One of the most important classes of functions, the integrals of which are expressed through elementary functions, is a class of rational functions.

Definition 1. Function of the form where
- polynomials of degreesnAndmcalled rational. Whole rational function, i.e. polynomial, integrates directly. The integral of a fractional-rational function can be found by decomposing into terms, which are converted in a standard way to the main tabular integrals.

Definition 2. Fraction
is called correct if the degree of the numeratornless denominator powers m. A fraction in which the degree of the numerator is greater than or equal to the degree of the denominator is called improper.

Any improper fraction can be represented as the sum of a polynomial and proper fraction. This is done by dividing a polynomial by a polynomial, like dividing numbers.

Example.

Let's imagine a fraction
as the sum of a polynomial and a proper fraction:

x - 1

First term
in the quotient it is obtained as a result of dividing the leading term
, divided by the leading term X divider Then we multiply
per divisor x-1 and the resulting result is subtracted from the dividend; The remaining terms of the incomplete quotient are found similarly.

Having divided the polynomials, we get:

This action is called selecting a whole part.

Definition 3. The simplest fractions are proper rational fractions of the following types:

II.
(K=2, 3, …).

III.
where is the square trinomial

IV.
where K=2, 3, …; quadratic trinomial
has no real roots.

a) expand the denominator
into the simplest real factors (according to the fundamental theorem of algebra, this expansion can contain linear binomials of the form
and quadratic trinomials
, having no roots);

b) write a diagram of the decomposition of a given fraction into a sum of simple fractions. Moreover, each factor of the form
corresponds k components of types I and II:

to each factor of the form
corresponds to e terms of types III and IV:

Example.

Write down the fraction expansion scheme
to the sum of the simplest.

c) perform the addition of the simplest fractions obtained. Write down the equality of the numerators of the resulting and original fractions;

d) find the coefficients of the corresponding expansion:
(solution methods will be discussed below);

e) substitute the found values of the coefficients into the decomposition scheme.

Integrating any proper rational fraction after decomposition into its simplest terms reduces to finding integrals of one of the following types:

(k And e =2, 3, …).

Calculation of the integral reduces to formula III:

integral - to formula II:

integral can be found by the rule specified in the theory of integration of functions containing a quadratic trinomial; - through the transformations shown below in example 4.

Example 1.

a) factor the denominator:

b) write a diagram for decomposing the integrand into terms:

c) perform the addition of simple fractions:

Let us write down the equality of the numerators of the fractions:

d) there are two methods for finding unknown coefficients A, B, C.

Two polynomials are equal if and only if their coefficients are equal equal degrees X, so you can create the corresponding system of equations. This is one of the solution methods.

Coefficients at

free members (coefficient at ):4A=8.

Having solved the system, we get A=2, B=1, C= - 10.

Another method - private values - will be discussed in the following example;

e) substitute the found values into the decomposition scheme:

Substituting the resulting sum under the integral sign and integrating each term separately, we find:

Example 2.

Identity is an equality that is valid for any values of the unknowns included in it. Based on this private value method. Can be given X any values. It is more convenient for calculations to take those values that make any terms on the right side of the equality vanish.

Let x = 0. Then 1 = A 0(0+2)+V 0 (0-1)+С (0-1)(0+2).

Similarly for x = - 2 we have 1= - 2V*(-3), at x = 1 we have 1 = 3A.

Hence,

Example 3.

d) first we use the partial value method.

Let x = 0, Then 1 = A 1, A = 1.

At x = - 1 we have - 1+4+2+1 = - B(1+1+1) or 6 = - 3V, B = - 2.

To find the coefficients C and D, you need to create two more equations. To do this, you can take any other values X, For example x = 1 And x = 2. You can use the first method, i.e. equate coefficients at any identical powers X, for example when And . We get

1 = A+B+C and 4 = C +D- IN.

Knowing A = 1, B = -2, we'll find C = 2, D = 0 .

Thus, both methods can be combined when calculating coefficients.

Last integral we find separately according to the rule specified in the method of specifying a new variable. Let's highlight perfect square in the denominator:

let's say
Then
We get:

Substituting into the previous equality, we find

Example 4.

Find

Integrating, we have:

Let us transform the first integral to formula III:

Let us transform the second integral to formula II:

In the third integral we replace the variable:

(When performing the transformations, we used the trigonometry formula

Find the integrals:

51.

52.

53.

54.

55.

56.

57.

58.

Self-test questions.

Which of these rational fractions are correct:

2. Is the diagram for decomposing a fraction into the sum of simple fractions written correctly?

Here we present detailed solutions three examples of integrating the following rational fractions:
, , .

