## Problem Statement

Calculate the indefinite integral:

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

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

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

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

We get:

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

We will use the method of undetermined coefficients:

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

Then we get:

$$
=\int\left(x^{3}+x^{2}\right) d x+\int\left(-\frac{1}{x}+\frac{1}{x-1}\right) d x=\frac{x^{4}}{4}+\frac{x^{3}}{3}-\ln |x|+\ln |x-1|+C
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+5-15$ »

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

Ukrainian Banner Network

- Last modified: 12:34, 8 November 2009.
- Content is available under CC-BY-SA 3.0.

Created by GeeTeatoo

## Problem Kuznetsov Integrals 5-16

## Material from PlusPi