Here we provide detailed solutions to 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 the 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 . 1 :

From here
.
Divide by x -
.
Solving a quadratic equation.
The roots of the equation are: , .
.

3. Then

.

Let's break down the fraction into its simplest form.
.
So we found:

Let's integrate.

Answer

Calculate the integral:
.

Solution

Example 2 Here the numerator of the fraction is a polynomial of degree zero ( 1 = x 0 0 < 3 ). The denominator is a polynomial of the third degree. Because the

1. , then the fraction is correct. Let's break it down into simple fractions.
.
Let's factorize the denominator of the fraction. To do this, you need to solve the third degree equation: 3 Let's assume that it has at least one whole root. Then it is a divisor of the number
1, 3, -1, -3 .
Let's substitute x = 1 :
.

(member without x). That is, the whole root can be one of the numbers: 1 So, we have found one root x = . Divide x 1 :

3 + 2 x - 3
.

on x -
So, Solving the quadratic equation:.
x 2 + x + 3 = 0 Find the discriminant: D =< 0 1 2 - 4 3 = -11
.

2.
.
.:
(2.1) .
Let's substitute x = 1 Since D 1 = 0 ,
.

, then the equation has no real roots. Thus, we obtained the factorization of the denominator: (2.1) (x - 1)(x 2 + x + 3) 0 :
.;
.

Then x - (2.1) Let's substitute in 2 :
;
x =;
.

.

3. So we found:
(2.2) .
1 = 3 A - C

;
;
.

Let's equate to 2 .


.
coefficients for x Solving the quadratic equation: 0 = A + B To calculate the second integral, we isolate the derivative of the denominator in the numerator and reduce the denominator to the sum of squares. Calculate I

Since the equation x (2.2) :
.

Let's integrate.

has no real roots, then x

Calculate the integral:
.

Solution

2 + x + 3 > 0 3 . 4 Therefore, the modulus sign can be omitted. 3 < 4 We deliver to

1. Let's factorize the denominator of the fraction. To do this, you need to solve the fourth degree equation:
.
Let's factorize the denominator of the fraction. To do this, you need to solve the third degree equation: 2 Let's assume that it has at least one whole root. Then it is a divisor of the number
1, 2, -1, -2 .
Let's substitute x = -1 :
.

(member without x). That is, the whole root can be one of the numbers: -1 . (-1) = x + 1:


3 + 2 x - 3
.

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 Let's assume that it has at least one whole root. Then it is a divisor of the number
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: 2 + 2 = 0 Since the equation x
.

2. has no real roots, then we get the factorization of the denominator:
.
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:
(3.1) .
Let's substitute x = -1 (x + 1) 2 (x 2 + 2) 1 = 0 ,
.

. (3.1) :

;

.
Let's substitute x = -1 Then x + 1 = 0 :
;
; .

, then the equation has no real roots. Thus, we obtained the factorization of the denominator: (3.1) (x - 1)(x 2 + x + 3) 0 :
Let's differentiate;
.

Then x - (3.1) Let's substitute in 3 :
;
and take into account that x +;
.

0 = 2 A + 2 B + D
.

3. So we found:


.

1 = B + C
So, we have found the decomposition into simple fractions:

Integration of a fractional-rational function. Uncertain coefficient method.

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 a 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 of 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 a typical algorithm for solving the integral of a fractional-rational function. Example 1 Step 1. The first thing we ALWAYS do when solving an integral of a fractional rational function is to clarify the following question:

is the fraction proper? This step is performed verbally, and now I will explain how: First we look at the numerator and find out

senior degree

polynomial: This step is performed verbally, and now I will explain how: The leading power of the numerator is two. Now we look at the denominator and find out denominator. The obvious way is to open the brackets and bring similar terms, but you can do it simpler, in

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 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:

The discriminant is greater than zero, which means that the trinomial really can be factorized:

General rule: EVERYTHING in the denominator CAN be factored - factored

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 turn our large fraction into several small ones. For example, like this:

The question arises, is it even possible to do this? Let us breathe a sigh of relief, the corresponding theorem of mathematical analysis states – 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 of a clear explanation, it is better to put the coefficients in brackets (although I personally never do this in order 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 on the right side there are no variables and/or a free term, then we put zeros on the right 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,...somehow I was 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 a 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 this case it is advantageous to express it 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 linearity properties of the indefinite integral and integrate. Please note that under each of the three integrals we have a “free” complex function; I talked about the features of its integration in the lesson 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

Immediately an example and a typical algorithm for solving the integral of a fractional-rational function. 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. The square trinomial cannot be expanded into a product 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 a linear function with undetermined coefficients (in our case with undetermined 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 you to solve on your own. Full solution and 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.

