## Problem Statement

Calculate the indefinite integral:

$$
\int \frac{x^{3}+6 x^{2}+14 x+10}{(x+1)(x+2)^{3}} d x
$$
## Solution

$$
\int \frac{x^{3}+6 x^{2}+14 x+10}{(x+1)(x+2)^{3}} d x=
$$

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

$$
\begin{aligned}
& \frac{x^{3}+6 x^{2}+14 x+10}{(x+1)(x+2)^{3}}=\frac{A}{x+1}+\frac{B_{1}}{x+2}+\frac{B_{2}}{(x+2)^{2}}+\frac{B_{3}}{(x+2)^{3}}= \\
& =\frac{A(x+2)^{3}+B_{1}(x+1)(x+2)^{2}+B_{2}(x+1)(x+2)+B_{3}(x+1)}{(x+1)(x+2)^{3}}= \\
& =\frac{A\left(x^{3}+6 x^{2}+12 x+8\right)+B_{1}\left(x^{3}+5 x^{2}+8 x+4\right)+B_{2}\left(x^{2}+3 x+2\right)+B_{3}(x+}{(x+1)(x+2)^{3}} \\
& =\frac{\left(A+B_{1}\right) x^{3}+\left(6 A+5 B_{1}+B_{2}\right) x^{2}+\left(12 A+8 B_{1}+3 B_{2}+B_{3}\right) x+\left(8 A+4 B_{1}+\right.}{(x+1)(x+2)^{3}} \\
& \left\{\begin{array}{l}
A+B_{1}=1 \\
6 A+5 B_{1}+B_{2}=6 \\
12 A+8 B_{1}+3 B_{2}+B_{3}=14 \\
8 A+4 B_{1}+2 B_{2}+B_{3}=10
\end{array}\right.
\end{aligned}
$$

Subtract the fourth equation from the third:

$$
\left\{\begin{array}{l}
A+B_{1}=1 \\
6 A+5 B_{1}+B_{2}=6 \\
4 A+4 B_{1}+B_{2}=4 \\
8 A+4 B_{1}+2 B_{2}+B_{3}=10
\end{array}\right.
$$

Add the first equation multiplied by -4 to the third equation:

$$
\begin{aligned}
& \left\{\begin{array}{l}
A+B_{1}=1 \\
6 A+5 B_{1}+B_{2}=6 \\
B_{2}=0 \\
8 A+4 B_{1}+2 B_{2}+B_{3}=10
\end{array}\right. \\
& \left\{\begin{array}{l}
A+B_{1}=1 \\
6 A+5 B_{1}=6 \\
B_{2}=0 \\
8 A+4 B_{1}+B_{3}=10
\end{array}\right.
\end{aligned}
$$

Add the first equation multiplied by -5 to the second equation:

$$
\begin{aligned}
& \left\{\begin{array}{l}
A+B_{1}=1 \\
A=1 \\
B_{2}=0 \\
8 A+4 B_{1}+B_{3}=10
\end{array}\right. \\
& \left\{\begin{array}{l}
B_{1}=0 \\
A=1 \\
B_{2}=0 \\
B_{3}=2
\end{array}\right. \\
& \frac{x^{3}+6 x^{2}+14 x+10}{(x+1)(x+2)^{3}}=\frac{1}{x+1}+\frac{2}{(x+2)^{3}}
\end{aligned}
$$

Then:

$$
\int \frac{x^{3}+6 x^{2}+14 x+10}{(x+1)(x+2)^{3}} d x=\int\left(\frac{1}{x+1}+\frac{2}{(x+2)^{3}}\right) d x=\ln |x+1|-\frac{1}{(x+2)^{2}}+C
$$

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%D1%82 $\% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} 6-4$ »

Categories: Kuznetsov Integral Problems 6 | Integrals

- Last modified: 14:52, 7 April 2009.
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## Problem Kuznetsov Integrals 6-5

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