## Problem Statement

Calculate the indefinite integral:

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

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

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

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

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

Then:

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

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%D1%82%D0%B5%D0%B3%D1%80%D0%B0%D0%BB%D1%8B_%D0%97%D0%B0%D0%B4%D0%B0%D1%87%D0%B0_6-14"

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

- Last modified: 08:30, April 9, 2009.
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## Problem Kuznetsov Integrals 6-15

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