## Problem Statement

Calculate the indefinite integral:

$$
\int \frac{-x^{5}+9 x^{3}+4}{x^{2}+3 x} d x
$$
## Solution

$$
\int \frac{-x^{5}+9 x^{3}+4}{x^{2}+3 x} d x=
$$

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

$$
\begin{array}{ll}
-x^{5}+9 x^{3}+4 \\
\frac{-x^{5}-3 x^{4}}{3 x^{4}+9 x^{3}+4} & \frac{\mid x^{2}+3 x}{-x^{3}+3 x^{2}} \\
\frac{3 x^{4}+9 x^{3}}{4} &
\end{array}
$$

We get:

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

Let's use the method of undetermined coefficients:

$$
\begin{aligned}
& \frac{4}{x^{2}+3 x}=\frac{4}{x(x+3)}=\frac{A}{x}+\frac{B}{x+3}=\frac{A(x+3)+B x}{x(x+3)}=\frac{(A+B) x+3 A}{x(x+3)} \\
& \left\{\begin{array}{l}
A+B=0 \\
3 A=4
\end{array}\right. \\
& \left\{\begin{array}{l}
B=-\frac{4}{3} \\
A=\frac{4}{3}
\end{array}\right. \\
& \frac{4}{x^{2}+3 x}=\frac{4}{3 x}-\frac{4}{3(x+3)}
\end{aligned}
$$

Then we get:

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

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

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+5-19$ »

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

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## Problem Kuznetsov Integrals 5-20

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