On this page you will find:

1. Actually, the table of antiderivatives - it can be downloaded in PDF format and printed;

2. Video on how to use this table;

3. A bunch of examples of calculating the antiderivative from various textbooks and tests.

In the video itself, we will analyze many problems where you need to calculate antiderivatives of functions, often quite complex, but most importantly, they are not power functions. All functions summarized in the table proposed above must be known by heart, like derivatives. Without them, further study of integrals and their application to solve practical problems is impossible.

Today we continue to study primitives and move on to a slightly more complex topic. If last time we looked at antiderivatives only of power functions and slightly more complex constructions, today we will look at trigonometry and much more.

As I said in the last lesson, antiderivatives, unlike derivatives, are never solved “right away” using any standard rules. Moreover, the bad news is that, unlike the derivative, the antiderivative may not be considered at all. If we write a completely random function and try to find its derivative, then with a very high probability we will succeed, but the antiderivative will almost never be calculated in this case. But there is good news: there is a fairly large class of functions called elementary functions, the antiderivatives of which are very easy to calculate. And all the other more complex structures that are given on all kinds of tests, independent tests and exams, in fact, are made up of these elementary functions through addition, subtraction and other simple actions. The prototypes of such functions have long been calculated and compiled into special tables. It is these functions and tables that we will work with today.

But we will begin, as always, with a repetition: let’s remember what an antiderivative is, why there are infinitely many of them, and how to determine their general appearance. To do this, I picked up two simple problems.

Solving easy examples

Example #1

Let us immediately note that $\frac(\text( )\!\!\pi\!\!\text( ))(6)$ and in general the presence of $\text( )\!\!\pi\!\!\ text( )$ immediately hints to us that the required antiderivative of the function is related to trigonometry. And, indeed, if we look at the table, we will find that $\frac(1)(1+((x)^(2)))$ is nothing more than $\text(arctg)x$. So let's write it down:

In order to find, you need to write down the following:

\[\frac(\pi )(6)=\text(arctg)\sqrt(3)+C\]

\[\frac(\text( )\!\!\pi\!\!\text( ))(6)=\frac(\text( )\!\!\pi\!\!\text( )) (3)+C\]

Example No. 2

We are also talking about trigonometric functions here. If we look at the table, then, indeed, this is what happens:

We need to find among the entire set of antiderivatives the one that passes through the indicated point:

\[\text( )\!\!\pi\!\!\text( )=\arcsin \frac(1)(2)+C\]

\[\text( )\!\!\pi\!\!\text( )=\frac(\text( )\!\!\pi\!\!\text( ))(6)+C\]

Let's finally write it down:

It's that simple. The only problem is that in order to calculate antiderivatives of simple functions, you need to learn a table of antiderivatives. However, after studying the derivative table for you, I think this will not be a problem.

Solving problems containing an exponential function

To begin with, let's write the following formulas:

\[((e)^(x))\to ((e)^(x))\]

\[((a)^(x))\to \frac(((a)^(x)))(\ln a)\]

Let's see how this all works in practice.

Example #1

If we look at the contents of the brackets, we will notice that in the table of antiderivatives there is no such expression for $((e)^(x))$ to be in a square, so this square must be expanded. To do this, we use the abbreviated multiplication formulas:

Let's find the antiderivative for each of the terms:

\[((e)^(2x))=((\left(((e)^(2)) \right))^(x))\to \frac(((\left(((e)^ (2)) \right))^(x)))(\ln ((e)^(2)))=\frac(((e)^(2x)))(2)\]

\[((e)^(-2x))=((\left(((e)^(-2)) \right))^(x))\to \frac(((\left(((e )^(-2)) \right))^(x)))(\ln ((e)^(-2)))=\frac(1)(-2((e)^(2x))) \]

Now let’s collect all the terms into a single expression and get the general antiderivative:

Example No. 2

This time the degree is larger, so the abbreviated multiplication formula will be quite complex. So let's open the brackets:

Now let’s try to take the antiderivative of our formula from this construction:

As you can see, there is nothing complicated or supernatural in the antiderivatives of the exponential function. All of them are calculated through tables, but attentive students will probably notice that the antiderivative $((e)^(2x))$ is much closer to simply $((e)^(x))$ than to $((a)^(x ))$. So, maybe there is some more special rule that allows, knowing the antiderivative $((e)^(x))$, to find $((e)^(2x))$? Yes, such a rule exists. And, moreover, it is an integral part of working with the table of antiderivatives. We will now analyze it using the same expressions that we just worked with as an example.