Immediately an example and a typical algorithm for solving the integral of a fractional-rational function. 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. 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)..

(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 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).).

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

(5) Take the third integral. Ready.

The material presented in this topic is based on the information presented in the topic "Rational fractions. Decomposition of rational fractions into elementary (simple) fractions". I highly recommend that you at least skim through this topic before moving on to reading this material. In addition, we will need a table of indefinite integrals.

Let me remind you of a couple of terms. They were discussed in the corresponding topic, so here I will limit myself to a brief formulation.

The ratio of two polynomials $\frac(P_n(x))(Q_m(x))$ is called a rational function or rational fraction. The rational fraction is called correct, if $n< m$, т.е. если степень многочлена, стоящего в числителе, меньше степени многочлена, стоящего в знаменателе. В противном случае (если $n ≥ m$) дробь называется wrong.

Elementary (simple) rational fractions are rational fractions of four types:

1. $\frac(A)(x-a)$;
2. $\frac(A)((x-a)^n)$ ($n=2,3,4, \ldots$);
3. $\frac(Mx+N)(x^2+px+q)$ ($p^2-4q< 0$);
4. $\frac(Mx+N)((x^2+px+q)^n)$ ($p^2-4q< 0$; $n=2,3,4,\ldots$).

Note (desirable for a more complete understanding of the text): show\hide

Why is the condition $p^2-4q needed?< 0$ в дробях третьего и четвертого типов? Рассмотрим квадратное уравнение $x^2+px+q=0$. Дискриминант этого уравнения $D=p^2-4q$. По сути, условие $p^2-4q < 0$ означает, что $D < 0$. Если $D < 0$, то уравнение $x^2+px+q=0$ не имеет действительных корней. Т.е. выражение $x^2+px+q$ неразложимо на множители. Именно эта неразложимость нас и интересует.

For example, for the expression $x^2+5x+10$ we get: $p^2-4q=5^2-4\cdot 10=-15$. Since $p^2-4q=-15< 0$, то выражение $x^2+5x+10$ нельзя разложить на множители.

By the way, for this check it is not at all necessary that the coefficient before $x^2$ be equal to 1. For example, for $5x^2+7x-3=0$ we get: $D=7^2-4\cdot 5 \cdot (-3)=$109. Since $D > 0$, the expression $5x^2+7x-3$ is factorizable.

Examples of rational fractions (proper and improper), as well as examples of decomposition of a rational fraction into elementary ones can be found. Here we will be interested only in questions of their integration. Let's start with the integration of elementary fractions. So, each of the four types of elementary fractions above is easy to integrate using the formulas below. Let me remind you that when integrating fractions of types (2) and (4), $n=2,3,4,\ldots$ are assumed. Formulas (3) and (4) require the fulfillment of the condition $p^2-4q< 0$.

\begin(equation) \int \frac(A)(x-a) dx=A\cdot \ln |x-a|+C \end(equation) \begin(equation) \int\frac(A)((x-a)^n )dx=-\frac(A)((n-1)(x-a)^(n-1))+C \end(equation) \begin(equation) \int \frac(Mx+N)(x^2 +px+q) dx= \frac(M)(2)\cdot \ln (x^2+px+q)+\frac(2N-Mp)(\sqrt(4q-p^2))\arctg\ frac(2x+p)(\sqrt(4q-p^2))+C \end(equation)

For $\int\frac(Mx+N)((x^2+px+q)^n)dx$ the substitution $t=x+\frac(p)(2)$ is made, after which the resulting interval is divided into two. The first will be calculated by entering under the differential sign, and the second will have the form $I_n=\int\frac(dt)((t^2+a^2)^n)$. This integral is taken using the recurrence relation

\begin(equation) I_(n+1)=\frac(1)(2na^2)\frac(t)((t^2+a^2)^n)+\frac(2n-1)(2na ^2)I_n,\; n\in N\end(equation)

The calculation of such an integral is discussed in example No. 7 (see the third part).