Example 1

Calculate the integral:
.

Solution

Here, under the integral sign there is a rational function, since integrand is a fraction of polynomials. Denominator polynomial degree ( 3 ) is less than the degree of the numerator polynomial ( 4 ). Therefore, first you need to select the whole part of the fraction.

1. Let's select the whole part of the fraction. Divide x 4 by x 3 - 6 x 2 + 11 x - 6:

From here
.

2. Let's factorize the denominator of the fraction. To do this, you need to solve the cubic equation:
.
6
1, 2, 3, 6, -1, -2, -3, -6 .
Let's substitute x = 1 :
.

1 . Divide by x - 1 :

From here
.
Let's decide quadratic equation.
.
The roots of the equation are: , .
Then
.

3. Let's break down the fraction into its simplest form.

.

So we found:
.
Let's integrate.

Answer

Example 2

Calculate the integral:
.

Solution

Here the numerator of the fraction is a polynomial of degree zero ( 1 = x 0). The denominator is a polynomial of the third degree. Since 0 < 3 , then the fraction is correct. Let's break it down into simple fractions.

1. Let's factorize the denominator of the fraction. To do this, you need to solve the third degree equation:
.
Let's assume that it has at least one whole root. Then it is a divisor of the number 3 (member without x). That is, the whole root can be one of the numbers:
1, 3, -1, -3 .
Let's substitute x = 1 :
.

So, we have found one root x = 1 . Divide x 3 + 2 x - 3 on x - 1 :

So,
.

Solving the quadratic equation:
x 2 + x + 3 = 0.
Find the discriminant: D = 1 2 - 4 3 = -11. Since D< 0 , then the equation has no real roots. Thus, we obtained the factorization of the denominator:
.

2.
.
(x - 1)(x 2 + x + 3):
(2.1) .
Let's substitute x = 1 . Then x - 1 = 0 ,
.

Let's substitute in (2.1) x = 0 :
1 = 3 A - C;
.

Let's equate to (2.1) coefficients for x 2 :
;
0 = A + B;
.

3. Let's integrate.
(2.2) .
To calculate the second integral, we select the derivative of the denominator in the numerator and reduce the denominator to the sum of squares.

;
;
.

Calculate I 2 .

.
Since the equation x 2 + x + 3 = 0 has no real roots, then x 2 + x + 3 > 0. Therefore, the modulus sign can be omitted.

We deliver to (2.2) :
.

Answer

Example 3

Calculate the integral:
.

Solution

Here under the integral sign there is a fraction of polynomials. Therefore, the integrand is a rational function. The degree of the polynomial in the numerator is equal to 3 . The degree of the polynomial of the denominator of the fraction is equal to 4 . Since 3 < 4 , then the fraction is correct. Therefore, it can be decomposed into simple fractions. But to do this you need to factorize the denominator.

1. Let's factorize the denominator of the fraction. To do this, you need to solve the fourth degree equation:
.
Let's assume that it has at least one whole root. Then it is a divisor of the number 2 (member without x). That is, the whole root can be one of the numbers:
1, 2, -1, -2 .
Let's substitute x = -1 :
.

So, we have found one root x = -1 . Divide by x - (-1) = x + 1:

So,
.

Now we need to solve the third degree equation:
.
If we assume that this equation has an integer root, then it is a divisor of the number 2 (member without x). That is, the whole root can be one of the numbers:
1, 2, -1, -2 .
Let's substitute x = -1 :
.

So, we found another root x = -1 . It would be possible, as in the previous case, to divide the polynomial by , but we will group the terms:
.

Since the equation x 2 + 2 = 0 has no real roots, then we get the factorization of the denominator:
.

2. Let's break down the fraction into its simplest form. We are looking for an expansion in the form:
.
We get rid of the denominator of the fraction, multiply by (x + 1) 2 (x 2 + 2):
(3.1) .
Let's substitute x = -1 . Then x + 1 = 0 ,
.

Let's differentiate (3.1) :

;

.
Let's substitute x = -1 and take into account that x + 1 = 0 :
;
; .

Let's substitute in (3.1) x = 0 :
0 = 2 A + 2 B + D;
.

Let's equate to (3.1) coefficients for x 3 :
;
1 = B + C;
.

So, we have found the decomposition into simple fractions:
.

3. Let's integrate.

.

Integration of a fractional-rational function.
Method uncertain coefficients

We continue to work on integrating fractions. We have already looked at integrals of some types of fractions in the lesson, and this lesson in some sense can be considered a continuation. To successfully understand the material, basic integration skills are required, so if you have just started studying integrals, that is, you are a beginner, then you need to start with the article Indefinite integral. Examples of solutions.