Rules for working with the table of antiderivatives

Let's write our function again:

In the previous case, we used the following formula to solve:

\[((a)^(x))\to \frac(((a)^(x)))(\operatorname(lna))\]

But now let’s do it a little differently: let’s remember on what basis $((e)^(x))\to ((e)^(x))$. As I already said, because the derivative $((e)^(x))$ is nothing more than $((e)^(x))$, therefore its antiderivative will be equal to the same $((e) ^(x))$. But the problem is that we have $((e)^(2x))$ and $((e)^(-2x))$. Now let's try to find the derivative of $((e)^(2x))$:

\[((\left(((e)^(2x)) \right))^(\prime ))=((e)^(2x))\cdot ((\left(2x \right))^( \prime ))=2\cdot ((e)^(2x))\]

Let's rewrite our construction again:

\[((\left(((e)^(2x)) \right))^(\prime ))=2\cdot ((e)^(2x))\]

\[((e)^(2x))=((\left(\frac(((e)^(2x)))(2) \right))^(\prime ))\]

This means that when we find the antiderivative $((e)^(2x))$ we get the following:

\[((e)^(2x))\to \frac(((e)^(2x)))(2)\]

As you can see, we got the same result as before, but we did not use the formula to find $((a)^(x))$. Now this may seem stupid: why complicate the calculations when there is a standard formula? However, in slightly more complex expressions you will find that this technique is very effective, i.e. using derivatives to find antiderivatives.

As a warm-up, let's find the antiderivative of $((e)^(2x))$ in a similar way:

\[((\left(((e)^(-2x)) \right))^(\prime ))=((e)^(-2x))\cdot \left(-2 \right)\]

\[((e)^(-2x))=((\left(\frac(((e)^(-2x)))(-2) \right))^(\prime ))\]

When calculating, our construction will be written as follows:

\[((e)^(-2x))\to -\frac(((e)^(-2x)))(2)\]

\[((e)^(-2x))\to -\frac(1)(2\cdot ((e)^(2x)))\]

We got exactly the same result, but took a different path. It is this path, which now seems a little more complicated to us, that in the future will turn out to be more effective for calculating more complex antiderivatives and using tables.

Note! This is a very important point: antiderivatives, like derivatives, can be counted in many different ways. However, if all calculations and calculations are equal, then the answer will be the same. We have just seen this with the example of $((e)^(-2x))$ - on the one hand, we calculated this antiderivative “right through”, using the definition and calculating it using transformations, on the other hand, we remembered that $ ((e)^(-2x))$ can be represented as $((\left(((e)^(-2)) \right))^(x))$ and only then we used the antiderivative for the function $( (a)^(x))$. However, after all the transformations, the result was the same, as expected.

And now that we understand all this, it’s time to move on to something more significant. Now we will analyze two simple constructions, but the technique that will be used when solving them is a more powerful and useful tool than simply “running” between neighboring antiderivatives from the table.

Problem solving: finding the antiderivative of a function

Example #1

Let's break down the amount that is in the numerators into three separate fractions:

This is a fairly natural and understandable transition - most students do not have problems with it. Let's rewrite our expression as follows:

Now let's remember this formula:

In our case we will get the following:

To get rid of all these three-story fractions, I suggest doing the following:

Example No. 2

Unlike the previous fraction, the denominator is not a product, but a sum. In this case, we can no longer divide our fraction into the sum of several simple fractions, but we must somehow try to make sure that the numerator contains approximately the same expression as the denominator. In this case, it's quite simple to do it:

This notation, which in mathematical language is called “adding a zero,” will allow us to again divide the fraction into two pieces:

Now let's find what we were looking for:

That's all the calculations. Despite the apparent greater complexity than in the previous problem, the amount of calculations turned out to be even smaller.

Nuances of the solution

And this is where the main difficulty of working with tabular antiderivatives lies, this is especially noticeable in the second task. The fact is that in order to select some elements that are easily calculated through the table, we need to know what exactly we are looking for, and it is in the search for these elements that the entire calculation of antiderivatives consists.

In other words, it is not enough just to memorize the table of antiderivatives - you need to be able to see something that does not yet exist, but what the author and compiler of this problem meant. That is why many mathematicians, teachers and professors constantly argue: “What is taking antiderivatives or integration - is it just a tool or is it a real art?” In fact, in my personal opinion, integration is not an art at all - there is nothing sublime in it, it is just practice and more practice. And to practice, let's solve three more serious examples.

