The method of integration by parts is used mainly when the integrand consists of the product of two factors of a certain type. The formula for integration by parts is:

It makes it possible to reduce the calculation of a given integral
to the calculation of the integral
, which turns out to be simpler than the given one.

Most of the integrals calculated by the method of integration by parts can be divided into three groups:

1. Integrals of the form
,
,
, Where
- polynomial,
- number not equal to zero

In this case, through denote polynomial

.

2. Integrals of the form
,
,
,
,
, Where
is a polynomial.

In this case, through
designate
, and the rest of the integrand through :

3. Integrals of the form
,
, Where
- numbers.

In this case, through designate
and apply the integration-by-parts formula twice, returning as a result to the original integral, after which the original integral is expressed from the equality.

Comment: In some cases, to find a given integral, the integration-by-parts formula must be applied several times. Also, the method of integration by parts is combined with other methods.

Example 26.

Find integrals by the method by parts: a)
; b)
.

Solution.

3.1.4. Integration of fractional rational functions

Fractional-rational function(rational fraction) is a function equal to the ratio of two polynomials:
, Where
is a degree polynomial
,
is a degree polynomial .

The rational fraction is called correct, if the degree of the polynomial in the numerator is less than the degree of the polynomial in the denominator, i.e.
otherwise (if
) a rational fraction is called wrong.

Any improper rational fraction can be represented as the sum of a polynomial
and a proper rational fraction by dividing the numerator by the denominator according to the rule of dividing polynomials:

Where
is the integer part of the division, is a proper rational fraction,
- remainder of the division.

Proper rational fractions of the form:

I. ;

II.
;

III.
;

IV.
,

Where ,,
,
,,,
are real numbers and
(that is, the square trinomial in the denominator III and IV of fractions has no roots - the discriminant is negative) are called the simplest rational fractions I, II, III and IV types.

Integration of simple fractions

Integrals from the simplest fractions of four types are calculated as follows.

I)
.

II) ,
.

III) To integrate the simplest fraction of type III, a full square is selected in the denominator, a replacement is made
. The integral after substitution is divided into two integrals. The first integral is calculated by extracting the derivative of the denominator in the numerator, which gives a tabular integral, and the second integral is transformed to the form
, because
, which also gives a table integral.

;

IV) To integrate the simplest fraction of type IV, a full square is selected in the denominator, a replacement is made
. The integral after substitution is divided into two integrals. The first integral is calculated by substituting
, and the second with the help of recurrence relations.

Example 27.

Find integrals of simple fractions:

A)
; b)
; V)
.

Solution.

A)
.

Any proper rational fraction whose denominator can be factorized can be represented as a sum of simple fractions. The expansion into the sum of simple fractions is carried out by the method of indefinite coefficients. It is as follows:

corresponds to one fraction of the form ;

- each multiplier of the denominator
corresponding amount fractions of the form

corresponds to a fraction of the form
;

- to each square factor of the denominator
corresponding amount fractions of the form

where are undefined coefficients.

To find indefinite coefficients, the right side in the form of the sum of simple fractions is reduced to a common denominator and converted. The result is a fraction with the same denominator as on the left side of the equation. Then discard the denominators and equate the numerators. The result is an identity equality in which the left side is a polynomial with known coefficients, and the right side is a polynomial with indefinite coefficients.

There are two ways to determine unknown coefficients: the method of uncertain coefficients and the method of partial values.

Method of indefinite coefficients.

Because polynomials are identically equal, then the coefficients are equal at the same powers . Equating the coefficients at the same powers in the polynomials of the left and right parts, we obtain a system of linear equations. Solving the system, we determine the uncertain coefficients.

Partial value method.

Because polynomials are identically equal, then, substituting instead of to the left and right sides of any number, we obtain the correct equality, which is linear with respect to the unknown coefficients. Substituting so many values , how many unknown coefficients, we get a system of linear equations. Instead of any numbers can be substituted in the left and right parts, however, it is more convenient to substitute the roots of the denominators of fractions.

After finding the values of the unknown coefficients, the original fraction is written as the sum of the simplest fractions in the integrand, and the previously considered integration over each simple fraction is carried out.

