Problem 1322

Solution.

$$\frac{6 x^{3}+5 x^{2}+3 x-4}{x^{2}+4}=\frac{6 x\left(x^{2}+4\right)+5 x^{2}-24 x+3 x-4}{x^{2}+4}=$$

we added and subtracted $24 x$ in the numerator and grouped $6 x^{3}+$ $+24 x$, then, after grouping, we factored out the common factor $6 x$. To divide a sum by any expression, it is necessary to divide each term of the sum by this expression. Therefore, continuing the started transformations, we have:

$$=\frac{6 x\left(x^{2}+4\right)}{x^{2}+4}+\frac{5 x^{2}-21 x-4}{x^{2}+4}=6 x+\frac{5 x^{2}-21 x-4}{x^{2}+4}=$$

The remaining fraction is also improper. Continuing the equality and dividing the numerator by the denominator (as on the first step of division), we get:

$$\begin{gathered} =6 x+\frac{5\left(x^{2}+4\right)-21 x-20-4}{x^{2}+4}=6 x+\frac{5\left(x^{2}+4\right)}{x^{2}+4}-\frac{21 x+24}{x^{2}+4}= \\ =6 x+5-\frac{21 x+24}{x^{2}+4} \end{gathered}$$

The remaining fraction is already a proper one, and the division process ends here. Thus,

$$\frac{6 x^{3}+5 x^{2}+3 x-4}{x^{2}+4}=6 x+5-\frac{21 x+24}{x^{2}+4}$$

The above scheme for dividing a polynomial by a polynomial can be conveniently formalized by representing the division algorithm as a long division scheme. Within this scheme, the division considered above looks as follows:

!

$$\begin{aligned} & \frac{5 x^{2}+20}{-21 x-24} \end{aligned}$$

Since integrating a polynomial is not particularly difficult, integrating rational fractions reduces to integrating proper rational fractions (hereafter referred to as a rational fraction).

The main idea of integrating rational fractions is to represent them as a sum of the simplest fractions of the following types:

1) $\frac{A}{x-a}$
2) $\frac{A}{(x-a)^{k}}, k>1(k-$ a positive integer);
3) $\frac{M x+N}{x^{2}+p x+q}$, where $D=p^{2}-4 q1$ (a positive integer);

$a, A, M, q, p$ - constants.

Each of the above simplest fractions is easily integrable: Consider the integration of fractions of types 1 and 2:

$$\begin{aligned} & \int \frac{A}{x-a} d x=A \int \frac{d(x-a)}{x-a}=A \ln |x-a|+C \\ & \int \frac{A}{(x-a)^{k}} d x=A \int(x-a)^{-k} d(x-a)= \\ & \quad=A \frac{(x-a)^{-k+1}}{-k+1}+C=\frac{A}{(1-k)(x-a)^{k-1}}+C . \end{aligned}$$

The calculation of integrals of fractions of type 3 is carried out by completing the square in the quadratic trinomial $x^{2}+p x+q$, that is,

$$x^{2}+p x+q=x^{2}+2 \frac{p}{2} x+\frac{p^{2}}{4}+q-\frac{p^{2}}{4}=\left(x+\frac{p}{2}\right)^{2}+q-\frac{p^{2}}{4}$$

and then introducing a new variable of integration $t$, related to the previous one by the equation $t=x+\frac{p}{2}$. Then $x^{2}+p x+q=t^{2}+a^{2}$, where $a=\sqrt{q-\frac{p^{2}}{4}},\left(q-\frac{p^{2}}{4}>0\right)$ and $d t=d\left(x+\frac{p}{2}\right)=\left(x+\frac{p}{2}\right)^{\prime} d x=d x$.

As a result, we get

$$$$
$\text{M x+N x 2 +p}$ x+q} d $x=\int \frac{M(t-p / 2)+N}{t^{2}+a^{2}}$ d t= $\\ =\int \frac{M t}{t^{2}+a^{2}}$ d t+(N-M p / 2) $\int \frac{d t}{t^{2}+a^{2}}= \\ =\frac{M}{2} \int \frac{d\left(t^{2}+a^{2}\right)}{t^{2}+a^{2}}+(N-M$ p / 2) $\text{1 a arctg t a = = M}$