We train in integration in practice

Task No. 1

Let's write the following formulas:

\[((x)^(n))\to \frac(((x)^(n+1)))(n+1)\]

\[\frac(1)(x)\to \ln x\]

\[\frac(1)(1+((x)^(2)))\to \text(arctg)x\]

Let's write the following:

Problem No. 2

Let's rewrite it as follows:

The total antiderivative will be equal to:

Problem No. 3

The difficulty of this task is that, unlike the previous functions above, there is no variable $x$ at all, i.e. it is not clear to us what to add or subtract in order to get at least something similar to what is below. However, in fact, this expression is considered even simpler than any of the previous expressions, because this function can be rewritten as follows:

You may now ask: why are these functions equal? Let's check:

Let's rewrite it again:

Let's transform our expression a little:

And when I explain all this to my students, almost always the same problem arises: with the first function everything is more or less clear, with the second you can also figure it out with luck or practice, but what kind of alternative consciousness do you need to have in order to solve the third example? Actually, don't be scared. The technique that we used when calculating the last antiderivative is called “decomposition of a function into its simplest”, and this is a very serious technique, and a separate video lesson will be devoted to it.

In the meantime, I propose to return to what we just studied, namely, to exponential functions and somewhat complicate the problems with their content.

More complex problems for solving antiderivative exponential functions

Task No. 1

Let's note the following:

\[((2)^(x))\cdot ((5)^(x))=((\left(2\cdot 5 \right))^(x))=((10)^(x) )\]

To find the antiderivative of this expression, simply use the standard formula - $((a)^(x))\to \frac(((a)^(x)))(\ln a)$.

In our case, the antiderivative will be like this:

Of course, compared to the design we just solved, this one looks simpler.

Problem No. 2

Again, it's easy to see that this function can easily be divided into two separate terms - two separate fractions. Let's rewrite:

It remains to find the antiderivative of each of these terms using the formula described above:

Despite the apparent greater complexity of exponential functions compared to power functions, the overall volume of calculations and calculations turned out to be much simpler.

Of course, for knowledgeable students, what we have just discussed (especially against the backdrop of what we have discussed before) may seem like elementary expressions. However, when choosing these two problems for today's video lesson, I did not set myself the goal of telling you another complex and sophisticated technique - all I wanted to show you is that you should not be afraid to use standard algebra techniques to transform original functions.

Using a "secret" technique

In conclusion, I would like to look at another interesting technique, which, on the one hand, goes beyond what we mainly discussed today, but, on the other hand, it is, firstly, not at all complicated, i.e. Even beginner students can master it, and, secondly, it is quite often found in all kinds of tests and independent work, i.e. knowledge of it will be very useful in addition to knowledge of the table of antiderivatives.

Task No. 1

Obviously, we have something very similar to a power function. What should we do in this case? Let's think about it: $x-5$ is not that much different from $x$ - they just added $-5$. Let's write it like this:

\[((x)^(4))\to \frac(((x)^(5)))(5)\]

\[((\left(\frac(((x)^(5)))(5) \right))^(\prime ))=\frac(5\cdot ((x)^(4))) (5)=((x)^(4))\]

Let's try to find the derivative of $((\left(x-5 \right))^(5))$:

\[((\left(((\left(x-5 \right))^(5)) \right))^(\prime ))=5\cdot ((\left(x-5 \right)) ^(4))\cdot ((\left(x-5 \right))^(\prime ))=5\cdot ((\left(x-5 \right))^(4))\]

This implies:

\[((\left(x-5 \right))^(4))=((\left(\frac(((\left(x-5 \right))^(5)))(5) \ right))^(\prime ))\]

There is no such value in the table, so we have now derived this formula ourselves using the standard antiderivative formula for a power function. Let's write the answer like this:

Problem No. 2

Many students who look at the first solution may think that everything is very simple: just replace $x$ in the power function with a linear expression, and everything will fall into place. Unfortunately, everything is not so simple, and now we will see this.