Integration scheme rational fractions:

1. If the integrand is incorrect, then it is necessary to represent it as the sum of a polynomial and a proper rational fraction (i.e., divide the numerator polynomial by the denominator polynomial with a remainder). If the integrand is correct, we immediately go to the second paragraph of the scheme.

2. Factor out the denominator of a proper rational fraction, if possible.

3. Decompose a proper rational fraction into the sum of simple rational fractions using the method of indefinite coefficients.

4. Integrate the resulting sum of the polynomial and simple fractions.

Example 28.

Find integrals of rational fractions:

A)
; b)
; V)
.

Solution.

A)
.

Because integrand is an improper rational fraction, then we select the integer part, i.e. represent it as the sum of a polynomial and a proper rational fraction. Divide the polynomial in the numerator by the polynomial in the denominator by a corner.

The original integral will take the form:
.

We expand a proper rational fraction into a sum of simple fractions using the method of indefinite coefficients:

, we get:

Solving a system of linear equations, we obtain the values of uncertain coefficients: A = 1; IN = 3.

Then the desired expansion has the form:
.

=
.

b)
.

We discard the denominators and equate the left and right parts:

Equating the coefficients at the same powers , we get the system:

Solving a system of five linear equations, we find uncertain coefficients:

Let's find the original integral, taking into account the resulting expansion:

V)
.

We expand the integrand (proper rational fraction) into the sum of simple fractions using the method of indefinite coefficients. We are looking for a decomposition in the form:

Reducing to a common denominator, we get:

We discard the denominators and equate the left and right parts:

To find uncertain coefficients, we use the method of partial values. Let's give partial values at which the factors vanish, i.e. we substitute these values into the last expression and get three equations:

;
;

;
.

Then the desired expansion has the form:

Let's find the original integral, taking into account the resulting expansion:

definite integral from a continuous function f(x) on the finite interval [ a, b] (where ) is the increment of some of its primitive on this segment. (In general, understanding will be noticeably easier if you repeat the topic indefinite integral) In this case, we use the notation

As can be seen in the graphs below (the increment of the antiderivative function is indicated by ), The definite integral can be either positive or negative.(It is calculated as the difference between the value of the antiderivative in the upper limit and its value in the lower limit, i.e. as F(b) - F(a)).

Numbers a And b are called the lower and upper limits of integration, respectively, and the interval [ a, b] is the segment of integration.

Thus, if F(x) is some antiderivative function for f(x), then, according to the definition,

(38)

Equality (38) is called Newton-Leibniz formula . Difference F(b) – F(a) is briefly written like this:

Therefore, the Newton-Leibniz formula will be written as follows:

(39)

Let us prove that the definite integral does not depend on which antiderivative of the integrand is taken when calculating it. Let F(x) and F( X) are arbitrary antiderivatives of the integrand. Since these are antiderivatives of the same function, they differ by a constant term: Ф( X) = F(x) + C. That's why

Thus, it is established that on the segment [ a, b] increments of all antiderivatives of the function f(x) match up.

Thus, to calculate the definite integral, it is necessary to find any antiderivative of the integrand, i.e. First you need to find the indefinite integral. Constant WITH excluded from subsequent calculations. Then the Newton-Leibniz formula is applied: the value of the upper limit is substituted into the antiderivative function b , further - the value of the lower limit a and calculate the difference F(b) - F(a) . The resulting number will be a definite integral..

At a = b accepted by definition

Example 1

Solution. Let's find the indefinite integral first:

Applying the Newton-Leibniz formula to the antiderivative

(at WITH= 0), we get

However, when calculating a definite integral, it is better not to find the antiderivative separately, but immediately write the integral in the form (39).

Example 2 Calculate a definite integral

Solution. Using the formula

Properties of the Definite Integral

Theorem 2.The value of the definite integral does not depend on the designation of the integration variable, i.e.

(40)

Let F(x) is antiderivative for f(x). For f(t) the antiderivative is the same function F(t), in which the independent variable is denoted differently. Hence,

Based on formula (39), the last equality means the equality of the integrals

Theorem 3.The constant factor can be taken out of the sign of a definite integral, i.e.

