TOPIC: Integration of rational fractions.
Attention! When studying one of the basic methods of integration: the integration of rational fractions, it is required to consider polynomials in the complex domain to carry out rigorous proofs. Therefore it is necessary study in advance some properties of complex numbers and operations on them.
Integration of simple rational fractions.
If P(z) And Q(z) are polynomials in the complex domain, then they are rational fractions. It is called correct, if degree P(z) less degree Q(z) , And wrong, if degree R no less than a degree Q.
Any improper fraction can be represented as: 
,
P(z) = Q(z) S(z) + R(z),
a R(z) – polynomial whose degree is less than the degree Q(z).
Thus, the integration of rational fractions comes down to the integration of polynomials, that is, power functions, and proper fractions, since it is a proper fraction.
Definition 5. The simplest (or elementary) fractions are the following types of fractions:
1) , 2) , 3) , 4) 
.
Let's find out how they integrate.
3) 
(studied earlier).
Theorem 5. Every proper fraction can be represented as a sum of simple fractions (without proof).
Corollary 1. If is a proper rational fraction, and if among the roots of the polynomial there are only simple real roots, then in the decomposition of the fraction into the sum of simple fractions there will be only simple fractions of the 1st type:
Example 1.
Corollary 2. If is a proper rational fraction, and if among the roots of the polynomial there are only multiple real roots, then in the decomposition of the fraction into the sum of simple fractions there will be only simple fractions of the 1st and 2nd types:
Example 2.
Corollary 3. If is a proper rational fraction, and if among the roots of the polynomial there are only simple complex conjugate roots, then in the decomposition of the fraction into the sum of simple fractions there will be only simple fractions of the 3rd type:
Example 3.
Corollary 4. If is a proper rational fraction, and if among the roots of the polynomial there are only multiple complex conjugate roots, then in the decomposition of the fraction into the sum of simple fractions there will be only simple fractions of the 3rd and 4th types:
To determine the unknown coefficients in the given expansions proceed as follows. The left and right sides of the expansion containing unknown coefficients are multiplied by The equality of two polynomials is obtained. From it, equations for the required coefficients are obtained using:
1. equality is true for any values of X (partial value method). In this case, any number of equations are obtained, any m of which allow one to find the unknown coefficients.
2. the coefficients coincide for the same degrees of X (method of indefinite coefficients). In this case, a system of m - equations with m - unknowns is obtained, from which the unknown coefficients are found.
3. combined method.
Example 5. Expand a fraction 
to the simplest.
Solution:
Let's find the coefficients A and B.
Method 1 - private value method:
Method 2 – method of undetermined coefficients:
Answer: 
Integrating rational fractions.
Theorem 6. The indefinite integral of any rational fraction on any interval on which its denominator is not equal to zero exists and is expressed through elementary functions, namely rational fractions, logarithms and arctangents.
Proof.
Let's imagine a rational fraction in the form: 
. In this case, the last term is a proper fraction, and according to Theorem 5 it can be represented as a linear combination of simple fractions. Thus, the integration of a rational fraction is reduced to the integration of a polynomial S(x)
and simple fractions, the antiderivatives of which, as has been shown, have the form indicated in the theorem.
Comment. The main difficulty in this case is the decomposition of the denominator into factors, that is, the search for all its roots.
Example 1. Find the integral
2.,
5.
,
3.
,
6.
.
In integrals 1-3 as u
accept 
. Then, after n-multiple application of formula (19) we arrive at one of the table integrals
,
,
.
In integrals 4-6, when differentiating, the transcendental factor will be simplified 
,
or 
, which should be taken as u.
Calculate the following integrals.
Example 7.
Example 8. 
Reducing integrals to themselves
If the integrand 
has the form:
,
,
and so on,
then after integrating twice by parts we obtain an expression containing the original integral 
:
,
Where 
- some constant.
Resolving the resulting equation for 
, we obtain a formula for calculating the original integral:
.
This case of applying the method of integration by parts is called " bringing the integral to itself».
Example 9. Calculate integral
.