By analogy with the first expression, we write the following:

\[((x)^(9))\to \frac(((x)^(10)))(10)\]

\[((\left(((\left(4-3x \right))^(10)) \right))^(\prime ))=10\cdot ((\left(4-3x \right)) ^(9))\cdot ((\left(4-3x \right))^(\prime ))=\]

\[=10\cdot ((\left(4-3x \right))^(9))\cdot \left(-3 \right)=-30\cdot ((\left(4-3x \right)) ^(9))\]

Returning to our derivative, we can write:

\[((\left(((\left(4-3x \right))^(10)) \right))^(\prime ))=-30\cdot ((\left(4-3x \right) )^(9))\]

\[((\left(4-3x \right))^(9))=((\left(\frac(((\left(4-3x \right))^(10)))(-30) \right))^(\prime ))\]

This immediately follows:

Nuances of the solution

Please note: if nothing essentially changed last time, then in the second case, instead of $-10$, $-30$ appeared. What is the difference between $-10$ and $-30$? Obviously, by a factor of $-3$. Question: where did it come from? If you look closely, you can see that it was taken as a result of calculating the derivative of a complex function - the coefficient that stood at $x$ appears in the antiderivative below. This is a very important rule, which I initially did not plan to discuss at all in today’s video lesson, but without it the presentation of tabular antiderivatives would be incomplete.

So let's do it again. Let there be our main power function:

\[((x)^(n))\to \frac(((x)^(n+1)))(n+1)\]

Now, instead of $x$, let's substitute the expression $kx+b$. What will happen then? We need to find the following:

\[((\left(kx+b \right))^(n))\to \frac(((\left(kx+b \right))^(n+1)))(\left(n+ 1\right)\cdot k)\]

On what basis do we claim this? Very simple. Let's find the derivative of the construction written above:

\[((\left(\frac(((\left(kx+b \right))^(n+1)))(\left(n+1 \right)\cdot k) \right))^( \prime ))=\frac(1)(\left(n+1 \right)\cdot k)\cdot \left(n+1 \right)\cdot ((\left(kx+b \right))^ (n))\cdot k=((\left(kx+b \right))^(n))\]

This is the same expression that originally existed. Thus, this formula is also correct, and it can be used to supplement the table of antiderivatives, or it is better to simply memorize the entire table.

Conclusions from the “secret: technique:

• Both functions that we just looked at can, in fact, be reduced to the antiderivatives indicated in the table by expanding the degrees, but if we can more or less somehow cope with the fourth degree, then I wouldn’t do the ninth degree at all dared to reveal.
• If we were to expand the degrees, we would end up with such a volume of calculations that a simple task would take us an inappropriately large amount of time.
• That is why such problems, which contain linear expressions, do not need to be solved “headlong”. As soon as you come across an antiderivative that differs from the one in the table only by the presence of the expression $kx+b$ inside, immediately remember the formula written above, substitute it into your table antiderivative, and everything will turn out much faster and easier.

Naturally, due to the complexity and seriousness of this technique, we will return to its consideration many times in future video lessons, but that’s all for today. I hope this lesson will really help those students who want to understand antiderivatives and integration.

Integration is one of the main operations in mathematical analysis. Tables of known antiderivatives can be useful, but now, after the advent of computer algebra systems, they are losing their significance. Below is a list of the most common primitives.

Table of basic integrals

Another, compact option

Table of integrals of trigonometric functions

From rational functions

From irrational functions

Integrals of transcendental functions

"C" is an arbitrary integration constant, which is determined if the value of the integral at any point is known. Each function has an infinite number of antiderivatives.

Most schoolchildren and students have problems calculating integrals. This page contains integral tables from trigonometric, rational, irrational and transcendental functions that will help in the solution. A table of derivatives will also help you.

Video - how to find integrals

If you don't quite understand this topic, watch the video, which explains everything in detail.

Table of antiderivatives ("integrals"). Table of integrals. Tabular indefinite integrals. (The simplest integrals and integrals with a parameter). Formulas for integration by parts. Newton-Leibniz formula.

Table of antiderivatives ("integrals"). Tabular indefinite integrals. (The simplest integrals and integrals with a parameter).
Integral of a power function.	Integral of a power function.
An integral that reduces to the integral of a power function if x is driven under the differential sign.
	Integral of an exponential, where a is a constant number.
Integral of a complex exponential function.	Integral of an exponential function.
An integral equal to the natural logarithm.	Integral: "Long logarithm".
	Integral: "Long logarithm".
Integral: "High logarithm".	An integral, where x in the numerator is placed under the differential sign (the constant under the sign can be either added or subtracted), is ultimately similar to an integral equal to the natural logarithm.
Integral: "High logarithm".
Cosine integral.	Sine integral.
Integral equal to tangent.	Integral equal to cotangent.
Integral equal to both arcsine and arccosine
An integral equal to both arcsine and arccosine.	An integral equal to both arctangent and arccotangent.
Integral equal to cosecant.	Integral equal to secant.
Integral equal to arcsecant.	Integral equal to arccosecant.
Integral equal to arcsecant.	Integral equal to arcsecant.
Integral equal to the hyperbolic sine.	Integral equal to hyperbolic cosine.
Integral equal to the hyperbolic sine, where sinhx is the hyperbolic sine in the English version.	Integral equal to the hyperbolic cosine, where sinhx is the hyperbolic sine in the English version.
Integral equal to the hyperbolic tangent.	Integral equal to the hyperbolic cotangent.
Integral equal to the hyperbolic secant.	Integral equal to the hyperbolic cosecant.