(41)

Theorem 4.The definite integral of the algebraic sum of a finite number of functions is equal to the algebraic sum of the definite integrals of these functions, i.e.

(42)

Theorem 5.If the integration segment is divided into parts, then the definite integral over the entire segment is equal to the sum of the definite integrals over its parts, i.e. If

(43)

Theorem 6.When rearranging the limits of integration, the absolute value of the definite integral does not change, but only its sign changes, i.e.

(44)

Theorem 7(mean value theorem). The definite integral is equal to the product of the length of the integration segment and the value of the integrand at some point inside it, i.e.

(45)

Theorem 8.If the upper integration limit is greater than the lower one and the integrand is non-negative (positive), then the definite integral is also non-negative (positive), i.e. If

Theorem 9.If the upper limit of integration is greater than the lower limit and the functions and are continuous, then the inequality

can be integrated term by term, i.e.

(46)

The properties of the definite integral allow us to simplify the direct calculation of integrals.

Example 5 Calculate a definite integral

Using Theorems 4 and 3, and when finding antiderivatives - tabular integrals(7) and (6), we get

Definite integral with variable upper limit

Let f(x) is continuous on the interval [ a, b] function, and F(x) is its prototype. Consider the definite integral

(47)

and through t the integration variable is denoted so as not to confuse it with the upper bound. When it changes X the definite integral (47) also changes, i.e., it is a function of the upper limit of integration X, which we denote by F(X), i.e.

(48)

Let us prove that the function F(X) is antiderivative for f(x) = f(t). Indeed, differentiating F(X), we get

because F(x) is antiderivative for f(x), A F(a) is a constant value.

Function F(X) is one of the infinite set of antiderivatives for f(x), namely the one that x = a goes to zero. This statement is obtained if in equality (48) we put x = a and use Theorem 1 of the previous section.

Calculation of definite integrals by the method of integration by parts and the method of change of variable

where, by definition, F(x) is antiderivative for f(x). If in the integrand we make the change of variable

then, in accordance with formula (16), we can write

In this expression

antiderivative function for

Indeed, its derivative, according to the rule of differentiation of a complex function, is equal to

Let α and β be the values of the variable t, for which the function

takes respectively the values a And b, i.e.

But, according to the Newton-Leibniz formula, the difference F(b) – F(a) There is

Definite integral by parts examples. Solving integrals online

Previously, for a given function, guided by various formulas and rules, we found its derivative. The derivative has numerous applications: it is the speed of movement (or, more generally, the speed of any process); slope of the tangent to the graph of the function; using the derivative, you can investigate the function for monotonicity and extrema; It helps to solve optimization problems.

But along with the problem of finding the speed from a known law of motion, there is also an inverse problem - the problem of restoring the law of motion from a known speed. Let's consider one of these problems.

Example 1 A material point moves along a straight line, the speed of its movement at time t is given by the formula v=gt. Find the law of motion.
Solution. Let s = s(t) be the desired law of motion. It is known that s"(t) = v(t). So, to solve the problem, you need to choose a function s = s(t), whose derivative is equal to gt. It is easy to guess that $s(t) = \frac(gt^ 2)(2) $ Indeed
$s"(t) = \left(\frac(gt^2)(2) \right)" = \frac(g)(2)(t^2)" = \frac(g)(2) \ cdot 2t=gt$
Answer: $s(t) = \frac(gt^2)(2) $

We note right away that the example is solved correctly, but incompletely. We got $s(t) = \frac(gt^2)(2) $. In fact, the problem has infinitely many solutions: any function of the form $s(t) = \frac(gt^2)(2) + C $, where C is an arbitrary constant, can serve as a law of motion, since $\left (\frac(gt^2)(2) +C \right)" = gt $

To make the problem more specific, we had to fix the initial situation: indicate the coordinate of the moving point at some point in time, for example, at t = 0. If, say, s(0) = s 0 , then from the equality s(t) = (gt 2)/2 + C we get: s(0) = 0 + C, i.e. C = s 0 . Now the law of motion is uniquely defined: s(t) = (gt 2)/2 + s 0 .

