## Problem Statement

Calculate the indefinite integral:

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

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

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

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

We get:

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

Let's decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:

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

Then we get:

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

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

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

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- Last modified: 17:19, 3 April 2009.
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