Formulas for integration by parts. Integration rules.

Formulas for integration by parts. Newton-Leibniz formula. Rules of integration.
Integrating a product (function) by a constant:
Integrating the sum of functions:
indefinite integrals:
Formula for integration by parts definite integrals:
Newton-Leibniz formula definite integrals:	Where F(a),F(b) are the values ​​of the antiderivatives at points b and a, respectively.

Table of derivatives. Tabular derivatives. Derivative of the product. Derivative of the quotient. Derivative of a complex function.

If x is an independent variable, then:

Table of derivatives. Tabular derivatives."table derivative" - ​​yes, unfortunately, this is exactly how they are searched for on the Internet
	Derivative of a power function
	Derivative of the exponent
Derivative of a complex exponential function	Derivative of exponential function
Derivative of a logarithmic function	Derivative of the natural logarithm
	Derivative of the natural logarithm of a function
Derivative of sine	Derivative of cosine
Derivative of cosecant	Derivative of a secant
Derivative of arcsine	Derivative of arc cosine
Derivative of arcsine	Derivative of arc cosine
Tangent derivative	Derivative of cotangent
Derivative of arctangent	Derivative of arc cotangent
Derivative of arctangent	Derivative of arc cotangent
Derivative of arcsecant	Derivative of arccosecant
Derivative of arcsecant	Derivative of arccosecant
Derivative of the hyperbolic sine Derivative of the hyperbolic sine in the English version	Derivative of hyperbolic cosine Derivative of hyperbolic cosine in English version
Derivative of hyperbolic tangent	Derivative of hyperbolic cotangent
Derivative of the hyperbolic secant	Derivative of the hyperbolic cosecant

Rules of differentiation. Derivative of the product. Derivative of the quotient. Derivative of a complex function.
Derivative of a product (function) by a constant:
Derivative of sum (functions):
Derivative of product (functions):
Derivative of the quotient (of functions):
Derivative of a complex function:

Properties of logarithms. Basic formulas for logarithms. Decimal (lg) and natural logarithms (ln).

Basic logarithmic identity
Let's show how any function of the form a b can be made exponential. Since a function of the form e x is called exponential, then
Any function of the form a b can be represented as a power of ten

Natural logarithm ln (logarithm to base e = 2.718281828459045...) ln(e)=1; ln(1)=0

Taylor series. Taylor series expansion of a function.

It turns out that the majority practically encountered mathematical functions can be represented with any accuracy in the vicinity of a certain point in the form of power series containing powers of a variable in increasing order. For example, in the vicinity of the point x=1:

When using series called Taylor's rows mixed functions containing, say, algebraic, trigonometric and exponential functions can be expressed as purely algebraic functions. Using series, you can often quickly perform differentiation and integration.

The Taylor series in the neighborhood of point a has the form:

1) , where f(x) is a function that has derivatives of all orders at x = a. R n - the remainder term in the Taylor series is determined by the expression

2)

The k-th coefficient (at x k) of the series is determined by the formula

3) A special case of the Taylor series is the Maclaurin (=McLaren) series (the expansion occurs around the point a=0)

at a=0

members of the series are determined by the formula

Conditions for using Taylor series.

1. In order for the function f(x) to be expanded into a Taylor series on the interval (-R;R), it is necessary and sufficient that the remainder term in the Taylor (Maclaurin (=McLaren)) formula for this function tends to zero as k →∞ on the specified interval (-R;R).

2. It is necessary that there are derivatives for a given function at the point in the vicinity of which we are going to construct the Taylor series.

Properties of Taylor series.

If f is an analytic function, then its Taylor series at any point a in the domain of definition of f converges to f in some neighborhood of a.

There are infinitely differentiable functions whose Taylor series converges, but at the same time differs from the function in any neighborhood of a. For example:

Taylor series are used in approximation (approximation is a scientific method that consists of replacing some objects with others, in one sense or another close to the original ones, but simpler) of a function by polynomials. In particular, linearization ((from linearis - linear), one of the methods of approximate representation of closed nonlinear systems, in which the study of a nonlinear system is replaced by the analysis of a linear system, in some sense equivalent to the original one.) equations occurs by expanding into a Taylor series and cutting off all terms above first order.

