The simplest fraction of type 2 has the form. Integrating simple fractions

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:

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

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 rational fractions(regular and improper), as well as examples of decomposition of rational fractions 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):

If the integrand is elementary, then apply formulas (1)-(4).
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:

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

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.

As I already noted, in integral calculus there is no convenient formula for integrating a fraction. And therefore, there is a sad trend: the more sophisticated the fraction, the more difficult it is to find its integral. In this regard, you have to resort to various tricks, which I will now tell you about. Prepared readers can immediately take advantage of table of contents:

Method of subsuming the differential sign for simple fractions

Artificial numerator conversion method

Example 1

By the way, the considered integral can also be solved by the change of variable method, denoting , but writing the solution will be much longer.

Example 2

Find indefinite integral. Perform check.

This is an example for independent decision. It should be noted that the variable replacement method will no longer work here.

Attention, important! Examples No. 1, 2 are typical and occur frequently. In particular, such integrals often arise during the solution of other integrals, in particular, when integrating irrational functions (roots).

The considered technique also works in the case if the highest degree of the numerator is greater than the highest degree of the denominator.

Example 3

Find the indefinite integral. Perform check.

We begin to select the numerator.

The algorithm for selecting the numerator is something like this:

1) In the numerator I need to organize , but there . What to do? I put it in brackets and multiply by: .

2) Now I try to open these brackets, what happens? . Hmm... that’s better, but there’s no two in the numerator initially. What to do? You need to multiply by:

3) I open the brackets again: . And here is the first success! It turned out just right! But the problem is that an extra term has appeared. What to do? To prevent the expression from changing, I must add the same to my construction:
. Life has become easier. Is it possible to organize again in the numerator?

4) It is possible. Let's try: . Open the brackets of the second term:
. Sorry, but in the previous step I actually had , not . What to do? You need to multiply the second term by:

5) Again, to check, I open the brackets in the second term:
. Now it's normal: derived from the final construction of point 3! But again there is a small “but”, an extra term has appeared, which means I must add to my expression:

If everything is done correctly, then when we open all the brackets we should get the original numerator of the integrand. We check:
Hood.

Thus:

Ready. In the last term, I used the method of subsuming a function under a differential.

If we find the derivative of the answer and reduce the expression to a common denominator, then we will get exactly the original integrand function. The considered method of decomposition into a sum is nothing more than the reverse action of bringing an expression to a common denominator.

The algorithm for selecting the numerator in such examples is best done in draft form. With some skills it will work mentally. I remember a record-breaking case when I was performing a selection for the 11th power, and the expansion of the numerator took up almost two lines of Verd.

Example 4

Find the indefinite integral. Perform check.

This is an example for you to solve on your own.

Method of subsuming the differential sign for simple fractions

Let's move on to consider the next type of fractions.
, , , (coefficients and are not equal to zero).

In fact, a couple of cases with arcsine and arctangent have already been mentioned in the lesson Variable change method in indefinite integral. Such examples are solved by subsuming the function under the differential sign and further integrating using a table. Here are more typical examples with long and high logarithms:

Example 5

Example 6

Here it is advisable to pick up a table of integrals and see what formulas and How transformation takes place. Note, how and why The squares in these examples are highlighted. In particular, in Example 6 we first need to represent the denominator in the form , then bring it under the differential sign. And all this needs to be done in order to use the standard tabular formula .

Why look, try to solve examples No. 7, 8 yourself, especially since they are quite short:

Example 7

Example 8

Find the indefinite integral:

If you also manage to check these examples, then great respect - your differentiation skills are excellent.

Full square selection method

Integrals of the form (coefficients and are not equal to zero) are solved complete square extraction method, which has already appeared in the lesson Geometric transformations of graphs.

In fact, such integrals reduce to one of the four tabular integrals we just looked at. And this is achieved using familiar abbreviated multiplication formulas:

The formulas are applied precisely in this direction, that is, the idea of the method is to artificially organize the expressions either in the denominator, and then convert them accordingly to either.

Example 9

Find the indefinite integral

This simplest example, in which with the term – unit coefficient(and not some number or minus).

Let's look at the denominator, here the whole matter clearly comes down to chance. Let's start converting the denominator:

Obviously, you need to add 4. And, so that the expression does not change, subtract the same four:

Now you can apply the formula:

After the conversion is complete ALWAYS It is advisable to perform the reverse move: everything is fine, there are no errors.

The final design of the example in question should look something like this:

Ready. Subsuming a “free” complex function under the differential sign: , in principle, could be neglected

Example 10

Find the indefinite integral:

This is an example for you to solve on your own, the answer is at the end of the lesson

Example 11

Find the indefinite integral:

What to do when there is a minus in front? In this case, we need to take the minus out of brackets and arrange the terms in the order we need: . Constant(“two” in this case) don't touch!

Now we add one in parentheses. Analyzing the expression, we come to the conclusion that we need to add one outside the brackets:

Here we get the formula, apply:

ALWAYS We check the draft:
, which was what needed to be checked.

The clean example looks something like this:

Making the task more difficult

Example 12

Find the indefinite integral:

Here the term is no longer a unit coefficient, but a “five”.

(1) If there is a constant at, then we immediately take it out of brackets.

(2) In general, it is always better to move this constant outside the integral so that it does not get in the way.

(3) Obviously, everything will come down to the formula. We need to understand the term, namely, get the “two”

(4) Yeah, . This means that we add to the expression and subtract the same fraction.

(5) Now select a complete square. In the general case, we also need to calculate , but here we have the formula for a long logarithm , and there is no point in performing the action; why will become clear below.

(6) Actually, we can apply the formula , only instead of “X” we have , which does not negate the validity of the table integral. Strictly speaking, one step was missed - before integration, the function should have been subsumed under the differential sign: , but, as I have repeatedly noted, this is often neglected.

(7) In the answer under the root, it is advisable to expand all the brackets back:

Difficult? This is not the most difficult part of integral calculus. Although, the examples under consideration are not so much complex as they require good computing techniques.

Example 13

Find the indefinite integral:

This is an example for you to solve on your own. The answer is at the end of the lesson.

There are integrals with roots in the denominator, which, using a substitution, are reduced to integrals of the type considered; you can read about them in the article Complex integrals, but it is designed for very prepared students.

Subsuming the numerator under the differential sign

This is the final part of the lesson, however, integrals of this type are quite common! If you're tired, maybe it's better to read tomorrow? ;)

The integrals that we will consider are similar to the integrals of the previous paragraph, they have the form: or (coefficients , and are not equal to zero).

That is, in the numerator we have linear function. How to solve such integrals?

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:

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

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.

.