## Problem Statement

Find the indefinite integral:

$$
\int \frac{x^{3}+9 x^{2}+21 x+21}{(x+3)^{2}\left(x^{2}+3\right)} d x
$$
## Solution

$$
\int \frac{x^{3}+9 x^{2}+21 x+21}{(x+3)^{2}\left(x^{2}+3\right)} d x=
$$

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

$$
\begin{aligned}
& \frac{x^{3}+9 x^{2}+21 x+21}{(x+3)^{2}\left(x^{2}+3\right)}=\frac{A_{1}}{x+3}+\frac{A_{2}}{(x+3)^{2}}+\frac{B x+C}{x^{2}+3}= \\
& =\frac{A_{1}(x+3)\left(x^{2}+3\right)+A_{2}\left(x^{2}+3\right)+(B x+C)(x+3)^{2}}{(x+3)^{2}\left(x^{2}+3\right)}= \\
& =\frac{A_{1}\left(x^{3}+3 x^{2}+3 x+9\right)+A_{2}\left(x^{2}+3\right)+B\left(x^{3}+6 x^{2}+9 x\right)+C\left(x^{2}+6 x+9\right)}{(x+3)^{2}\left(x^{2}+3\right)}= \\
& =\frac{\left(A_{1}+B\right) x^{3}+\left(3 A_{1}+A_{2}+6 B+C\right) x^{2}+\left(3 A_{1}+9 B+6 C\right) x+\left(9 A_{1}+3 A_{2}+9 C\right)}{(x+3)^{2}\left(x^{2}+3\right)} \\
& \left\{\begin{array}{l}
A_{1}+B=1 \\
3 A_{1}+A_{2}+6 B+C=9 \\
3 A_{1}+9 B+6 C=21 \\
9 A_{1}+3 A_{2}+9 C=21
\end{array}\right. \\
& \left\{\begin{array}{l}
B=1-A_{1} \\
3 A_{1}+A_{2}+6-6 A_{1}+C=9 \\
3 A_{1}+9-9 A_{1}+6 C=21 \\
9 A_{1}+3 A_{2}+9 C=21
\end{array}\right. \\
& \left\{\begin{array}{l}
B=1-A_{1} \\
-3 A_{1}+A_{2}+C=3 \\
-6 A_{1}+6 C=12 \\
9 A_{1}+3 A_{2}+9 C=21
\end{array}\right. \\
& \left\{\begin{array}{l}
B=1-A_{1} \\
-3 A_{1}+A_{2}+C=3 \\
C=2+A_{1} \\
9 A_{1}+3 A_{2}+9 C=21
\end{array}\right.
\end{aligned}
$$

$$
\begin{aligned}
& \left\{\begin{array}{l}
B=1-A_{1} \\
-3 A_{1}+A_{2}+2+A_{1}=3 \\
C=2+A_{1} \\
9 A_{1}+3 A_{2}+18+9 A_{1}=21
\end{array}\right. \\
& \left\{\begin{array}{l}
B=1-A_{1} \\
-2 A_{1}+A_{2}=1 \\
C=2+A_{1} \\
18 A_{1}+3 A_{2}=3
\end{array}\right. \\
& \left\{\begin{array}{l}
B=1-A_{1} \\
-2 A_{1}+A_{2}=1 \\
C=2+A_{1} \\
6 A_{1}+A_{2}=1
\end{array}\right. \\
& \left\{\begin{array}{l}
B=1-A_{1} \\
A_{2}=1+2 A_{1} \\
C=2+A_{1} \\
6 A_{1}+1+2 A_{1}=1
\end{array}\right. \\
& \begin{array}{l}
B=1 \\
A_{2}=1 \\
C=2 \\
A_{1}=0
\end{array} \\
& \frac{x^{3}+9 x^{2}+21 x+21}{(x+3)^{2}\left(x^{2}+3\right)}=\frac{1}{(x+3)^{2}}+\frac{x+2}{x^{2}+3}
\end{aligned}
$$

Then

$$
\begin{aligned}
& \int \frac{x^{3}+9 x^{2}+21 x+21}{(x+3)^{2}\left(x^{2}+3\right)} d x=\int\left(\frac{1}{(x+3)^{2}}+\frac{x+2}{x^{2}+3}\right) d x= \\
& =\int \frac{1}{(x+3)^{2}} d x+\frac{1}{2} \cdot \int \frac{2 x+4}{x^{2}+3} d x= \\
& =-\frac{1}{x+3}+\frac{1}{2} \cdot \int \frac{2 x}{x^{2}+3} d x+\frac{1}{2} \cdot \int \frac{4}{x^{2}+3} d x= \\
& =-\frac{1}{x+3}+\frac{1}{2} \cdot \int \frac{d\left(x^{2}+3\right)}{x^{2}+3}+\frac{2}{\sqrt{3}} \cdot \operatorname{arctg} \frac{x}{\sqrt{3}}= \\
& =-\frac{1}{x+3}+\frac{1}{2} \cdot \ln \left(x^{2}+3\right)+\frac{2}{\sqrt{3}} \cdot \operatorname{arctg} \frac{x}{\sqrt{3}}+C
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\% $\% \mathrm{D} 0 \% 9 \mathrm{~A} \% \mathrm{D} 1 \% 83 \% \mathrm{D} 0 \% \mathrm{~B} 7 \% \mathrm{D} 0 \% \mathrm{BD} \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 1 \% 86 \% \mathrm{D} 0 \% \mathrm{BE} \% \mathrm{D} 0 \% \mathrm{~B} 2 \% \mathrm{D} 0 \% 98 \% \mathrm{D} 0 \% \mathrm{BD} \% \mathrm{D} 1 \% 82$ $\% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+7-8$ "

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

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## Problem Kuznetsov Integrals 7-9

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