Thus, almost any function can be represented as a polynomial with a given accuracy.

Examples of some common expansions of power functions in Maclaurin series (=McLaren, Taylor in the vicinity of point 0) and Taylor in the vicinity of point 1. The first terms of expansions of the main functions in Taylor and McLaren series.

Examples of some common expansions of power functions in Maclaurin series (=McLaren, Taylor in the vicinity of point 0)

Examples of some common Taylor series expansions in the vicinity of point 1

Let us list the integrals of elementary functions, which are sometimes called tabular:

Any of the above formulas can be proven by taking the derivative of the right-hand side (the result will be the integrand).

Integration methods

Let's look at some basic integration methods. These include:

1. Decomposition method(direct integration).

This method is based on the direct use of tabular integrals, as well as on the use of properties 4 and 5 of the indefinite integral (i.e., taking the constant factor out of brackets and/or representing the integrand as a sum of functions - decomposition of the integrand into terms).

Example 1. For example, to find(dx/x 4) you can directly use the table integral forx n dx. In fact,(dx/x 4) =x -4 dx=x -3 /(-3) +C= -1/3x 3 +C.

Example 2. To find it, we use the same integral:

Example 3. To find it you need to take

Example 4. To find, we represent the integrand function in the form and use the table integral for the exponential function:

Let's consider the use of bracketing a constant factor.

Example 5.Let's find, for example . Considering that, we get

Example 6. We'll find it. Because the , let's use the table integral We get

In the following two examples, you can also use bracketing and table integrals:

Example 7.

(we use and );

Example 8.

(we use And ).

Let's look at more complex examples that use the sum integral.

Example 9. For example, let's find
. To apply the expansion method in the numerator, we use the sum cube formula , and then divide the resulting polynomial by the denominator, term by term.

=((8x 3/2 + 12x+ 6x 1/2 + 1)/(x 3/2))dx=(8 + 12x -1/2 + 6/x+x -3/2)dx= 8 dx+ 12x -1/2 dx+ + 6dx/x+x -3/2 dx=

It should be noted that at the end of the solution one common constant C is written (and not separate ones when integrating each term). In the future, it is also proposed to omit the constants from the integration of individual terms in the solution process as long as the expression contains at least one indefinite integral (we will write one constant at the end of the solution).

Example 10. We'll find . To solve this problem, let's factorize the numerator (after this we can reduce the denominator).

Example 11. We'll find it. Trigonometric identities can be used here.

Sometimes, in order to decompose an expression into terms, you have to use more complex techniques.

Example 12. We'll find . In the integrand we select the whole part of the fraction . Then

Example 13. We'll find

2. Variable replacement method (substitution method)

The method is based on the following formula: f(x)dx=f((t))`(t)dt, where x =(t) is a function differentiable on the interval under consideration.

Proof. Let's find the derivatives with respect to the variable t from the left and right sides of the formula.

Note that on the left side there is a complex function whose intermediate argument is x = (t). Therefore, to differentiate it with respect to t, we first differentiate the integral with respect to x, and then take the derivative of the intermediate argument with respect to t.

( f(x)dx)` t = ( f(x)dx)` x *x` t = f(x) `(t)

Derivative from the right side:

(f((t))`(t)dt)` t =f((t))`(t) =f(x)`(t)

Since these derivatives are equal, by corollary to Lagrange’s theorem, the left and right sides of the formula being proved differ by a certain constant. Since the indefinite integrals themselves are defined up to an indefinite constant term, this constant can be omitted from the final notation. Proven.

A successful change of variable allows you to simplify the original integral, and in the simplest cases, reduce it to a tabular one. In the application of this method, a distinction is made between linear and nonlinear substitution methods.

a) Linear substitution method Let's look at an example.

Example 1.
. Let t= 1 – 2x, then

dx=d(½ - ½t) = - ½dt

It should be noted that the new variable does not need to be written out explicitly. In such cases, they talk about transforming a function under the differential sign or about introducing constants and variables under the differential sign, i.e. O implicit variable replacement.

Example 2. For example, let's findcos(3x + 2)dx. By the properties of the differential dx = (1/3)d(3x) = (1/3)d(3x + 2), thencos(3x + 2)dx =(1/3)cos(3x + 2)d (3x + + 2) = (1/3)cos(3x + 2)d(3x + 2) = (1/3)sin(3x + 2) +C.

