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 (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 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 that the condition $p^2-4q is met< 0$. 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.

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:

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 simpler, 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 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 for 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...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 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

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. 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.

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.

As we will see below, not every elementary function has an integral expressed in elementary functions. Therefore, it is very important to identify classes of functions whose integrals are expressed in terms of elementary functions. The simplest of these classes is the class of rational functions.

Any rational function can be represented as a rational fraction, that is, as a ratio of two polynomials:

Without limiting the generality of the argument, we will assume that the polynomials do not have common roots.

If the degree of the numerator is lower than the degree of the denominator, then the fraction is called proper, otherwise the fraction is called improper.

If the fraction is improper, then by dividing the numerator by the denominator (according to the rule for dividing polynomials), you can represent this fraction as the sum of a polynomial and some proper fraction:

here is a polynomial, and a is a proper fraction.

Example t. Let an improper rational fraction be given

Dividing the numerator by the denominator (using the rule for dividing polynomials), we get

Since integrating polynomials is not difficult, the main difficulty in integrating rational fractions is integrating proper rational fractions.

Definition. Proper rational fractions of the form

are called simple fractions of types I, II, III and IV.

Integrating the simplest fractions of types I, II and III is not very difficult, so we will carry out their integration without any additional explanation:

More complex calculations require the integration of simple fractions of type IV. Let us be given an integral of this type:

Let's make the transformations:

The first integral is taken by substitution

The second integral - we denote it by writing it in the form

By assumption, the roots of the denominator are complex, and therefore, Next we proceed as follows:

Let's transform the integral:

Integrating by parts, we have

Substituting this expression into equality (1), we obtain

The right-hand side contains an integral of the same type as but the exponent of the denominator of the integrand is one lower; thus, we expressed it through . Continuing along the same path, we reach the well-known integral.

Let us remind you that fractional-rational are called functions of the form $$ f(x) = \frac(P_n(x))(Q_m(x)), $$ in the general case being the ratio of two polynomials %%P_n(x)%% and %%Q_m(x)% %.

If %%m > n \geq 0%%, then the rational fraction is called correct, otherwise - incorrect. Using the rule for dividing polynomials, an improper rational fraction can be represented as the sum of a polynomial %%P_(n - m)%% of degree %%n - m%% and some proper fraction, i.e. $$ \frac(P_n(x))(Q_m(x)) = P_(n-m)(x) + \frac(P_l(x))(Q_n(x)), $$ where the degree %%l%% of the polynomial %%P_l(x)%% is less than the degree %%n%% of the polynomial %%Q_n(x)%%.

Thus, the indefinite integral of a rational function can be represented as the sum of the indefinite integrals of a polynomial and a proper rational fraction.

Integrals from simple rational fractions

Among proper rational fractions, there are four types, which are classified as simple rational fractions:

1. %%\displaystyle \frac(A)(x - a)%%,
2. %%\displaystyle \frac(A)((x - a)^k)%%,
3. %%\displaystyle \frac(Ax + B)(x^2 + px + q)%%,
4. %%\displaystyle \frac(Ax + B)((x^2 + px + q)^k)%%,

where %%k > 1%% is an integer and %%p^2 - 4q< 0%%, т.е. квадратные уравнения не имеют действительных корней.

Calculation of indefinite integrals of fractions of the first two types

Calculating indefinite integrals of fractions of the first two types does not cause difficulties: $$ \begin(array)(ll) \int \frac(A)(x - a) \mathrm(d)x &= A\int \frac(\mathrm (d)(x - a))(x - a) = A \ln |x - a| + C, \\ \\ \int \frac(A)((x - a)^k) \mathrm(d)x &= A\int \frac(\mathrm(d)(x - a))(( x - a)^k) = A \frac((x-a)^(-k + 1))(-k + 1) + C = \\ &= -\frac(A)((k-1)(x-a )^(k-1)) + C. \end(array) $$

Calculation of indefinite integrals of fractions of the third type

We first transform the third type of fraction by highlighting the perfect square in the denominator: $$ \frac(Ax + B)(x^2 + px + q) = \frac(Ax + B)((x + p/2)^2 + q - p^2/4), $$ since %%p^2 - 4q< 0%%, то %%q - p^2/4 >0%%, which we denote as %%a^2%%. Also replacing %%t = x + p/2, \mathrm(d)t = \mathrm(d)x%%, we transform the denominator and write the integral of the third type fraction in the form $$ \begin(array)(ll) \ int \frac(Ax + B)(x^2 + px + q) \mathrm(d)x &= \int \frac(Ax + B)((x + p/2)^2 + q - p^2 /4) \mathrm(d)x = \\ &= \int \frac(A(t - p/2) + B)(t^2 + a^2) \mathrm(d)t = \int \frac (At + (B - A p/2))(t^2 + a^2) \mathrm(d)t. \end(array) $$

Using the linearity of the indefinite integral, we represent the last integral as a sum of two and in the first of them we introduce %%t%% under the differential sign: $$ \begin(array)(ll) \int \frac(At + (B - A p /2))(t^2 + a^2) \mathrm(d)t &= A\int \frac(t \mathrm(d)t)(t^2 + a^2) + \left(B - \frac(pA)(2)\right)\int \frac(\mathrm(d)t)(t^2 + a^2) = \\ &= \frac(A)(2) \int \frac( \mathrm(d)\left(t^2 + a^2\right))(t^2 + a^2) + - \frac(2B - pA)(2)\int \frac(\mathrm(d) t)(t^2 + a^2) = \\ &= \frac(A)(2) \ln \left| t^2 + a^2\right| + \frac(2B - pA)(2a) \text(arctg)\frac(t)(a) + C. \end(array) $$

Returning to the original variable %%x%%, as a result, for a fraction of the third type we obtain $$ \int \frac(Ax + B)(x^2 + px + q) \mathrm(d)x = \frac(A)( 2) \ln \left| x^2 + px + q\right| + \frac(2B - pA)(2a) \text(arctg)\frac(x + p/2)(a) + C, $$ where %%a^2 = q - p^2 / 4 > 0% %.

Calculating a type 4 integral is difficult and is therefore not covered in this course.

Examples of integrating rational functions (fractions) with detailed solutions are considered.

Content

See also: Roots of a quadratic equation

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

Example 1

Calculate the integral:
.

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 . Divide by x - 1 :

From here
.
Solving a 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.

Example 2

Calculate the integral:
.

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

Example 3

Calculate the integral:
.

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 . Because the 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.


.

See also: