## Problem Statement

Calculate the indefinite integral:

$$
\int \frac{x^{3}+6 x^{2}+13 x+8}{x(x+2)^{3}} d x
$$
## Solution

$$
\int \frac{x^{3}+6 x^{2}+13 x+8}{x(x+2)^{3}} d x=
$$

Decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:

$$
\begin{aligned}
& \frac{x^{3}+6 x^{2}+13 x+8}{x(x+2)^{3}}=\frac{A}{x}+\frac{B_{1}}{x+2}+\frac{B_{2}}{(x+2)^{2}}+\frac{B_{3}}{(x+2)^{3}}= \\
& =\frac{A(x+2)^{3}+B_{1} x(x+2)^{2}+B_{2} x(x+2)+B_{3} x}{x(x+2)^{3}}= \\
& =\frac{A\left(x^{3}+6 x^{2}+12 x+8\right)+B_{1}\left(x^{3}+4 x^{2}+4 x\right)+B_{2}\left(x^{2}+2 x\right)+B_{3} x}{x(x+2)^{3}}= \\
& =\frac{\left(A+B_{1}\right) x^{3}+\left(6 A+4 B_{1}+B_{2}\right) x^{2}+\left(12 A+4 B_{1}+2 B_{2}+B_{3}\right) x+8 A}{x(x+2)^{3}} \\
& \left\{\begin{array}{l}
A+B_{1}=1 \\
6 A+4 B_{1}+B_{2}=6 \\
12 A+4 B_{1}+2 B_{2}+B_{3}=13 \\
8 A=8
\end{array}\right. \\
& \left\{\begin{array}{l}
B_{1}=0 \\
4 B_{1}+B_{2}=0 \\
4 B_{1}+2 B_{2}+B_{3}=1 \\
A=1
\end{array}\right. \\
& \left\{\begin{array}{l}
B_{1}=0 \\
B_{2}=0 \\
2 B_{2}+B_{3}=1 \\
A=1
\end{array}\right.
\end{aligned}
$$

$$
\begin{aligned}
& \left\{\begin{array}{l}
B_{1}=0 \\
B_{2}=0 \\
B_{3}=1 \\
A=1
\end{array}\right. \\
& \frac{x^{3}+6 x^{2}+13 x+8}{x(x+2)^{3}}=\frac{1}{x}+\frac{1}{(x+2)^{3}}
\end{aligned}
$$

Then:

$$
\int \frac{x^{3}+6 x^{2}+13 x+8}{x(x+2)^{3}} d x=\int\left(\frac{1}{x}+\frac{1}{(x+2)^{3}}\right) d x=\ln |x|-\frac{1}{2(x+2)^{2}}+C
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}$ 6-2»

Categories: Kuznetsov's Problem Book Integrals Problem 6 | Integrals

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## Problem Kuznetsov Integrals 6-3

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