In mathematics, mutually inverse operations are assigned different names, come up with special notations, for example: squaring (x 2) and extracting the square root ($\sqrt(x) $), sine (sin x) and arcsine (arcsin x) and etc. The process of finding the derivative with respect to a given function is called differentiation, and the inverse operation, i.e. the process of finding a function by a given derivative, - integration.

The term "derivative" itself can be justified "in a worldly way": the function y \u003d f (x) "produces into the world" a new function y" \u003d f "(x). The function y \u003d f (x) acts as if as a “parent”, but mathematicians, of course, do not call it a “parent” or “producer”, they say that this is, in relation to the function y "= f" (x) , the primary image, or antiderivative.

Definition. A function y = F(x) is called an antiderivative for a function y = f(x) on an interval X if $x \in X $ satisfies the equality F"(x) = f(x)

In practice, the interval X is usually not specified, but implied (as the natural domain of the function).

Let's give examples.
1) The function y \u003d x 2 is an antiderivative for the function y \u003d 2x, since for any x the equality (x 2)" \u003d 2x is true
2) The function y \u003d x 3 is an antiderivative for the function y \u003d 3x 2, since for any x the equality (x 3)" \u003d 3x 2 is true
3) The function y \u003d sin (x) is an antiderivative for the function y \u003d cos (x), since for any x the equality (sin (x)) "= cos (x) is true

When finding antiderivatives, as well as derivatives, not only formulas are used, but also some rules. They are directly related to the corresponding rules for computing derivatives.

We know that the derivative of a sum is equal to the sum of the derivatives. This rule generates the corresponding rule for finding antiderivatives.

Rule 1 The antiderivative of a sum is equal to the sum of antiderivatives.

We know that the constant factor can be taken out of the sign of the derivative. This rule generates the corresponding rule for finding antiderivatives.

Rule 2 If F(x) is an antiderivative for f(x), then kF(x) is an antiderivative for kf(x).

Theorem 1. If y = F(x) is the antiderivative for the function y = f(x), then the antiderivative for the function y = f(kx + m) is the function $y=\frac(1)(k)F(kx+m) $

Theorem 2. If y = F(x) is an antiderivative for a function y = f(x) on an interval X, then the function y = f(x) has infinitely many antiderivatives, and they all have the form y = F(x) + C.

Integration methods

Variable replacement method (substitution method)

The substitution integration method consists in introducing a new integration variable (that is, a substitution). In this case, the given integral is reduced to a new integral, which is tabular or reducible to it. There are no general methods for selecting substitutions. The ability to correctly determine the substitution is acquired by practice.
Let it be required to calculate the integral $\textstyle \int F(x)dx $. Let's make a substitution $x= \varphi(t) $ where $\varphi(t) $ is a function that has a continuous derivative.
Then $dx = \varphi " (t) \cdot dt $ and based on the invariance property of the indefinite integral integration formula, we obtain the substitution integration formula:
$\int F(x) dx = \int F(\varphi(t)) \cdot \varphi " (t) dt $

Integration of expressions like $\textstyle \int \sin^n x \cos^m x dx $

If m is odd, m > 0, then it is more convenient to make the substitution sin x = t.
If n is odd, n > 0, then it is more convenient to make the substitution cos x = t.
If n and m are even, then it is more convenient to make the substitution tg x = t.

Integration by parts

Integration by parts - applying the following formula for integration:
$\textstyle \int u \cdot dv = u \cdot v - \int v \cdot du $
or:
$\textstyle \int u \cdot v" \cdot dx = u \cdot v - \int v \cdot u" \cdot dx $

Table of indefinite integrals (antiderivatives) of some functions

$$ \int 0 \cdot dx = C $$ $$ \int 1 \cdot dx = x+C $$ $$ \int x^n dx = \frac(x^(n+1))(n+1 ) +C \;\; (n \neq -1) $$ $$ \int \frac(1)(x) dx = \ln |x| +C $$ $$ \int e^x dx = e^x +C $$ $$ \int a^x dx = \frac(a^x)(\ln a) +C \;\; (a>0, \;\; a \neq 1) $$ $$ \int \cos x dx = \sin x +C $$ $$ \int \sin x dx = -\cos x +C $$ $ $ \int \frac(dx)(\cos^2 x) = \text(tg) x +C $$ $$ \int \frac(dx)(\sin^2 x) = -\text(ctg) x +C $$ $$ \int \frac(dx)(\sqrt(1-x^2)) = \text(arcsin) x +C $$ $$ \int \frac(dx)(1+x^2 ) = \text(arctg) x +C $$ $$ \int \text(ch) x dx = \text(sh) x +C $$ $$ \int \text(sh) x dx = \text(ch )x+C $$