Scheme for calculating integrals of rational functions (rational fractions):

1. If the integrand is elementary, then apply formulas (1)-(4).
2. If the integrand is not elementary, then represent it as a sum of elementary fractions, and then integrate using formulas (1)-(4).

The above algorithm for integrating rational fractions has an undeniable advantage - it is universal. Those. using this algorithm you can integrate any rational fraction. That is why almost all changes of variables in an indefinite integral (Euler, Chebyshev, universal trigonometric substitution) are made in such a way that after this change we obtain a rational fraction under the interval. And then apply the algorithm to it. We will analyze the direct application of this algorithm using examples, after making a small note.

$$ \int\frac(7dx)(x+9)=7\ln|x+9|+C. $$

In principle, this integral is easy to obtain without mechanical application of the formula. If we take the constant $7$ out of the integral sign and take into account that $dx=d(x+9)$, we get:

$$ \int\frac(7dx)(x+9)=7\cdot \int\frac(dx)(x+9)=7\cdot \int\frac(d(x+9))(x+9 )=|u=x+9|=7\cdot\int\frac(du)(u)=7\ln|u|+C=7\ln|x+9|+C. $$

For detailed information, I recommend looking at the topic. It explains in detail how such integrals are solved. By the way, the formula is proved by the same transformations that were applied in this paragraph when solving it “manually”.

2) Again, there are two ways: use the ready-made formula or do without it. If you apply the formula, then you should take into account that the coefficient in front of $x$ (number 4) will have to be removed. To do this, let’s simply take this four out of brackets:

$$ \int\frac(11dx)((4x+19)^8)=\int\frac(11dx)(\left(4\left(x+\frac(19)(4)\right)\right)^ 8)= \int\frac(11dx)(4^8\left(x+\frac(19)(4)\right)^8)=\int\frac(\frac(11)(4^8)dx) (\left(x+\frac(19)(4)\right)^8). $$

Now it’s time to apply the formula:

$$ \int\frac(\frac(11)(4^8)dx)(\left(x+\frac(19)(4)\right)^8)=-\frac(\frac(11)(4 ^8))((8-1)\left(x+\frac(19)(4) \right)^(8-1))+C= -\frac(\frac(11)(4^8)) (7\left(x+\frac(19)(4) \right)^7)+C=-\frac(11)(7\cdot 4^8 \left(x+\frac(19)(4) \right )^7)+C. $$

You can do without using the formula. And even without taking the constant $4$ out of brackets. If we take into account that $dx=\frac(1)(4)d(4x+19)$, we get:

$$ \int\frac(11dx)((4x+19)^8)=11\int\frac(dx)((4x+19)^8)=\frac(11)(4)\int\frac( d(4x+19))((4x+19)^8)=|u=4x+19|=\\ =\frac(11)(4)\int\frac(du)(u^8)=\ frac(11)(4)\int u^(-8)\;du=\frac(11)(4)\cdot\frac(u^(-8+1))(-8+1)+C= \\ =\frac(11)(4)\cdot\frac(u^(-7))(-7)+C=-\frac(11)(28)\cdot\frac(1)(u^7 )+C=-\frac(11)(28(4x+19)^7)+C. $$

Detailed explanations for finding such integrals are given in the topic “Integration by substitution (substitution under the differential sign)”.

3) We need to integrate the fraction $\frac(4x+7)(x^2+10x+34)$. This fraction has the structure $\frac(Mx+N)(x^2+px+q)$, where $M=4$, $N=7$, $p=10$, $q=34$. However, to make sure that this is really an elementary fraction of the third type, you need to check the fulfillment of the condition $p^2-4q< 0$. Так как $p^2-4q=10^2-4\cdot 34=-16 < 0$, то мы действительно имеем дело с интегрированием элементарной дроби третьего типа. Как и в предыдущих пунктах есть два пути для нахождения $\int\frac{4x+7}{x^2+10x+34}dx$. Первый путь - банально использовать формулу . Подставив в неё $M=4$, $N=7$, $p=10$, $q=34$ получим:

$$ \int\frac(4x+7)(x^2+10x+34)dx = \frac(4)(2)\cdot \ln (x^2+10x+34)+\frac(2\cdot 7-4\cdot 10)(\sqrt(4\cdot 34-10^2)) \arctg\frac(2x+10)(\sqrt(4\cdot 34-10^2))+C=\\ = 2\cdot \ln (x^2+10x+34)+\frac(-26)(\sqrt(36)) \arctg\frac(2x+10)(\sqrt(36))+C =2\cdot \ln (x^2+10x+34)+\frac(-26)(6) \arctg\frac(2x+10)(6)+C=\\ =2\cdot \ln (x^2+10x +34)-\frac(13)(3) \arctg\frac(x+5)(3)+C. $$

Let's solve the same example, but without using a ready-made formula. Let's try to isolate the derivative of the denominator in the numerator. What does this mean? We know that $(x^2+10x+34)"=2x+10$. It is the expression $2x+10$ that we have to isolate in the numerator. So far the numerator contains only $4x+7$, but this will not last long. Let's apply the following transformation to the numerator:

$$ 4x+7=2\cdot 2x+7=2\cdot (2x+10-10)+7=2\cdot(2x+10)-2\cdot 10+7=2\cdot(2x+10) -13. $$

Now the required expression $2x+10$ appears in the numerator. And our integral can be rewritten as follows:

$$ \int\frac(4x+7)(x^2+10x+34) dx= \int\frac(2\cdot(2x+10)-13)(x^2+10x+34)dx. $$

Let's split the integrand into two. Well, and, accordingly, the integral itself is also “bifurcated”:

$$ \int\frac(2\cdot(2x+10)-13)(x^2+10x+34)dx=\int \left(\frac(2\cdot(2x+10))(x^2 +10x+34)-\frac(13)(x^2+10x+34) \right)\; dx=\\ =\int \frac(2\cdot(2x+10))(x^2+10x+34)dx-\int\frac(13dx)(x^2+10x+34)=2\cdot \int \frac((2x+10)dx)(x^2+10x+34)-13\cdot\int\frac(dx)(x^2+10x+34). $$

Let's first talk about the first integral, i.e. about $\int \frac((2x+10)dx)(x^2+10x+34)$. Since $d(x^2+10x+34)=(x^2+10x+34)"dx=(2x+10)dx$, then the numerator of the integrand contains the differential of the denominator. In short, instead of the expression $( 2x+10)dx$ we write $d(x^2+10x+34)$.

Now let's say a few words about the second integral. Let's select a complete square in the denominator: $x^2+10x+34=(x+5)^2+9$. In addition, we take into account $dx=d(x+5)$. Now the sum of integrals we obtained earlier can be rewritten in a slightly different form:

$$ 2\cdot\int \frac((2x+10)dx)(x^2+10x+34)-13\cdot\int\frac(dx)(x^2+10x+34) =2\cdot \int \frac(d(x^2+10x+34))(x^2+10x+34)-13\cdot\int\frac(d(x+5))((x+5)^2+ 9). $$

If in the first integral we make the replacement $u=x^2+10x+34$, then it will take the form $\int\frac(du)(u)$ and can be obtained by simply applying the second formula from . As for the second integral, the change $u=x+5$ is feasible for it, after which it will take the form $\int\frac(du)(u^2+9)$. This is the purest eleventh formula from the table of indefinite integrals. So, returning to the sum of integrals, we have:

$$ 2\cdot\int \frac(d(x^2+10x+34))(x^2+10x+34)-13\cdot\int\frac(d(x+5))((x+ 5)^2+9) =2\cdot\ln(x^2+10x+34)-\frac(13)(3)\arctg\frac(x+5)(3)+C. $$

We received the same answer as when applying the formula, which, strictly speaking, is not surprising. In general, the formula is proved by the same methods that we used to find this integral. I believe that the attentive reader may have one question here, so I will formulate it:

Question No. 1

If we apply the second formula from the table of indefinite integrals to the integral $\int \frac(d(x^2+10x+34))(x^2+10x+34)$, then we get the following:

$$ \int \frac(d(x^2+10x+34))(x^2+10x+34)=|u=x^2+10x+34|=\int\frac(du)(u) =\ln|u|+C=\ln|x^2+10x+34|+C. $$

Why was there no module in the solution?