On the right side is the original integral 
. Moving it to the left side, we get:
.
Example 10. Calculate integral
.
4.5. Integrating the simplest proper rational fractions
Definition.The simplest proper fractions I , II And III types The following fractions are called:
I.
;
II.
;
(
- positive integer);
III.
; 
.
(the roots of the denominator are complex, that is:
I.
;
(20)
II. ; (21)
III.
;
Let's consider integrals of simple fractions. 
We transform the numerator of the fraction in such a way as to isolate the term in the numerator
, equal to the derivative of the denominator.
Let us consider the first of the two integrals obtained and make a change in it:
In the second integral we add the denominator to a perfect square:
=
+
.
(22)
Finally, the integral of a fraction of the third type is equal to:
Thus, the integral of the simplest fractions of type I is expressed through logarithms, type II - through rational functions, type III - through logarithms and arctangents.
4.6.Integration of fractional-rational functions
One of the classes of functions that have an integral expressed in terms of elementary functions is the class of algebraic rational functions, that is, functions resulting from a finite number of algebraic operations on an argument. 
Every rational function 
can be represented as a ratio of two polynomials 
:
.
(23)
And
We will assume that the polynomials do not have common roots. correct A fraction of the form (23) is called , if the degree of the numerator is less than the degree of the denominator, that is,< n m wrong.
. Otherwise -
,
(24)
Where 
If the fraction is improper, then by dividing the numerator by the denominator (according to the rule for dividing polynomials), we present the fraction as the sum of a polynomial and a proper fraction: 
- polynomial, 
- a proper fraction, and the degree of the polynomial n-1).
- not higher than degree ( 
Example.
Since the integration of a polynomial is reduced to the sum of tabulated integrals of a power function, the main difficulty in integrating rational fractions lies in the integration of proper rational fractions. 
It has been proven in algebra that every proper fraction decomposes into the sum of the above protozoa 
.
fractions, the form of which is determined by the roots of the denominator Let's consider three special cases. 
Here and further we will assume that the coefficient 
at the highest degree of the denominator 
equal to one=1, that is
.
reduced polynomial Case 1. 
The roots of the denominator, that is, the roots 
equations
=0, are valid and distinct. Then we represent the denominator as a product of linear factors:
,
(26)
Where 
and the proper fraction is decomposed into the simplest fractions of the I-gotype:
– some constant numbers that are found by the method of indefinite coefficients.
To do this you need:
2. Equate the coefficients for the same degrees of identical polynomials in the numerator of the left and right sides. We obtain a system of linear equations to determine 
.
3. Solve the resulting system and find the undetermined coefficients 
.
Then the integral of the fractional-rational function (26) will be equal to the sum of the integrals of the simplest fractions of the I-type, calculated by formula (20).
- not higher than degree ( Calculate integral 
.
Solution. Let's factorize the denominator using Vieta's theorem:
Then, the integrand function is decomposed into a sum of simple fractions:
.
X:
Let us write a system of three equations to find 
X on the left and right sides:
.
Let us indicate a simpler way of finding uncertain coefficients, called partial value method.
Assuming in equality (27) 
we get 
, where 
. Believing 
we get 
. Finally, believing 
we get 
.
.
Case 2. Root of the denominator 
are valid, but among them there are multiple (equal) roots. Then we represent the denominator as a product of linear factors included in the product to the extent that the multiplicity of the corresponding root is:
Where 
.
Proper fraction 
the sum of fractions of types I and II will be decomposed. Let, for example, 
- root of the denominator of the multiplicity k, and everyone else ( n-
k) roots are different.
Then the expansion will look like:
Likewise, if there are other multiple roots. For non-multiple roots, expansion (28) includes the simplest fractions of the first type.
Example. Calculate integral 
.
Solution. Let's imagine the fraction as a sum of the simplest fractions of the first and second kind with undetermined coefficients:
.
Let's bring the right side to a common denominator and equate the polynomials in the numerators of the left and right sides:
On the right side we present similar ones with the same degrees X:
Let us write a system of four equations to find 
can be represented as a ratio of two polynomials 
. To do this, we equate the coefficients at the same powers X on the left and right side
.
