## Problem Statement

Calculate the indefinite integral:

$$
\int \frac{2 x^{3}+6 x^{2}+7 x+2}{x(x+1)^{3}} d x
$$
## Solution

$$
\int \frac{2 x^{3}+6 x^{2}+7 x+2}{x(x+1)^{3}} d x=
$$

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

$$
\begin{aligned}
& \frac{2 x^{3}+6 x^{2}+7 x+2}{x(x+1)^{3}}=\frac{A}{x}+\frac{B_{1}}{x+1}+\frac{B_{2}}{(x+1)^{2}}+\frac{B_{3}}{(x+1)^{3}}= \\
& =\frac{A(x+1)^{3}+B_{1} x(x+1)^{2}+B_{2} x(x+1)+B_{3} x}{x(x+1)^{3}}= \\
& =\frac{A\left(x^{3}+3 x^{2}+3 x+1\right)+B_{1}\left(x^{3}+2 x^{2}+x\right)+B_{2}\left(x^{2}+x\right)+B_{3} x}{x(x+1)^{3}}= \\
& =\frac{\left(A+B_{1}\right) x^{3}+\left(3 A+2 B_{1}+B_{2}\right) x^{2}+\left(3 A+B_{1}+B_{2}+B_{3}\right) x+A}{x(x+2)^{3}} \\
& \left\{\begin{array}{l}
A+B_{1}=2 \\
3 A+2 B_{1}+B_{2}=6 \\
3 A+B_{1}+B_{2}+B_{3}=7 \\
A=2
\end{array}\right. \\
& \left\{\begin{array}{l}
B_{1}=0 \\
2 B_{1}+B_{2}=0 \\
B_{1}+B_{2}+B_{3}=1 \\
A=2
\end{array}\right. \\
& \left\{\begin{array}{l}
B_{1}=0 \\
B_{2}=0 \\
B_{2}+B_{3}=1 \\
A=2
\end{array}\right.
\end{aligned}
$$

$$
\begin{aligned}
& \left\{\begin{array}{l}
B_{1}=0 \\
B_{2}=0 \\
B_{3}=1 \\
A=2
\end{array}\right. \\
& \frac{2 x^{3}+6 x^{2}+7 x+2}{x(x+1)^{3}}=\frac{2}{x}+\frac{1}{(x+1)^{3}}
\end{aligned}
$$

Then:

$$
\int \frac{2 x^{3}+6 x^{2}+7 x+2}{x(x+1)^{3}} d x=\int\left(\frac{2}{x}+\frac{1}{(x+1)^{3}}\right) d x=2 \cdot \ln |x|-\frac{1}{2(x+1)^{2}}+C
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+6-9$ " Categories: Kuznetsov's Problem Book Integrals Problem 6 | Integrals

- Last edited: 18:28, 28 May 2009.
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## Problem Kuznetsov Integrals 6-10

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