## Problem Statement

Calculate the indefinite integral:

$$
\int \frac{3 x^{5}-12 x^{3}-7}{x^{2}+2 x} d x
$$
## Solution

$$
\int \frac{3 x^{5}-12 x^{3}-7}{x^{2}+2 x} d x=
$$

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

$$
\begin{aligned}
& \begin{array}{ll}
3 x^{5}-12 x^{3}-7 \\
\frac{3 x^{5}+6 x^{4}}{-6 x^{4}}-12 x^{3}-7
\end{array} \quad \frac{\mid x^{2}+2 x}{3 x^{3}-6 x^{2}} \\
& \frac{-6 x^{4}-12 x^{3}}{-7}
\end{aligned}
$$

We get:

$$
=\int\left(3 x^{3}-6 x^{2}\right) d x-\int \frac{7}{x^{2}+2 x} d x=
$$

Let's use the method of undetermined coefficients:

$$
\begin{aligned}
& \frac{7}{x^{2}+2 x}=\frac{7}{x(x+2)}=\frac{A}{x}+\frac{B}{x+2}=\frac{A(x+2)+B x}{x(x+2)}=\frac{(A+B) x+2 A}{x(x+2)} \\
& \left\{\begin{array}{l}
A+B=0 \\
2 A=7
\end{array}\right. \\
& \left\{\begin{array}{l}
B=-\frac{7}{2} \\
A=\frac{7}{2}
\end{array}\right. \\
& \frac{7}{x^{2}+2 x}=\frac{7}{2 x}-\frac{7}{2(x+2)}
\end{aligned}
$$

Then we get:

$$
=\int\left(3 x^{3}-6 x^{2}\right) d x-\int\left(\frac{7}{2 x}-\frac{7}{2(x+2)}\right) d x=
$$

$$
=\frac{3 x^{4}}{4}-2 x^{3}-\frac{7}{2} \cdot \ln |x|+\frac{7}{2} \cdot \ln |x+2|+C
$$

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}+5-18 »

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

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## Problem Kuznetsov Integrals 5-19

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