Integration by parts. Solution examples

Hello again. Today in the lesson we will learn how to integrate by parts. The method of integration by parts is one of the cornerstones of integral calculus. At the test, exam, the student is almost always offered to solve integrals of the following types: the simplest integral (see article) or an integral to change the variable (see article) or the integral just on method of integration by parts.

As always, on hand should be: Table of integrals And Derivative table. If you still do not have them, then please visit the storeroom of my site: Mathematical formulas and tables. I will not get tired of repeating - it is better to print everything. I will try to present all the material in a consistent, simple and accessible way; there are no particular difficulties in integrating by parts.

What problem does integration by parts solve? The method of integration by parts solves a very important problem, it allows you to integrate some functions that are not in the table, work functions, and in some cases - and private. As we remember, there is no convenient formula: . But there is this one: is the formula for integration by parts in person. I know, I know, you are the only one - with her we will work the whole lesson (it’s already easier).

And immediately the list in the studio. Integrals of the following types are taken by parts:

1) , , - logarithm, logarithm multiplied by some polynomial.

2) ,is an exponential function multiplied by some polynomial. This also includes integrals like - an exponential function multiplied by a polynomial, but in practice it is 97 percent, a pretty letter “e” flaunts under the integral. ... the article turns out to be something lyrical, oh yes ... spring has come.

3) , , are trigonometric functions multiplied by some polynomial.

4) , - inverse trigonometric functions (“arches”), “arches”, multiplied by some polynomial.

Also, some fractions are taken in parts, we will also consider the corresponding examples in detail.

Integrals of logarithms

Example 1

Classic. From time to time, this integral can be found in tables, but it is undesirable to use a ready-made answer, since the teacher has beriberi in the spring and he will scold a lot. Because the integral under consideration is by no means tabular - it is taken in parts. We decide:

We interrupt the solution for intermediate explanations.

We use the formula for integration by parts:

The formula is applied from left to right

We look at the left side:. Obviously, in our example (and in all the others that we will consider), something needs to be denoted by , and something by .

In integrals of the type under consideration, we always denote the logarithm.

Technically, the design of the solution is implemented as follows, we write in the column:

That is, for we denoted the logarithm, and for - the remaining part integrand.

Next step: find the differential:

The differential is almost the same as the derivative, we have already discussed how to find it in previous lessons.

Now we find the function . In order to find the function it is necessary to integrate right side lower equality :

Now we open our solution and construct the right side of the formula: .
By the way, here is an example of a final solution with a few notes:

The only moment in the product, I immediately rearranged and, since it is customary to write the multiplier before the logarithm.

As you can see, applying the integration-by-parts formula essentially reduced our solution to two simple integrals.

Please note that in some cases right after application of the formula, a simplification is necessarily carried out under the remaining integral - in the example under consideration, we reduced the integrand by "x".

Let's do a check. To do this, you need to take the derivative of the answer:

The original integrand has been obtained, which means that the integral has been solved correctly.

During the verification, we used the product differentiation rule: . And this is no coincidence.

Integration by parts formula and formula These are two mutually inverse rules.

Example 2

Find the indefinite integral.

The integrand is the product of the logarithm and the polynomial.
We decide.

I will once again describe in detail the procedure for applying the rule, in the future the examples will be made out more briefly, and if you have any difficulties in solving it yourself, you need to go back to the first two examples of the lesson.

As already mentioned, for it is necessary to designate the logarithm (the fact that it is in a degree does not matter). We denote the remaining part integrand.

We write in a column:

First we find the differential:

Here we use the rule of differentiation of a complex function . It is no coincidence that at the very first lesson of the topic Indefinite integral. Solution examples I focused on the fact that in order to master the integrals, you need to "get your hand" on the derivatives. Derivatives will have to face more than once.