Case 3. Among the roots of the denominator 
there are complex single roots. That is, the expansion of the denominator includes factors of the second degree 
, not decomposable into real linear factors, and they are not repeated.
Then, in the decomposition of a fraction, each such factor will correspond to a simplest fraction of type III. Linear factors correspond to the simplest fractions of types I and II.
Example. Calculate integral 
.
Solution.
.
.
.
Integration of rational functions Fractional - rational function The simplest rational fractions Decomposition of a rational fraction into simple fractions Integration of simple fractions General rule for the integration of rational fractions
polynomial of degree n. Fractional - rational function A fractional - rational function is a function equal to the ratio of two polynomials: A rational fraction is called proper if the degree of the numerator is less than the degree of the denominator, that is, m< n , в противном случае дробь называется неправильной. многочлен степени m Всякую неправильную рациональную дробь можно, путем деления числителя на знаменатель, представить в виде суммы многочлена L(x) и правильной рациональной дроби:)()()(x. Q x. P xf n m)()()(x. Q x. R x. L x. Q x. P
Fractional - rational function Reduce an improper fraction to the correct form: 2 95 4 x xx 95 4 xx 2 x 3 x 34 2 xx 952 3 xx 2 2 x 23 42 xx 954 2 xx x 4 xx 84 2 93 x 3 63 x 15 2 95 4 x xx 342 23 xxx 2 15 x
Simplest rational fractions Proper rational fractions of the form: They are called simplest rational fractions of types. ax A); 2(Nkk ax A k)04(2 2 qp qpxx NMx); 2; 04(2 2 Nkkqp qpxx NMx k V V,
Decomposition of a rational fraction into simple fractions Theorem: Any proper rational fraction, the denominator of which is factorized: can be represented, moreover, in a unique way in the form of a sum of simple fractions: s k qxpxxxxxx. Q)()()(22 2 11 2 21)()(x. Q x. P 1 xx A k k xx B)()(2 2 2 1 11 2 qxpx DCx 2 22 22 2 11)(qxpx Nx. M s ss qxpx Nx)
Decomposition of a rational fraction into simple fractions Let us explain the formulation of the theorem using the following examples: To find the uncertain coefficients A, B, C, D..., two methods are used: the method of comparing coefficients and the method of partial values of a variable. Let's look at the first method using an example. 3 2)3)(2(4 xx x 2 x A 3 3 2 21)3()3(3 x B x B 1 2 x DCx 22 22 2 11)1(1 xx Nx. M)1(3 22 3 xx x 2 21 x A 22 2)1)(4(987 xxx xx 4 x
Decomposition of a rational fraction into simple fractions Present the fraction as a sum of simple fractions: Let's bring the simplest fractions to a common denominator Equate the numerators of the resulting and original fractions Equate the coefficients at the same powers x)52)(1(332 2 2 xxx xx 1 x A 52 2 xx CBx )52)(1()1)(()52(2 2 xxx x. CBxxx. A 33252 222 xx. CBx. Cx. Bx. AAx. Ax 35 32 2 0 1 2 CAx BAx 2 3 1 C B A 52 23 1 1 2 xx x x
Integration of the simplest fractions Let's find the integrals of the simplest rational fractions: Let's look at the integration of type 3 fractions using an example. dx ax A k dx qpxx NMx 2 ax axd A)(Cax. Aln)(axdax. A k C k ax. A k
Integration of simple fractionsdx xx x 102 13 2 dx xx x 9)12(13 2 dx x x 9)1(13 2 dtdx tx tx 1 1 dt t t 9 1)1(3 2 dt t t 9 23 2 9 322 t dtt 9 9 2 3 2 2 t td 33 2 t arctg. C t arctgt 33 2 9 ln 2 32 C x arctgxx 3 1 3 2 102 ln
Integration of simple fractions An integral of this type using substitution: is reduced to the sum of two integrals: The first integral is calculated by introducing t under the differential sign. The second integral is calculated using the recurrence formula: dx qpxx NMx k 2 V t p x 2 kk at dt N at dtt M 22122 1221222))(1(222 321 kkkk atk t k k aat dt
Integration of simple fractions a = 1; k = 3 323)1(t dt tarctg t dt 1 21)1)(12(2222 322 1 21222 t t t dt)1(22 1 2 t t tarctg 2223)1)(13(2232 332 t t C t t tarctg 222)1 (4)1(
General rule for integrating rational fractions If the fraction is improper, then represent it as the sum of a polynomial and a proper fraction. Having factorized the denominator of a proper rational fraction, represent it as a sum of simple fractions with indefinite coefficients. Find indefinite coefficients by the method of comparing coefficients or by the method of partial values of a variable. Integrate the polynomial and the resulting sum of simple fractions.