Oddly enough, now we will be engaged not so much in finding integrals, but... in solving systems linear equations. In this regard urgently I recommend attending the lesson. Namely, you need to be well versed in substitution methods (“the school” method and the method of term-by-term addition (subtraction) of system equations).

What is a fractional rational function? In simple words, a fractional-rational function is a fraction whose numerator and denominator contain polynomials or products of polynomials. Moreover, the fractions are more sophisticated than those discussed in the article Integrating Some Fractions.

Integrating a Proper Fractional-Rational Function

Immediately an example and standard algorithm solutions to the integral of a fractional rational function.

Example 1

Step 1. The first thing we ALWAYS do when solving the integral of a fractional rational function is to find out next question: is the fraction proper? This step is done orally, and now I will explain how:

First we look at the numerator and find out senior degree polynomial:

The leading power of the numerator is two.

Now we look at the denominator and find out senior degree denominator. The obvious way is to open the brackets and bring similar terms, but you can do it easier, in each find the highest degree in brackets

and mentally multiply: - thus, the highest degree of the denominator is equal to three. It is quite obvious that if we actually open the brackets, we will not get a degree greater than three.

Conclusion: Major degree of numerator STRICTLY is less than the highest power of the denominator, which means the fraction is proper.

If in in this example the numerator contained the polynomial 3, 4, 5, etc. degrees, then the fraction would be wrong.

Now we will consider only the correct fractional rational functions. The case when the degree of the numerator is greater than or equal to the degree of the denominator will be discussed at the end of the lesson.

Step 2. Let's factorize the denominator. Let's look at our denominator:

Generally speaking, this is already a product of factors, but, nevertheless, we ask ourselves: is it possible to expand something else? The object of torture will undoubtedly be the square trinomial. Solving the quadratic equation:

Discriminant greater than zero, which means that the trinomial really can be factorized:

General rule: EVERYTHING that CAN be factored in the denominator - we factor it

Let's begin to formulate a solution:

Step 3. Using the method of indefinite coefficients, we expand the integrand into a sum of simple (elementary) fractions. Now it will be clearer.

Let's look at our integrand function:

And, you know, somehow an intuitive thought pops up that it would be nice to have our large fraction turn into several small ones. For example, like this:

The question arises, is it even possible to do this? Let's breathe a sigh of relief, the corresponding theorem mathematical analysis asserts - IT IS POSSIBLE. Such a decomposition exists and is unique.

There's just one catch, the odds are Bye We don’t know, hence the name – the method of indefinite coefficients.

As you guessed, subsequent body movements are like that, don’t cackle! will be aimed at just RECOGNIZING them - to find out what they are equal to.

Be careful, I will explain in detail only once!

So, let's start dancing from:

On the left side we reduce the expression to a common denominator:

Now we can safely get rid of the denominators (since they are the same):

On the left side we open the brackets, but do not touch the unknown coefficients for now:

At the same time, we repeat the school rule of multiplying polynomials. When I was a teacher, I learned to pronounce this rule with a straight face: In order to multiply a polynomial by a polynomial, you need to multiply each term of one polynomial by each term of the other polynomial.

From the point of view clear explanation It’s better to put the coefficients in brackets (although I personally never do this to save time):

We compose a system of linear equations.
First we look for senior degrees:

And we write the corresponding coefficients into the first equation of the system:

Remember the following point well. What would happen if there were no s on the right side at all? Let's say, would it just show off without any square? In this case, in the equation of the system it would be necessary to put a zero on the right: . Why zero? But because on the right side you can always assign this same square with zero: If there are no variables on the right side and/or free member, then we put zeros on the right-hand sides of the corresponding equations of the system.

We write the corresponding coefficients into the second equation of the system:

And finally, mineral water, we select free members.

Eh...I was kind of joking. Jokes aside - mathematics is a serious science. In our institute group, no one laughed when the assistant professor said that she would scatter the terms along the number line and choose the largest ones. Let's get serious. Although... whoever lives to see the end of this lesson will still smile quietly.

The system is ready:

We solve the system:

(1) From the first equation we express and substitute it into the 2nd and 3rd equations of the system. In fact, it was possible to express (or another letter) from another equation, but in in this case it is advantageous to express precisely from the 1st equation, since there the smallest odds.

(2) We present similar terms in the 2nd and 3rd equations.

(3) We add the 2nd and 3rd equations term by term, obtaining the equality , from which it follows that

(4) We substitute into the second (or third) equation, from where we find that

(5) Substitute and into the first equation, obtaining .