Now we find the function , for this we integrate right side lower equality :

For integration, we applied the simplest tabular formula

Now you are ready to apply the formula . We open it with an "asterisk" and "design" the solution in accordance with the right side:

Under the integral, we again have a polynomial on the logarithm! Therefore, the solution is interrupted again and the rule of integration by parts is applied a second time. Do not forget that for in similar situations the logarithm is always denoted.

It would be nice if at this point you were able to find the simplest integrals and derivatives orally.

(1) Do not get confused in the signs! Very often a minus is lost here, also note that the minus applies to all bracket , and these brackets need to be opened correctly.

(2) Expand the brackets. We simplify the last integral.

(3) We take the last integral.

(4) “Combing” the answer.

The need to apply the rule of integration by parts twice (or even thrice) is not uncommon.

And now a couple of examples for an independent solution:

Example 3

Find the indefinite integral.

This example is solved by the change of variable method (or subsuming under the differential sign)! And why not - you can try to take it in parts, you get a funny thing.

Example 4

Find the indefinite integral.

But this integral is integrated by parts (the promised fraction).

These are examples for self-solving, solutions and answers at the end of the lesson.

It seems that in examples 3,4 the integrands are similar, but the solution methods are different! This is precisely the main difficulty in mastering integrals - if you choose the wrong method for solving the integral, then you can fiddle with it for hours, like with a real puzzle. Therefore, the more you solve various integrals, the better, the easier the test and the exam will be. In addition, in the second year there will be differential equations, and without experience in solving integrals and derivatives there is nothing to do there.

By logarithms, perhaps more than enough. For a snack, I can also remember that tech students call female breasts logarithms =). By the way, it is useful to know by heart the graphs of the main elementary functions: sine, cosine, arc tangent, exponent, polynomials of the third, fourth degree, etc. No, of course, a condom on a globe
I will not pull, but now you will remember a lot from the section Graphs and functions =).

Integrals of the exponent multiplied by the polynomial

General rule:

Example 5

Find the indefinite integral.

Using a familiar algorithm, we integrate by parts:

If you have any difficulties with the integral, then you should return to the article Variable change method in indefinite integral.

The only other thing to do is to "comb" the answer:

But if your calculation technique is not very good, then leave the most profitable option as an answer. or even

That is, the example is considered solved when the last integral is taken. It won’t be a mistake, it’s another matter that the teacher may ask to simplify the answer.

Example 6

Find the indefinite integral.

This is a do-it-yourself example. This integral is integrated twice by parts. Particular attention should be paid to the signs - it is easy to get confused in them, we also remember that - a complex function.

There is not much more to say about the exhibitor. I can only add that the exponential and the natural logarithm are mutually inverse functions, this is me on the topic of entertaining graphs of higher mathematics =) Stop-stop, don't worry, the lecturer is sober.

Integrals of trigonometric functions multiplied by a polynomial

General rule: always stands for the polynomial

Example 7

Find the indefinite integral.

Integrating by parts:

Hmmm... and nothing to comment on.

Example 8

Find the indefinite integral

This is an example for a do-it-yourself solution

Example 9

Find the indefinite integral

Another example with a fraction. As in the two previous examples, a polynomial is denoted by.

Integrating by parts:

If you have any difficulties or misunderstanding with finding the integral, then I recommend attending the lesson Integrals of trigonometric functions.

Example 10

Find the indefinite integral

This is a do-it-yourself example.

Hint: before using the integration by parts method, you should apply some trigonometric formula that turns the product of two trigonometric functions into one function. The formula can also be used in the course of applying the method of integration by parts, to whom it is more convenient.

That, perhaps, is all in this paragraph. For some reason, I recalled a line from the anthem of the Physics and Mathematics Department “And the sine graph wave after wave runs along the abscissa axis” ....

Integrals of inverse trigonometric functions.
Integrals of inverse trigonometric functions multiplied by a polynomial

General rule: always stands for the inverse trigonometric function.

I remind you that the inverse trigonometric functions include arcsine, arccosine, arctangent and arccotangent. For the sake of brevity, I will refer to them as "arches"