## Problem Statement

Calculate the indefinite integral:

$$
\int \frac{-x^{5}+25 x^{3}+1}{x^{2}+5 x} d x
$$
## Solution

$$
\int \frac{-x^{5}+25 x^{3}+1}{x^{2}+5 x} d x=
$$

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

$$
\begin{aligned}
& \begin{array}{ll}
-x^{5}+25 x^{3}+1 & \frac{\mid x^{2}+5 x}{-x^{3}+5 x^{2}} \\
\frac{-x^{5}-5 x^{4}}{5 x^{4}+25 x^{3}+1} &
\end{array} \\
& 5 x^{4}+25 x^{3} \\
& 1
\end{aligned}
$$

We get:

$$
=\int\left(-x^{3}+5 x^{2}\right) d x+\int \frac{1}{x^{2}+5 x} d x=
$$

Let's use the method of undetermined coefficients:

$$
\begin{aligned}
& \frac{1}{x^{2}+5 x}=\frac{1}{x(x+5)}=\frac{A}{x}+\frac{B}{x+5}=\frac{A(x+5)+B x}{x(x+5)}=\frac{(A+B) x+5 A}{x(x+5)} \\
& \left\{\begin{array}{l}
A+B=0 \\
5 A=1
\end{array}\right. \\
& \left\{\begin{array}{l}
B=-\frac{1}{5} \\
A=\frac{1}{5}
\end{array}\right. \\
& \frac{1}{x^{2}+5 x}=\frac{1}{5 x}-\frac{1}{5(x+5)}
\end{aligned}
$$

Then we get:

$$
=\int\left(-x^{3}+5 x^{2}\right) d x+\int\left(\frac{1}{5 x}-\frac{1}{5(x+5)}\right) d x=
$$

$$
=-\frac{x^{4}}{4}+\frac{5 x^{3}}{3}+\frac{1}{5} \cdot \ln |x|-\frac{1}{5} \cdot \ln |x+5|+C
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+5-20$ »

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

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## Problem Kuznetsov Integrals $5-21$

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