Indefinite integral of a fraction. Integration of a fractional-rational function. Method of indefinite coefficients. Integration of the correct fractional-rational function

Integration of a fractional-rational function.
Method of undetermined coefficients

We continue to work on integrating fractions. We have already considered 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 teapot, then you need to start with the article Indefinite integral. Solution examples.

Oddly enough, now we will deal not so much with finding integrals as ... solving systems of linear equations. In this connection strongly I recommend visiting the lesson Namely, you need to be well versed in the 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 in the numerator and denominator of which are polynomials or products of polynomials. At the same time, fractions are more sophisticated than those discussed in the article. Integration of some fractions.

Integration of the correct 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 rational-fractional function is ask the following question: is the fraction correct? This step is done orally, and now I will explain how:

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

The highest power of the numerator is two.

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

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

Conclusion: Highest power of the numerator STRICTLY less than the highest power of the denominator, then the fraction is correct.

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

Now we will consider only proper fractional-rational functions. The case when the degree of the numerator is greater than or equal to the degree of the denominator, we will analyze at the end of the lesson.

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

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

The discriminant is greater than zero, which means that the trinomial is indeed factorized:

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

Let's start making a decision:

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, an intuitive thought somehow slips through 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's breathe a sigh of relief, the corresponding theorem of mathematical analysis states - IT IS POSSIBLE. Such a decomposition exists and is unique.

There is only one catch, the coefficients we bye we do not know, hence the name - the method of indefinite coefficients.

You guessed it, the subsequent gestures so, do not cackle! will be aimed at just LEARNING them - to find out what they are equal to.

Be careful, I explain in detail once!

So, let's start dancing from:

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

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

On the left side, we open the brackets, while we do not touch the unknown coefficients yet:

At the same time, we repeat the school rule of multiplication of polynomials. When I was a teacher, I learned to say 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 in the first equation of the system:

Well remember the following nuance. What would happen if the right side did not exist at all? Say, would it just show off without any square? In this case, in the equation of the system, it would be necessary to put zero on the right: . Why zero? And because on the right side you can always attribute this very square with zero: If there are no variables or (and) a free term on the right side, then we put zeros on the right sides of the corresponding equations of the system.

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

And, finally, mineral water, we select free members.

Eh, ... 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 members along a number line and choose the largest of them. Let's get serious. Although ... whoever lives to see the end of this lesson will still smile quietly.

System 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, while obtaining the equality , from which it follows that

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

(5) We substitute and into the first equation, getting .

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

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

Almost arrived. The coefficients are found, while:

A clean 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 the linearity of the indefinite integral and integrate. I draw your attention to the fact that under each of the three integrals we have a “free” complex function, I spoke about the features of its integration in the lesson Variable change method in indefinite integral.

Check: Differentiate the answer:

The original integrand was obtained, which means that the integral was found correctly.
During the verification, it was necessary to bring the expression to a common denominator, and this is not accidental. The method of indefinite coefficients and bringing the expression to a common denominator are mutually inverse actions.

Example 2

Find the indefinite integral.

Let's go back to the fraction from the first example: . It is easy to see that in the denominator all factors are DIFFERENT. The question arises, what to do if, for example, such a fraction is given: ? Here we have degrees in the denominator, or, in mathematical terms, multiple factors. In addition, there is an indecomposable square trinomial (it is easy to verify that the discriminant of the equation is negative, so the trinomial cannot be factored in any way). What to do? The expansion into a sum of elementary fractions will look like with unknown coefficients at the top or some other way?

Example 3

Submit a function

Step 1. Checking if we have a correct fraction
Highest power of the numerator: 2
Highest denominator: 8
, so the fraction is correct.

Step 2 Can anything be factored in the denominator? Obviously not, everything is already laid out. The square trinomial does not expand into a product for the above reasons. Good. Less work.

Step 3 Let us represent a fractional-rational function as a sum of elementary fractions.
In this case, the decomposition 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 in the first degree (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 contains multiple multiplier, then you need to decompose as follows:
- that is, sequentially sort 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 decomposition I have given and make sure that they are decomposed exactly according to this rule.

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

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

Example 4

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

This is a do-it-yourself example. Full solution and answer at the end of the lesson.
Strictly follow the algorithm!

If you have figured out the principles by which you need to decompose a fractional-rational function into a sum, then you can crack almost any integral of the type under consideration.

Example 5

Find the indefinite integral.

Step 1. Obviously, the fraction is correct:

Step 2 Can anything be factored in the denominator? Can. Here is the sum of cubes . Factoring 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:

Note that the polynomial is indecomposable (check that the discriminant is negative), so at the top we put a linear function with unknown coefficients, and not just a single letter.

