## Problem Statement

Find the indefinite integral:

$$
\int \frac{4 x^{3}+24 x^{2}+20 x-28}{(x+3)^{2}\left(x^{2}+2 x+2\right)} d x
$$
## Solution

$$
\int \frac{4 x^{3}+24 x^{2}+20 x-28}{(x+3)^{2}\left(x^{2}+2 x+2\right)} d x=
$$

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

$$
\begin{aligned}
& \frac{4 x^{3}+24 x^{2}+20 x-28}{(x+3)^{2}\left(x^{2}+2 x+2\right)}=\frac{A_{1}}{x+3}+\frac{A_{2}}{(x+3)^{2}}+\frac{B x+C}{x^{2}+2 x+2}= \\
& =\frac{A_{1}(x+3)\left(x^{2}+2 x+2\right)+A_{2}\left(x^{2}+2 x+2\right)+(B x+C)(x+3)^{2}}{(x+3)^{2}\left(x^{2}+2 x+2\right)}= \\
& =\frac{A_{1}\left(x^{3}+5 x^{2}+8 x+6\right)+A_{2}\left(x^{2}+2 x+2\right)+B\left(x^{3}+6 x^{2}+9 x\right)+C\left(x^{2}+6 x+9\right)}{(x+3)^{2}\left(x^{2}+2 x+2\right)}= \\
& =\frac{\left(A_{1}+B\right) x^{3}+\left(5 A_{1}+A_{2}+6 B+C\right) x^{2}+\left(8 A_{1}+2 A_{2}+9 B+6 C\right) x+\left(6 A_{1}+2 A_{2}+9 C\right)}{(x+3)^{2}\left(x^{2}+2 x+2\right)} \\
& \left\{\begin{array}{l}
A_{1}+B=4 \\
5 A_{1}+A_{2}+6 B+C=24 \\
8 A_{1}+2 A_{2}+9 B+6 C=20 \\
6 A_{1}+2 A_{2}+9 C=-28
\end{array}\right.
\end{aligned}
$$

Subtract the first equation multiplied by 6 from the second equation:

$$
\left\{\begin{array}{l}
A_{1}+B=4 \\
-A_{1}+A_{2}+C=0 \\
8 A_{1}+2 A_{2}+9 B+6 C=20 \\
6 A_{1}+2 A_{2}+9 C=-28
\end{array}\right.
$$

Subtract the first equation multiplied by 9 from the third equation:

$$
\left\{\begin{array}{l}
A_{1}+B=4 \\
-A_{1}+A_{2}+C=0 \\
-A_{1}+2 A_{2}+6 C=-16 \\
6 A_{1}+2 A_{2}+9 C=-28
\end{array}\right.
$$

Subtract the second equation from the third equation:

$$
\left\{\begin{array}{l}
A_{1}+B=4 \\
-A_{1}+A_{2}+C=0 \\
A_{2}+5 C=-16 \\
6 A_{1}+2 A_{2}+9 C=-28
\end{array}\right.
$$

Add the second equation multiplied by 6 to the fourth equation:

$$
\left\{\begin{array}{l}
A_{1}+B=4 \\
-A_{1}+A_{2}+C=0 \\
A_{2}+5 C=-16 \\
8 A_{2}+15 C=-28
\end{array}\right.
$$

Subtract the third equation multiplied by 3 from the fourth equation:

$$
\begin{aligned}
& \left\{\begin{array}{l}
A_{1}+B=4 \\
-A_{1}+C=-4 \\
5 C=-20 \\
A_{2}=4
\end{array}\right. \\
& \left\{\begin{array}{l}
A_{1}+B=4 \\
A_{1}=0 \\
C=-4 \\
A_{2}=4
\end{array}\right. \\
& \left\{\begin{array}{l}
B=4 \\
A_{1}=0 \\
C=-4 \\
A_{2}=4
\end{array}\right. \\
& \frac{4 x^{3}+24 x^{2}+20 x-28}{(x+3)^{2}\left(x^{2}+2 x+2\right)}=\frac{4}{(x+3)^{2}}+\frac{4 x-4}{x^{2}+2 x+2}
\end{aligned}
$$

Then:

$$
\begin{aligned}
& \int \frac{4 x^{3}+24 x^{2}+20 x-28}{(x+3)^{2}\left(x^{2}+2 x+2\right)} d x=\int\left(\frac{4}{(x+3)^{2}}+\frac{4 x-4}{x^{2}+2 x+2}\right) d x= \\
& =\int \frac{4}{(x+3)^{2}} d x+2 \cdot \int \frac{2 x+2-4}{x^{2}+2 x+2} d x= \\
& =-\frac{4}{x+3}+2 \cdot \int \frac{2 x+2}{x^{2}+2 x+2} d x-2 \cdot \int \frac{4}{x^{2}+2 x+2} d x= \\
& =-\frac{4}{x+3}+2 \cdot \int \frac{d\left(x^{2}+2 x+2\right)}{x^{2}+2 x+2}-8 \cdot \int \frac{1}{(x+1)^{2}+1} d x= \\
& =-\frac{4}{x+3}+2 \cdot \ln \left(x^{2}+2 x+2\right)-8 \cdot \operatorname{arctg}(x+1)+C
\end{aligned}
$$

Categories: Kuznetsov Integral Problems Problem 7 | Integrals

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## Problem Kuznetsov Integrals 7-16

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