If you have any difficulties with the methods of solving the system, practice them in class How to solve a system of linear equations?

After solving the system, it is always useful to check - substitute the found values every equation of the system, as a result everything should “converge”.

Almost there. The coefficients were found, and:

The finished job should look something like this:

As you can see, the main difficulty of the task was to compose (correctly!) and solve (correctly!) a system of linear equations. And at the final stage everything is not so difficult: we use the properties of linearity indefinite integral and integrate. Please note that under each of the three integrals we have “free” complex function, I talked about the features of its integration in class Variable change method in indefinite integral.

Check: Differentiate the answer:

The original integrand function has been obtained, which means that the integral has been found correctly.
During the verification, we had to reduce the expression to a common denominator, and this is not accidental. The method of indefinite coefficients and reducing an expression to a common denominator are mutually inverse actions.

Example 2

Find the indefinite integral.

Let's return to the fraction from the first example: . It is easy to notice that in the denominator all the factors are DIFFERENT. The question arises, what to do if, for example, the following fraction is given: ? Here we have degrees in the denominator, or, mathematically, multiples. In addition, there is a quadratic trinomial that cannot be factorized (it is easy to verify that the discriminant of the equation is negative, so the trinomial cannot be factorized). What to do? The expansion into a sum of elementary fractions will look something like with unknown coefficients at the top or something else?

Example 3

Introduce a function

Step 1. Checking if we have a proper fraction
Major numerator: 2
Highest degree of denominator: 8
, which means the fraction is correct.

Step 2. Is it possible to factor something in the denominator? Obviously not, everything is already laid out. Square trinomial does not decompose into a work for the reasons stated above. Hood. Less work.

Step 3. Let's imagine a fractional-rational function as a sum of elementary fractions.
In this case, the expansion has the following form:

Let's look at our denominator:
When decomposing a fractional-rational function into a sum of elementary fractions, three fundamental points can be distinguished:

1) If the denominator contains a “lonely” factor to the first power (in our case), then we put an indefinite coefficient at the top (in our case). Examples No. 1, 2 consisted only of such “lonely” factors.

2) If the denominator has multiple multiplier, then you need to decompose it like this:
- that is, sequentially go through all the degrees of “X” from the first to the nth degree. In our example there are two multiple factors: and , take another look at the expansion I gave and make sure that they are expanded exactly according to this rule.

3) If the denominator contains an indecomposable polynomial of the second degree (in our case), then when decomposing in the numerator you need to write linear function with uncertain coefficients (in our case with uncertain coefficients and ).

In fact, there is another 4th case, but I will keep silent about it, since in practice it is extremely rare.

Example 4

Introduce a function as a sum of elementary fractions with unknown coefficients.

This is an example for independent decision. Complete solution and the answer at the end of the lesson.
Follow the algorithm strictly!

If you understand the principles by which you need to expand a fractional-rational function into a sum, you can chew through almost any integral of the type under consideration.

Example 5

Find the indefinite integral.

Step 1. Obviously the fraction is correct:

Step 2. Is it possible to factor something in the denominator? Can. Here is the sum of cubes . Factor the denominator using the abbreviated multiplication formula

Step 3. Using the method of indefinite coefficients, we expand the integrand into a sum of elementary fractions:

Please note that the polynomial cannot be factorized (check that the discriminant is negative), so at the top we put a linear function with unknown coefficients, and not just one letter.

We bring the fraction to a common denominator:

Let's compose and solve the system:

(1) We express from the first equation and substitute it into the second equation of the system (this is the most rational way).

(2) We present similar terms in the second equation.

(3) We add the second and third equations of the system term by term.

All further calculations are, in principle, oral, since the system is simple.

(1) We write down the sum of fractions in accordance with the found coefficients.

(2) We use the linearity properties of the indefinite integral. What happened in the second integral? You can familiarize yourself with this method in the last paragraph of the lesson. Integrating Some Fractions.

(3) Once again we use the properties of linearity. In the third integral we begin to isolate the complete square (penultimate paragraph of the lesson Integrating Some Fractions).

(4) We take the second integral, in the third we select the complete square.

(5) Take the third integral. Ready.

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 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's called correct, if degree P(z) less degree Q(z) , And wrong, if degree R no less than a degree Q.

I love it 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 factorization of the denominator, that is, the search for all its roots.

Example 1. Find the integral

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 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 right kind: 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 in following examples: To find the uncertain coefficients A, B, C, D..., two methods are used: the coefficient comparison method and the method of partial variable values. 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 Integral of this type using substitution: 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