method? The fact is that for a number of integrals it is not so easy to "fit" the function under the sign of the differential.

Example 6

Find the indefinite integral.

Let's make a replacement: (it's hard to think of another replacement here)

As you can see, as a result of the replacement, the original integral has been greatly simplified - reduced to an ordinary power function. This is the purpose of the replacement - to simplify the integral.

Lazy advanced people can easily solve this integral by bringing the function under the differential sign:

Another thing is that such a solution is not obvious to all students. In addition, already in this example, the use of the method of bringing a function under the differential sign significantly increases the risk of confusion in the decision.

Example 7

Find the indefinite integral. Run a check.

Example 8

Find the indefinite integral.

Replacement:
It remains to be seen what will become

Well, we have expressed, but what to do with the “X” remaining in the numerator?!
From time to time, in the course of solving integrals, the following trick occurs: we will express from the same replacement !

Example 9

Find the indefinite integral.

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

Example 10

Find the indefinite integral.

Surely some have noticed that my reference table does not have a variable substitution rule. It was done deliberately. The rule would confuse the explanation and understanding, since it does not appear explicitly in the above examples.

It's time to talk about the basic premise of using the variable substitution method: the integrand must contain some function and its derivative:(functions may not be in the product)

In this regard, when finding integrals, one often has to look into the table of derivatives.

In this example, we notice that the degree of the numerator is one less than the degree of the denominator. In the table of derivatives we find the formula, which just lowers the degree by one. And, therefore, if you designate for the denominator, then there are great chances that the numerator will turn into something good.