We bring the fraction to a common denominator:

Let's create and solve the system:

(1) From the first equation, we express and substitute 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, in principle, are 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 find this method in the last paragraph of the lesson. Integration of some fractions.

(3) Once again we use the properties of linearity. In the third integral, we begin to select a full square (the penultimate paragraph of the lesson Integration of some fractions).

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

(5) We 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 strongly advise you to at least skim through this topic before proceeding 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 relevant 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 a 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 (simplest) 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 better understanding of the text): show\hide

Why is the $p^2-4q condition necessary?< 0$ 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 necessary that the coefficient in front of $x^2$ equals 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 (regular and improper), as well as examples of the decomposition of a rational fraction into elementary ones, can be found. Here we are only interested in questions of their integration. Let's start with the integration of elementary fractions. So, each of the four types of the above elementary fractions is easy to integrate using the formulas below. Let me remind you that when integrating fractions of type (2) and (4) $n=2,3,4,\ldots$ is assumed. Formulas (3) and (4) require 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 replacement $t=x+\frac(p)(2)$ is made, after which the resulting integral is split into two. The first one will be calculated by inserting it under the differential sign, and the second one will look like $I_n=\int\frac(dt)((t^2+a^2)^n)$. 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 analyzed in example No. 7 (see the third part).

Scheme for calculating integrals from 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, one can integrate any rational fraction. That is why almost all replacements of variables in the indefinite integral (Euler, Chebyshev substitutions, universal trigonometric substitution) are done in such a way that after this replacement we get a rational fraction under the interval. And 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 $dx=d(x+9)$, then 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 to look 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 "manually".

2) Again, there are two ways: to apply a ready-made formula or to do without it. If you apply the formula, then you should take into account that the coefficient in front of $x$ (the number 4) will have to be removed. To do this, we simply take out the four of them in 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 putting the constant $4$ out of the brackets. If we take into account that $dx=\frac(1)(4)d(4x+19)$, then 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 on finding such integrals are given in the topic "Integration by substitution (introduction 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 indeed an elementary fraction of the third type, you need to check 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 the 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 is not for long. 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 $2x+10$ has appeared 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 break the integrand into two. Well, and, accordingly, the integral itself is also "split":

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

Let's talk about the first integral first, 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 denominator differential is located in the numerator of the integrand. 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 single out the full 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 obtained by us 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 we make the change $u=x^2+10x+34$ in the first integral, then it will take the form $\int\frac(du)(u)$ and is taken by simply applying the second formula from . As for the second integral, the replacement $u=x+5$ is feasible for it, after which it takes the form $\int\frac(du)(u^2+9)$. This is the purest water, the eleventh formula from the table of indefinite integrals. So, returning to the sum of integrals, we will 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 got the same answer as when applying the formula , which, in fact, is not surprising. In general, the formula is proved by the same methods that we used to find this integral. I believe that an attentive reader may have one question here, therefore I will formulate it:

Question #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 the module missing from the solution?

Answer to question #1

The question is completely legitimate. The modulus 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 $(x+5)^2+9 > 0$ . It is possible to judge in a different way, without involving the selection of a full 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 square inequalities). In any case, since $x^2+10x+34 > 0$, then $|x^2+10x+34|=x^2+10x+34$, i.e. you can use normal brackets instead of a module.

All points of example No. 1 are solved, it remains only 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 #2

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

At first glance, the integrand $\frac(7x+12)(3x^2-5x-2)$ is very similar to an elementary fraction of the third type, i.e. to $\frac(Mx+N)(x^2+px+q)$. It seems that the only difference is the coefficient $3$ in front of $x^2$, but it won't take long to remove the coefficient (out of brackets). However, this similarity is apparent. For the fraction $\frac(Mx+N)(x^2+px+q)$ the condition $p^2-4q< 0$, 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 in front of $x^2$ is not equal to one, so 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, $D >0$, so the expression $3x^2-5x-2$ can be factorized. And this means that the fraction $\frac(7x+12)(3x^2-5x-2)$ is not an elementary fraction of the third type, and apply to the integral $\int\frac(7x+12)(3x^2- 5x-2)dx$ formula is not allowed.

