Calculate by integration by parts. Complex integrals
The method of integration by parts is mainly used 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.
b) 
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 
– whole part from 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 of the simplest fractions of four types are calculated as follows.
I) 
.
II) , 
.
III) For integration the simplest fraction Type III in the denominator allocate a full square, make a replacement 
. The integral after substitution is divided into two integrals. The first integral is calculated by separating 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) 
; in) 
.
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 right part 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 the system 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) 
; in) 
.
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: BUT = 1; AT = 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:
.
in) 
.
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:
Examples of integration by parts of a similar composition are given to students of the 1st and 2nd courses. These tasks were assigned to control work at LNU I. Frank. So that the formulas in the tasks and answers do not repeat, we will not describe the tasks. According to the condition of the tasks, you need to either "Find the integral" or "Calculate the integral".
Example 8. We find the integral according to the rule of integration by parts int(u*dv)=u*v-int(v*du). The main thing here is to choose the right functions for the rule. (For yourself, remember that for dv, if possible, choose periodic functions or those that, when differentiated up to a factor, give themselves - the exponent). In this integral, you need to bring the sine under the differential
Further integration is quite simple and we will not dwell on the details. 
Example 9. Again, we need to apply the rule of integration by parts u*dv . Here we have the product of a periodic function by an exponent, so it is up to you to choose what is better to bring under the differential. You can use both the exponent and the cosine (in each case we get a recursive formula). 
Applying integration by parts again 
We arrived at a recursive formula. If we write the integral that we were looking for and the result of the calculations, then we get two similar terms
We group them and find the required integral
Example 10. We have a ready-made record of the integral under the rule u * dv. Find du and perform integration 
We reduce the second integral under the tabular formula and calculate it 
Example 11. Denote by the new variable cos(ln(x))=u і find du , then inserting under the differential 
To the integral, we again apply the rule of integration by parts 
Came to the recursive formula
with which we calculate the unknown integral
Example 12. To find the integral, we select a full square in the denominator. Further reducing the denominator to well-known formula integration we get the arc tangent 
Remember the order of alternation of multipliers. The unit divided by the root of the free term appears before the arc tangent, and this factor is also present in the arc tangent before the variable.
Example 13. We are dealing with a similar integral, only in the denominator the quadratic dependence is under the root. We select the full square and reduce it to the integration formula, which gives the logarithm 
These are examples on the control or tests. Remember the basic integration schemes well.
If you can not solve the integral yourself, then ask for help.
Solution of integrals - the task is easy but only for a select few. This article is for those who want to learn to understand integrals, but know little or nothing about them. Integral... Why is it needed? How to calculate it? What are definite and indefinite integrals? If the only use of the integral you know is to hook something useful out of hard-to-reach places then welcome! Learn how to solve integrals and why you can't do without it.
We study the concept of "integral"
Integration was already known in Ancient Egypt. Of course not in modern form, but still. Since then, mathematicians have written a great many books on the subject. Particularly distinguished Newton and Leibniz but the essence of things has not changed. How to understand integrals from scratch? No way! To understand this topic, you will still need a basic knowledge of the basics of mathematical analysis. Information about , which is also necessary for understanding integrals, is already in our blog.
Indefinite integral
Let's have some function f(x) .
The indefinite integral of the function f(x) such a function is called F(x) , whose derivative is equal to the function f(x) .
In other words, an integral is a reverse derivative or antiderivative. By the way, about how to read in our article.
An antiderivative exists for all continuous functions. Also, a constant sign is often added to the antiderivative, since the derivatives of functions that differ by a constant coincide. The process of finding an integral is called integration.
Simple example:
In order not to constantly calculate the antiderivatives of elementary functions, it is convenient to summarize them in a table and use ready-made values:
Definite integral
When dealing with the concept of an integral, we are dealing with infinitesimal quantities. The integral will help calculate the area of the figure, the mass of an inhomogeneous body, the path traveled during uneven movement, and much more. It should be remembered that the integral is the sum of an infinitely large number of infinitely small terms.
As an example, imagine a graph of some function. How to find the area of a figure bounded by a graph of a function?
With the help of an integral! Let's break the curvilinear trapezoid, bounded by the coordinate axes and the graph of the function, into infinitesimal segments. Thus, the figure will be divided into thin columns. The sum of the areas of the columns will be the area of the trapezoid. But remember that such a calculation will give an approximate result. However, the smaller and narrower the segments, the more accurate the calculation will be. If we reduce them to such an extent that the length tends to zero, then the sum of the areas of the segments will tend to the area of the figure. This is the definite integral, which is written as follows:
The points a and b are called the limits of integration.
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Rules for Calculating Integrals for Dummies
Properties of the indefinite integral
How to solve indefinite integral? Here we will consider the properties of the indefinite integral, which will be useful in solving examples.
- The derivative of the integral is equal to the integrand:
- The constant can be taken out from under the integral sign:
- The integral of the sum is equal to the sum of the integrals. Also true for the difference:
Properties of the Definite Integral
- Linearity:
- The sign of the integral changes if the limits of integration are reversed:
- At any points a, b and With:
We have already found out that the definite integral is the limit of the sum. But how to get a specific value when solving an example? For this, there is the Newton-Leibniz formula:
Examples of solving integrals
Below we consider several examples of finding indefinite integrals. We suggest that you independently understand the intricacies of the solution, and if something is not clear, ask questions in the comments.
To consolidate the material, watch a video on how integrals are solved in practice. Do not despair if the integral is not given immediately. Turn to a professional student service, and any triple or curvilinear integral over a closed surface will be within your power.
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definite integral from continuous function f(x) on the finite interval [ a, b] (where ) is the increment of some of its antiderivatives on this segment. (In general, understanding will be noticeably easier if you repeat the topic of the indefinite integral) In this case, 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 number (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.
Thus, to calculate the definite integral, it is necessary to find any antiderivative of the integrand, i.e. must first be found indefinite integral. Constant FROM 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 FROM= 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. Consequently,
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). Definite integral is equal to the product the length of the segment of integration by 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 obtain
Definite integral with variable upper limit
Let f(x) is continuous on the segment [ 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