Answer to question #1

The question is completely natural. The module was missing only because the expression $x^2+10x+34$ for any $x\in R$ is greater than zero. This is quite easy to show in several ways. For example, since $x^2+10x+34=(x+5)^2+9$ and $(x+5)^2 ≥ 0$, then $(x+5)^2+9 > 0$ . You can think differently, without using the selection of a complete square. Since $10^2-4\cdot 34=-16< 0$, то $x^2+10x+34 >0$ for any $x\in R$ (if this logical chain is surprising, I advise you to look at the graphical method for solving quadratic inequalities). In any case, since $x^2+10x+34 > 0$, then $|x^2+10x+34|=x^2+10x+34$, i.e. Instead of a module, you can use regular parentheses.

All points of example No. 1 have been solved, all that remains is to write down the answer.

Answer:

1. $\int\frac(7dx)(x+9)=7\ln|x+9|+C$;
2. $\int\frac(11dx)((4x+19)^8)=-\frac(11)(28(4x+19)^7)+C$;
3. $\int\frac(4x+7)(x^2+10x+34)dx=2\cdot\ln(x^2+10x+34)-\frac(13)(3)\arctg\frac(x +5)(3)+C$.

Example No. 2

Find the integral $\int\frac(7x+12)(3x^2-5x-2)dx$.

At first glance, the integrand fraction $\frac(7x+12)(3x^2-5x-2)$ is very similar to an elementary fraction of the third type, i.e. by $\frac(Mx+N)(x^2+px+q)$. It seems that the only difference is the coefficient of $3$ in front of $x^2$, but it doesn’t take long to remove the coefficient (put it out of brackets). However, this similarity is apparent. For the fraction $\frac(Mx+N)(x^2+px+q)$ the condition $p^2-4q is mandatory< 0$, которое гарантирует, что знаменатель $x^2+px+q$ нельзя разложить на множители. Проверим, как обстоит дело с разложением на множители у знаменателя нашей дроби, т.е. у многочлена $3x^2-5x-2$.

Our coefficient before $x^2$ is not equal to one, therefore check the condition $p^2-4q< 0$ напрямую мы не можем. Однако тут нужно вспомнить, откуда взялось выражение $p^2-4q$. Это всего лишь дискриминант квадратного уравнения $x^2+px+q=0$. Если дискриминант меньше нуля, то выражение $x^2+px+q$ на множители не разложишь. Вычислим дискриминант многочлена $3x^2-5x-2$, расположенного в знаменателе нашей дроби: $D=(-5)^2-4\cdot 3\cdot(-2)=49$. Итак, $D >0$, therefore the expression $3x^2-5x-2$ can be factorized. This means that the fraction $\frac(7x+12)(3x^2-5x-2)$ is not an elemental fraction of the third type, and apply $\int\frac(7x+12)(3x^2-) to the integral 5x-2)dx$ formula is not possible.

Well, if the given rational fraction is not an elementary fraction, then it needs to be represented as a sum of elementary fractions and then integrated. In short, take advantage of the trail. How to decompose a rational fraction into elementary ones is written in detail. Let's start by factoring the denominator:

$$ 3x^2-5x-2=0;\\ \begin(aligned) & D=(-5)^2-4\cdot 3\cdot(-2)=49;\\ & x_1=\frac( -(-5)-\sqrt(49))(2\cdot 3)=\frac(5-7)(6)=\frac(-2)(6)=-\frac(1)(3); \\ & x_2=\frac(-(-5)+\sqrt(49))(2\cdot 3)=\frac(5+7)(6)=\frac(12)(6)=2.\ \\end(aligned)\\ 3x^2-5x-2=3\cdot\left(x-\left(-\frac(1)(3)\right)\right)\cdot (x-2)= 3\cdot\left(x+\frac(1)(3)\right)(x-2). $$

We present the subintercal fraction in this form:

$$ \frac(7x+12)(3x^2-5x-2)=\frac(7x+12)(3\cdot\left(x+\frac(1)(3)\right)(x-2) )=\frac(\frac(7)(3)x+4)(\left(x+\frac(1)(3)\right)(x-2)). $$