In both examples considered, linear substitution t=kx+b(k0) was used to find the integrals.

In the general case, the following theorem is valid.

Linear substitution theorem. Let F(x) be some antiderivative of the function f(x). Thenf(kx+b)dx= (1/k)F(kx+b) +C, where k and b are some constants,k0.

Proof.

By definition of the integral f(kx+b)d(kx+b) =F(kx+b) +C. Hod(kx+b)= (kx+b)`dx=kdx. Let's take the constant factor k out of the integral sign: kf(kx+b)dx=F(kx+b) +C. Now we can divide the left and right sides of the equality into two and obtain the statement to be proved up to the designation of the constant term.

This theorem states that if in the definition of the integral f(x)dx= F(x) + C instead of the argument x we ​​substitute the expression (kx+b), this will lead to the appearance of an additional factor 1/k in front of the antiderivative.

Using the proven theorem, we solve the following examples.

Example 3.

We'll find . Here kx+b= 3 –x, i.e. k= -1,b= 3. Then

Example 4.

We'll find it. Herekx+b= 4x+ 3, i.e. k= 4,b= 3. Then

Example 5.

We'll find . Here kx+b= -2x+ 7, i.e. k= -2,b= 7. Then

.

Example 6. We'll find
. Here kx+b= 2x+ 0, i.e. k= 2,b= 0.

.

Let us compare the result obtained with example 8, which was solved by the decomposition method. Solving the same problem using a different method, we got the answer
. Let's compare the results: Thus, these expressions differ from each other by a constant term , i.e. The answers received do not contradict each other.

Example 7. We'll find
. Let's select a perfect square in the denominator.

In some cases, changing a variable does not reduce the integral directly to a tabular one, but can simplify the solution, making it possible to use the expansion method at a subsequent step.

Example 8. For example, let's find . Replace t=x+ 2, then dt=d(x+ 2) =dx. Then

,

where C = C 1 – 6 (when substituting the expression (x+ 2) instead of the first two terms we get ½x 2 -2x– 6).

Example 9. We'll find
. Let t= 2x+ 1, then dt= 2dx;dx= ½dt;x= (t– 1)/2.

Let's substitute the expression (2x+ 1) for t, open the brackets and give similar ones.

Note that in the process of transformations we moved to another constant term, because the group of constant terms could be omitted during the transformation process.

b) Nonlinear substitution method Let's look at an example.

Example 1.
. Lett= -x 2. Next, one could express x in terms of t, then find an expression for dx and implement a change of variable in the desired integral. But in this case it’s easier to do things differently. Let's finddt=d(-x 2) = -2xdx. Note that the expression xdx is a factor of the integrand of the desired integral. Let us express it from the resulting equalityxdx= - ½dt. Then

=  (- ½)e t dt = (- ½) e t dt = (- ½)e t + C = (- ½)
+C

Let's look at a few more examples.

Example 2. We'll find . Let t= 1 -x 2 . Then

Example 3. We'll find . Lett=. Then

;

Example 4. In the case of nonlinear substitution, it is also convenient to use implicit variable substitution.

For example, let's find
. Let's write xdx= = (-1/4)d(3 - 2x 2) (implicitly replaced by the variable t= 3 - 2x 2). Then

Example 5. We'll find . Here we also introduce a variable under the differential sign: (implicit replacement = 3 + 5x 3). Then

Example 6. We'll find . Because the ,

Example 7. We'll find it. Since then

Let's look at a few examples in which it becomes necessary to combine various substitutions.

Example 8. We'll find
. Lett= 2x+ 1, thenx= (t– 1)/2;dx= ½dt.

Example 9. We'll find
. Lett=x- 2, thenx=t+ 2;dx=dt.

Direct integration using the table of antiderivatives (table of indefinite integrals)

Table of antiderivatives

We can find the antiderivative from a known differential of a function if we use the properties of the indefinite integral. From the table of basic elementary functions, using the equalities ∫ d F (x) = ∫ F " (x) d x = ∫ f (x) d x = F (x) + C and ∫ k f (x) d x = k ∫ f (x) d x we ​​can make a table of antiderivatives.

Let's write the table of derivatives in the form of differentials.

