## Problem Statement

Calculate the indefinite integral:

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

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

Under the integral is an improper fraction. Let's separate the integer part:

$$
\begin{aligned}
& x^{5}+3 x^{3}-1 \\
& \frac{x^{5}+x^{4}}{-x^{4}+3 x^{3}-1} \\
& \frac{-x^{4}-x^{3}}{4 x^{3}-1} \\
& \frac{4 x^{3}+4 x^{2}}{-4 x^{2}}-1 \\
& \quad \frac{-4 x^{2}-4 x}{4 x}-1
\end{aligned}
$$

We get:

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

Let's use the method of undetermined coefficients:

$$
\begin{aligned}
& \frac{4 x-1}{x^{2}+x}=\frac{4 x-1}{x(x+1)}=\frac{A}{x}+\frac{B}{x+1}=\frac{A(x+1)+B x}{x(x+1)}=\frac{(A+B) x+A}{x(x+1)} \\
& \left\{\begin{array}{l}
A+B=4 \\
A=-1
\end{array}\right. \\
& \left\{\begin{array}{l}
B=5 \\
A=-1
\end{array}\right. \\
& \frac{4 x-1}{x^{2}+x}=-\frac{1}{x}+\frac{5}{x+1}
\end{aligned}
$$

Then we get:

$$
\begin{aligned}
& =\int\left(x^{3}-x^{2}+4 x-4\right) d x+\int\left(-\frac{1}{x}+\frac{5}{x+1}\right) d x= \\
& =\frac{x^{4}}{4}-\frac{x^{3}}{3}+2 x^{2}-4 x-\ln |x|+5 \cdot \ln |x+1|+C
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+5-16$ » Categories: Kuznetsov's Problem Book Integrals Problem 5 | Integrals

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## Problem Kuznetsov Integrals 5-17

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