Well, if the given rational fraction is not elementary, then it must be represented as a sum of elementary fractions, and then integrated. In short, trail take advantage of . 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 represent the subinternal fraction in the following 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 expand 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 indeterminate coefficients and the method of substitution of partial values. Let's apply the partial value substitution method by 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, it remains only 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 version:

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

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 take out 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 #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 is a polynomial of the second degree, and the denominator is 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:

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

We just have to break the given integral into three, and apply the formula to each. I prefer to immediately take out 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: $\int\frac(x^2-38x+157)((x-1)(x+4)(x-9))dx=-3\ln|x-1|+5\ln|x+ 4|-\ln|x-9|+C$.

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

“A mathematician, like an artist or a poet, creates patterns. And if his patterns are more stable, it is only because they are made up of ideas ... The patterns of a mathematician, just like those of an artist or a poet, must be beautiful; ideas, just like colors or words, must match. 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 in terms of elementary functions. In this regard, those classes of functions acquire great practical importance, about which it can be said for sure that their antiderivatives are elementary functions. This class of functions includes rational functions, which is 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 the ratio of two algebraic polynomials:

where and are polynomials.

Recall that polynomial (polynomial, an entire rational function) nth degree is called a function of the form

where are real numbers. For example,

is a polynomial of the first degree;

is a 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 (integer part) and a proper fraction (fractional part). The selection of the integer and fractional parts of an improper fraction can be done according to the rule of dividing polynomials by a “corner”.

Example 2.1.1. Select the integer and fractional parts of the following improper rational fractions:

a) , b) .

Solution . a) Using the division algorithm "corner", we obtain

Thus, we get

.

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

As a result, we get

.

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

2.1.2. The simplest rational fractions and their integration

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

3) ,

4) ,

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

Integration of the simplest fractions of the 1st and 2nd type does not present great difficulties:

, (2.1.3)

. (2.1.4)

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

We start with integrals of the form

.

This integral is usually calculated by taking the full square in the denominator. The result is a table integral of the following form

or .

Example 2.1.2. Find integrals:

a) , b) .

Solution . a) We select a full square from a square trinomial:

From here we find

b) Selecting the full square from the square trinomial, we get:

In this way,

.

To find the integral

we can extract the derivative of the denominator in the numerator and expand the integral into the sum of two integrals: the first of them by substituting comes down to the form

,

and the second - to the above.

Example 2.1.3. Find integrals:

.

Solution . notice, that . We select the derivative of the denominator in the numerator:

The first integral is calculated using the substitution :

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

Finally, we get

2.1.3. Expansion of a proper rational fraction
the sum of simple fractions

Any proper rational fraction can be represented uniquely as a sum of simple fractions. To do this, the denominator must be decomposed into factors. It is known from higher algebra that every polynomial with real coefficients

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

Example 1

Calculate integral:
.

Solution

Here, under the integral sign there is a rational function, since the integrand is a fraction of polynomials. The degree of the denominator polynomial ( 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 take the integer part of the fraction. Divide x 4 on x 3 - 6 x 2 + 11 x - 6:

From here
.

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

1 . Divide by x - 1 :

From here
.
We solve a quadratic equation.
.
Equation roots: , .
Then
.

3. Let's decompose the fraction into simple ones.

.

So we found:
.
Let's integrate.

Answer

Example 2

Calculate integral:
.

Solution

Here in the numerator of the fraction is a polynomial of degree zero ( 1 = x0). The denominator is a third degree polynomial. Because the 0 < 3 , then the fraction is correct. Let's break it down into simple fractions.

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

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

So,
.

We solve the quadratic equation:
x 2 + x + 3 = 0.
Find the discriminant: D = 1 2 - 4 3 = -11. Because D< 0 , then the equation has no real roots. Thus, we have obtained the decomposition of the denominator into factors:
.

2.
.
(x - 1)(x 2 + x + 3):
(2.1) .
Substitute x = 1 . Then x- 1 = 0 ,
.

Substitute in (2.1) x= 0 :
1 = 3 A - C;
.

Equate in (2.1) coefficients at 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 module sign can be omitted.

We deliver to (2.2) :
.

Answer

Example 3

Calculate integral:
.

Solution

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

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

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


So,
.

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

So, we have 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 decompose the fraction into simple ones. We are looking for a decomposition in the form:
.
We get rid of the denominator of the fraction, multiply by (x + 1) 2 (x 2 + 2):
(3.1) .
Substitute x = -1 . Then x + 1 = 0 ,
.

Differentiate (3.1) :

;

.
Substitute x = -1 and take into account that x + 1 = 0 :
;
; .

Substitute in (3.1) x= 0 :
0 = 2A + 2B + D;
.

Equate in (3.1) coefficients at x 3 :
;
1=B+C;
.

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

3. Let's integrate.


.