Constant y = C C" = 0 Power function y = x p. (x p) " = p x p - 1	Constant y = C d (C) = 0 d x Power function y = x p. d (x p) = p x p - 1 d x
(a x) " = a x ln a	Exponential function y = a x. d (a x) = a x ln α d x In particular, for a = e we have y = e x d (e x) = e x d x
log a x " = 1 x ln a	Logarithmic functions y = log a x . d (log a x) = d x x ln a In particular, for a = e we have y = ln x d (ln x) = d x x
Trigonometric functions. sin x " = cos x (cos x) " = - sin x (t g x) " = 1 c o s 2 x (c t g x) " = - 1 sin 2 x	Trigonometric functions. d sin x = cos x · d x d (cos x) = - sin x · d x d (t g x) = d x c o s 2 x d (c t g x) = - d x sin 2 x
a r c sin x " = 1 1 - x 2 a r c cos x " = - 1 1 - x 2 a r c t g x " = 1 1 + x 2 a r c c t g x " = - 1 1 + x 2	Inverse trigonometric functions. d a r c sin x = d x 1 - x 2 d a r c cos x = - d x 1 - x 2 d a r c t g x = d x 1 + x 2 d a r c c t g x = - d x 1 + x 2

Let us illustrate the above with an example. Let's find the indefinite integral of the power function f (x) = x p.

According to the table of differentials d (x p) = p · x p - 1 · d x. By the properties of the indefinite integral we have ∫ d (x p) = ∫ p · x p - 1 · d x = p · ∫ x p - 1 · d x = x p + C . Therefore, ∫ x p - 1 · d x = x p p + C p , p ≠ 0. The second version of the entry is as follows: ∫ x p · d x = x p + 1 p + 1 + C p + 1 = x p + 1 p + 1 + C 1, p ≠ - 1.

Let us take it equal to - 1 and find the set of antiderivatives of the power function f (x) = x p: ∫ x p · d x = ∫ x - 1 · d x = ∫ d x x .

Now we need a table of differentials for the natural logarithm d (ln x) = d x x, x > 0, therefore ∫ d (ln x) = ∫ d x x = ln x. Therefore ∫ d x x = ln x , x > 0 .

Table of antiderivatives (indefinite integrals)

The left column of the table contains formulas that are called basic antiderivatives. The formulas in the right column are not basic, but can be used to find indefinite integrals. They can be checked by differentiation.

Direct integration

To perform direct integration, we will use tables of antiderivatives, integration rules ∫ f (k x + b) d x = 1 k F (k x + b) + C, as well as properties of indefinite integrals ∫ k f (x) d x = k · ∫ f (x) d x ∫ (f (x) ± g (x)) d x = ∫ f (x) d x ± ∫ g (x) d x

The table of basic integrals and properties of integrals can be used only after an easy transformation of the integrand.

Example 1

Let's find the integral ∫ 3 sin x 2 + cos x 2 2 d x

Solution

We remove coefficient 3 from under the integral sign:

∫ 3 sin x 2 + cos x 2 2 d x = 3 ∫ sin x 2 + cos x 2 2 d x

Using trigonometry formulas, we transform the integrand function:

3 ∫ sin x 2 + cos x 2 2 d x = 3 ∫ sin x 2 2 + 2 sin x 2 cos x 2 + cos x 2 2 d x = = 3 ∫ 1 + 2 sin x 2 cos x 2 d x = 3 ∫ 1 + sin x d x

Since the integral of the sum is equal to the sum of the integrals, then
3 ∫ 1 + sin x d x = 3 ∫ 1 d x + ∫ sin x d x

We use the data from the table of antiderivatives: 3 ∫ 1 d x + ∫ sin x d x = 3 (1 x + C 1 - cos x + C 2) = = empty 3 C 1 + C 2 = C = 3 x - 3 cos x + C

Answer:∫ 3 sin x 2 + cos x 2 2 d x = 3 x - 3 cos x + C .

Example 2

It is necessary to find the set of antiderivatives of the function f (x) = 2 3 4 x - 7 .

Solution

We use the table of antiderivatives for the exponential function: ∫ a x · d x = a x ln a + C . This means that ∫ 2 x · d x = 2 x ln 2 + C .

We use the integration rule ∫ f (k x + b) d x = 1 k F (k x + b) + C .

We get ∫ 2 3 4 x - 7 · d x = 1 3 4 · 2 3 4 x - 7 ln 2 + C = 4 3 · 2 3 4 x - 7 ln 2 + C .

Answer: f (x) = 2 3 4 x - 7 = 4 3 2 3 4 x - 7 ln 2 + C

Using the table of antiderivatives, properties and the rule of integration, we can find a lot of indefinite integrals. This is possible in cases where it is possible to transform the integrand.

To find the integral of the logarithm function, tangent and cotangent functions, and a number of others, special methods are used, which we will consider in the section “Basic methods of integration.”

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