Example Let's put the fraction in the correct form. dx xxx 23 35 2 442 35 xxxxxx 23 2 2 x 345 2 xxx 442 34 xxx x 2 234 242 xxx 4425 23 xxx xxx 23 35 2 442 xxx xx xx 23 2 2 2 48 52 5 xxx 5105 23 48 2 x x
Example Let's factorize the denominator of a proper fraction Let's represent the fraction as a sum of simple fractions Let's find the undetermined coefficients using the method of partial values of the variable xxx xx 23 2 2 48 2 2)1(48 xx xx 2)1(1 x C x B x A 2 2)1 ()1(xx Cxx. Bxx. A 48)1()1(22 xx. Cxx. Bxx. A 5241 31 40 CBAx Cx Ax 3 12 4 C B A xxx xx 23 2 2 48 2)1(3 1 124 xxx
Example dx xx 2 2)1(3 1 124 52 2 2)1(3 1 12452 x dx dxxdxdxx C x xxxx x 1 3 1 ln 12 ln
A test on the integration of functions, including rational fractions, is given to 1st and 2nd year students. Examples of integrals will mainly be of interest to mathematicians, economists, and statisticians. These examples were asked during the test at LNU. I. Frank. The conditions of the following examples are “Find the integral” or “Calculate the integral”, so to save space and your time they were not written out.
Example 15. We came to the integration of fractional-rational functions. They occupy a special place among integrals because they require a lot of time to calculate and help teachers test your knowledge not only of integration. To simplify the function under the integral, we add and subtract an expression in the numerator that will allow us to split the function under the integral into two simple ones
As a result, we find one integral quite quickly, in the second we need to expand the fraction into a sum of elementary fractions 
When reduced to a common denominator, we obtain the following numerals
Next, open the brackets and group
We equate the value for the same powers of “x” on the right and left. As a result, we arrive at a system of three linear equations (SLAE) with three unknowns. 
How to solve systems of equations is described in other articles on the site. In the final version you will receive the following SLAE solution
A=4; B=-9/2; C=-7/2.
We substitute constants into the expansion of fractions into simplest ones and perform integration
This concludes the example.
Example 16. Again we need to find the integral of a fractional rational function. To begin with, we will decompose the cubic equation contained in the denominator of the fraction into simple factors 
Next, we decompose the fraction into its simplest forms 
We reduce the right side to a common denominator and open the brackets in the numerator. 
We equate the coefficients for the same degrees of the variable. Let's come to the SLAE again with three unknowns 
We substitute the values of A, B, C into the expansion and calculate the integral 
The first two terms give the logarithm, the last one is also easy to find.
Example 17. In the denominator of the fractional rational function we have the difference of cubes. Using abbreviated multiplication formulas, we decompose it into two simple factors 
Next, we write the resulting fractional function into the sum of simple fractions and reduce them to a common denominator 
In the numerator we get the following expression.
From it we form a system of linear equations to calculate 3 unknowns 
A=1/3; B=-1/3; C=1/3.
We substitute A, B, C into the formula and perform integration. As a result, we arrive at the following answer: 
Here the numerator of the second integral was converted into a logarithm, and the remainder under the integral gives the arctangent.
There are a lot of similar examples on the integration of rational fractions on the Internet. You can find similar examples from the materials below.
