## Problem Statement

Calculate the indefinite integral:

$$
\int \frac{2 x^{5}-8 x^{3}+3}{x^{2}-2 x} d x
$$
## Solution

$$
\int \frac{2 x^{5}-8 x^{3}+3}{x^{2}-2 x} d x=
$$

Under the integral, we have an improper fraction. Let's separate the integer part:

$$
\begin{aligned}
& \begin{array}{ll}
2 x^{5}-8 x^{3}+3 & \frac{\mid x^{2}-2 x}{2 x^{3}+4 x^{2}} \\
\frac{2 x^{5}-4 x^{4}}{4 x^{4}-8 x^{3}+3}
\end{array} \\
& \frac{4 x^{4}-8 x^{3}}{3}
\end{aligned}
$$

We get:

$$
=\int\left(2 x^{3}+4 x^{2}\right) d x+\int \frac{3}{x^{2}-2 x} d x=
$$

Let's use the method of undetermined coefficients:

$$
\begin{aligned}
& \frac{3}{x^{2}-2 x}=\frac{3}{x(x-2)}=\frac{A}{x}+\frac{B}{x-2}=\frac{A(x-2)+B x}{x(x-2)}=\frac{(A+B) x-2 A}{x(x-2)} \\
& \left\{\begin{array}{l}
A+B=0 \\
-2 A=3
\end{array}\right. \\
& \left\{\begin{array}{l}
B=\frac{3}{2} \\
A=-\frac{3}{2}
\end{array}\right. \\
& \frac{3}{x^{2}-2 x}=-\frac{3}{2 x}+\frac{3}{2(x-2)}
\end{aligned}
$$

Then we get:

$$
=\int\left(2 x^{3}+4 x^{2}\right) d x+\int\left(-\frac{3}{2 x}+\frac{3}{2(x-2)}\right) d x=
$$

$$
=\frac{x^{4}}{2}+\frac{4 x^{3}}{3}-\frac{3}{2} \cdot \ln |x|+\frac{3}{2} \cdot \ln |x-2|+C
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+5-17$ »

Categories: Kuznetsov's Problem Book Integrals Problem 5 | Integrals

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## Problem Kuznetsov Integrals 5-18

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