Now let’s decompose the fraction $\frac(\frac(7)(3)x+4)(\left(x+\frac(1)(3)\right)(x-2))$ into elementary ones:

$$ \frac(\frac(7)(3)x+4)(\left(x+\frac(1)(3)\right)(x-2)) =\frac(A)(x+\frac( 1)(3))+\frac(B)(x-2)=\frac(A(x-2)+B\left(x+\frac(1)(3)\right))(\left(x+ \frac(1)(3)\right)(x-2));\\ \frac(7)(3)x+4=A(x-2)+B\left(x+\frac(1)( 3)\right). $$

To find the coefficients $A$ and $B$ there are two standard ways: the method of undetermined coefficients and the method of substitution of partial values. Let's apply the partial value substitution method, substituting $x=2$ and then $x=-\frac(1)(3)$:

$$ \frac(7)(3)x+4=A(x-2)+B\left(x+\frac(1)(3)\right).\\ x=2;\; \frac(7)(3)\cdot 2+4=A(2-2)+B\left(2+\frac(1)(3)\right); \; \frac(26)(3)=\frac(7)(3)B;\; B=\frac(26)(7).\\ x=-\frac(1)(3);\; \frac(7)(3)\cdot \left(-\frac(1)(3) \right)+4=A\left(-\frac(1)(3)-2\right)+B\left (-\frac(1)(3)+\frac(1)(3)\right); \; \frac(29)(9)=-\frac(7)(3)A;\; A=-\frac(29\cdot 3)(9\cdot 7)=-\frac(29)(21).\\ $$

Since the coefficients have been found, all that remains is to write down the finished expansion:

$$ \frac(\frac(7)(3)x+4)(\left(x+\frac(1)(3)\right)(x-2))=\frac(-\frac(29)( 21))(x+\frac(1)(3))+\frac(\frac(26)(7))(x-2). $$

In principle, you can leave this entry, but I like a more accurate option:

$$ \frac(\frac(7)(3)x+4)(\left(x+\frac(1)(3)\right)(x-2))=-\frac(29)(21)\ cdot\frac(1)(x+\frac(1)(3))+\frac(26)(7)\cdot\frac(1)(x-2). $$

Returning to the original integral, we substitute the resulting expansion into it. Then we divide the integral into two, and apply the formula to each. I prefer to immediately place the constants outside the integral sign:

$$ \int\frac(7x+12)(3x^2-5x-2)dx =\int\left(-\frac(29)(21)\cdot\frac(1)(x+\frac(1) (3))+\frac(26)(7)\cdot\frac(1)(x-2)\right)dx=\\ =\int\left(-\frac(29)(21)\cdot\ frac(1)(x+\frac(1)(3))\right)dx+\int\left(\frac(26)(7)\cdot\frac(1)(x-2)\right)dx =- \frac(29)(21)\cdot\int\frac(dx)(x+\frac(1)(3))+\frac(26)(7)\cdot\int\frac(dx)(x-2 )dx=\\ =-\frac(29)(21)\cdot\ln\left|x+\frac(1)(3)\right|+\frac(26)(7)\cdot\ln|x- 2|+C. $$

Answer: $\int\frac(7x+12)(3x^2-5x-2)dx=-\frac(29)(21)\cdot\ln\left|x+\frac(1)(3)\right| +\frac(26)(7)\cdot\ln|x-2|+C$.

Example No. 3

Find the integral $\int\frac(x^2-38x+157)((x-1)(x+4)(x-9))dx$.

