## Problem Statement

Find the indefinite integral:

$$
\int \frac{2 x^{3}+11 x^{2}+16 x+10}{(x+2)^{2}\left(x^{2}+2 x+3\right)} d x
$$
## Solution

$$
\int \frac{2 x^{3}+11 x^{2}+16 x+10}{(x+2)^{2}\left(x^{2}+2 x+3\right)} d x=
$$

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

$$
\begin{aligned}
& \frac{2 x^{3}+11 x^{2}+16 x+10}{(x+2)^{2}\left(x^{2}+2 x+3\right)}=\frac{A_{1}}{x+2}+\frac{A_{2}}{(x+2)^{2}}+\frac{B x+C}{x^{2}+2 x+3}= \\
& =\frac{A_{1}(x+2)\left(x^{2}+2 x+3\right)+A_{2}\left(x^{2}+2 x+3\right)+(B x+C)(x+2)^{2}}{(x+2)^{2}\left(x^{2}+2 x+3\right)}= \\
& =\frac{A_{1}\left(x^{3}+4 x^{2}+7 x+6\right)+A_{2}\left(x^{2}+2 x+3\right)+B\left(x^{3}+4 x^{2}+4 x\right)+C\left(x^{2}+4 x+4\right)}{(x+2)^{2}\left(x^{2}+2 x+3\right)}= \\
& =\frac{\left(A_{1}+B\right) x^{3}+\left(4 A_{1}+A_{2}+4 B+C\right) x^{2}+\left(7 A_{1}+2 A_{2}+4 B+4 C\right) x+\left(6 A_{1}+3 A_{2}+4 C\right)}{(x+2)^{2}\left(x^{2}+2 x+3\right)} \\
& \left\{\begin{array}{l}
A_{1}+B=2 \\
4 A_{1}+A_{2}+4 B+C=11 \\
7 A_{1}+2 A_{2}+4 B+4 C=16 \\
6 A_{1}+3 A_{2}+4 C=10
\end{array}\right.
\end{aligned}
$$

Subtract the first equation multiplied by 4 from the second equation:

$$
\left\{\begin{array}{l}
A_{1}+B=2 \\
A_{2}+C=3 \\
7 A_{1}+2 A_{2}+4 B+4 C=16 \\
6 A_{1}+3 A_{2}+4 C=10
\end{array}\right.
$$

Subtract the first equation multiplied by 4 from the third equation:

$$
\left\{\begin{array}{l}
A_{1}+B=2 \\
A_{2}+C=3 \\
3 A_{1}+2 A_{2}+4 C=8 \\
6 A_{1}+3 A_{2}+4 C=10
\end{array}\right.
$$

Subtract the third equation multiplied by 2 from the fourth equation:

$$
\left\{\begin{array}{l}
A_{1}+B=2 \\
A_{2}+C=3 \\
3 A_{1}+2 A_{2}+4 C=8 \\
-A_{2}-4 C=-6
\end{array}\right.
$$

Add the second equation to the fourth equation:

$$
\begin{aligned}
& \left\{\begin{array}{l}
A_{1}+B=2 \\
A_{2}=2 \\
3 A_{1}+2 A_{2}=4 \\
C=1
\end{array}\right. \\
& \left\{\begin{array}{l}
A_{1}+B=2 \\
A_{2}=2 \\
3 A_{1}=0 \\
C=1
\end{array}\right. \\
& \left\{\begin{array}{l}
B=2 \\
A_{2}=2 \\
A_{1}=0 \\
C=1
\end{array}\right. \\
& \frac{2 x^{3}+11 x^{2}+16 x+10}{(x+2)^{2}\left(x^{2}+2 x+3\right)}=\frac{2}{(x+2)^{2}}+\frac{2 x+1}{x^{2}+2 x+3}
\end{aligned}
$$

Then:

$$
\begin{aligned}
& \int \frac{2 x^{3}+11 x^{2}+16 x+10}{(x+2)^{2}\left(x^{2}+2 x+3\right)} d x=\int\left(\frac{2}{(x+2)^{2}}+\frac{2 x+1}{x^{2}+2 x+3}\right) d x= \\
& =\int \frac{2}{(x+2)^{2}} d x+\int \frac{2 x+2-1}{x^{2}+2 x+3} d x= \\
& =-\frac{2}{x+2}+\int \frac{2 x+2}{x^{2}+2 x+3} d x-\int \frac{1}{x^{2}+2 x+3} d x= \\
& =-\frac{2}{x+2}+\int \frac{d\left(x^{2}+2 x+3\right)}{x^{2}+2 x+3}-\int \frac{1}{(x+1)^{2}+2} d x= \\
& =-\frac{2}{x+2}+\ln \left(x^{2}+2 x+3\right)-\int \frac{d(x+1)}{(x+1)^{2}+2}= \\
& =-\frac{2}{x+2}+\ln \left(x^{2}+2 x+3\right)-\frac{1}{\sqrt{2}} \cdot \operatorname{arctg}\left(\frac{x+1}{\sqrt{2}}\right)+C
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_\�\� \%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD\%D1\%82\%D0\%B5\%D0\%B3 $\% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+7-6$ "

Categories: Kuznetsov's Problem Book Integrals Problem 7 | Integrals

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## Problem Kuznetsov Integrals $7-7$

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