We need to integrate the fraction $\frac(x^2-38x+157)((x-1)(x+4)(x-9))$. The numerator contains a polynomial of the second degree, and the denominator contains a polynomial of the third degree. Since the degree of the polynomial in the numerator is less than the degree of the polynomial in the denominator, i.e. $2< 3$, то подынтегральная дробь является правильной. Разложение этой дроби на элементарные (простейшие) было получено в примере №3 на странице, посвящённой разложению рациональных дробей на элементарные. Полученное разложение таково:

$$ \frac(x^2-38x+157)((x-1)(x+4)(x-9))=-\frac(3)(x-1)+\frac(5)(x +4)-\frac(1)(x-9). $$

All we have to do is split the given integral into three and apply the formula to each. I prefer to immediately place the constants outside the integral sign:

$$ \int\frac(x^2-38x+157)((x-1)(x+4)(x-9))dx=\int\left(-\frac(3)(x-1) +\frac(5)(x+4)-\frac(1)(x-9) \right)dx=\\=-3\cdot\int\frac(dx)(x-1)+ 5\cdot \int\frac(dx)(x+4)-\int\frac(dx)(x-9)=-3\ln|x-1|+5\ln|x+4|-\ln|x- 9|+C. $$

Answer: $\int\frac(x^2-38x+157)((x-1)(x+4)(x-9))dx=-3\ln|x-1|+5\ln|x+ 4|-\ln|x-9|+C$.

Continuation of the analysis of examples of this topic is located in the second part.

“A mathematician, just like an artist or poet, creates patterns. And if his patterns are more stable, it is only because they are composed of ideas... The patterns of a mathematician, just like the patterns of an artist or a poet, must be beautiful; Ideas, just like colors or words, must correspond to each other. Beauty is the first requirement: there is no place in the world for ugly mathematics».

G.H.Hardy

In the first chapter it was noted that there are antiderivatives of fairly simple functions that can no longer be expressed through elementary functions. In this regard, those classes of functions about which we can accurately say that their antiderivatives are elementary functions acquire enormous practical importance. This class of functions includes rational functions, representing the ratio of two algebraic polynomials. Many problems lead to the integration of rational fractions. Therefore, it is very important to be able to integrate such functions.

2.1.1. Fractional rational functions

Rational fraction(or fractional rational function) is called the relation of two algebraic polynomials:

where and are polynomials.

Let us recall that polynomial (polynomial, whole rational function) nth degree called a function of the form

Where – real numbers. For example,

– polynomial of the first degree;

– polynomial of the fourth degree, etc.

The rational fraction (2.1.1) is called correct, if the degree is lower than the degree , i.e. n<m, otherwise the fraction is called wrong.

Any improper fraction can be represented as the sum of a polynomial (the whole part) and a proper fraction (the fractional part). The separation of the whole and fractional parts of an improper fraction can be done according to the rule for dividing polynomials with a “corner”.

Example 2.1.1. Identify the whole and fractional parts of the following improper rational fractions:

A) , b) .

Solution . a) Using the “corner” division algorithm, we get

Thus, we get

.

b) Here we also use the “corner” division algorithm:

As a result, we get

.

Let's summarize. In the general case, the indefinite integral of a rational fraction can be represented as the sum of the integrals of the polynomial and the proper rational fraction. Finding antiderivatives of polynomials is not difficult. Therefore, in what follows we will mainly consider proper rational fractions.

2.1.2. The simplest rational fractions and their integration

Among proper rational fractions, there are four types, which are classified as the simplest (elementary) rational fractions:

where is an integer, , i.e. quadratic trinomial has no real roots.

Integrating simple fractions of type 1 and type 2 does not present much difficulty:

, (2.1.3)

. (2.1.4)

Let us now consider the integration of simple fractions of the 3rd type, but we will not consider fractions of the 4th type.

Let's start with integrals of the form

.

This integral is usually calculated by isolating the perfect square of the denominator. The result is a table integral of the following form

or .

Example 2.1.2. Find the integrals:

A) , b) .

Solution . a) Select a complete square from a quadratic trinomial:

From here we find

b) By isolating a complete square from a quadratic trinomial, we obtain:

Thus,

.

To find the integral

you can isolate the derivative of the denominator in the numerator and expand the integral into the sum of two integrals: the first of them by substitution comes down to appearance

,

and the second - to the one discussed above.

Example 2.1.3. Find the integrals:

.

Solution . notice, that . Let us isolate the derivative of the denominator in the numerator:

The first integral is calculated using the substitution :

In the second integral, we select the perfect square in the denominator

Finally, we get

2.1.3. Proper rational fraction expansion
for the sum of simple fractions

Any proper rational fraction can be represented in a unique way as a sum of simple fractions. To do this, the denominator must be factorized. From higher algebra it is known that every polynomial with real coefficients