## Problem Statement

Find the indefinite integral:

$$
\int \frac{3 x^{3}+x+46}{(x-1)^{2}\left(x^{2}+9\right)} d x
$$
## Solution

$$
\int \frac{3 x^{3}+x+46}{(x-1)^{2}\left(x^{2}+9\right)} d x=
$$

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

$$
\begin{aligned}
& \frac{3 x^{3}+x+46}{(x-1)^{2}\left(x^{2}+9\right)}=\frac{A_{1}}{x-1}+\frac{A_{2}}{(x-1)^{2}}+\frac{B x+C}{x^{2}+9}= \\
& =\frac{A_{1}(x-1)\left(x^{2}+9\right)+A_{2}\left(x^{2}+9\right)+(B x+C)(x-1)^{2}}{(x-1)^{2}\left(x^{2}+9\right)}= \\
& =\frac{A_{1}\left(x^{3}-x^{2}+9 x-9\right)+A_{2}\left(x^{2}+9\right)+B\left(x^{3}-2 x^{2}+x\right)+C\left(x^{2}-2 x+1\right)}{(x-1)^{2}\left(x^{2}+9\right)}= \\
& =\frac{\left(A_{1}+B\right) x^{3}+\left(-A_{1}+A_{2}-2 B+C\right) x^{2}+\left(9 A_{1}+B-2 C\right) x+\left(-9 A_{1}+9 A_{2}+C\right)}{(x-1)^{2}\left(x^{2}+9\right)} \\
& \left\{\begin{array}{l}
A_{1}+B=3 \\
-A_{1}+A_{2}-2 B+C=0 \\
9 A_{1}+B-2 C=1 \\
-9 A_{1}+9 A_{2}+C=46
\end{array}\right.
\end{aligned}
$$

Add the second equation multiplied by -9 to the fourth equation:

$$
\begin{aligned}
& \left\{\begin{array}{l}
A_{1}+B=3 \\
-A_{1}+A_{2}-2 B+C=0 \\
9 A_{1}+B-2 C=1 \\
18 B-8 C=46
\end{array}\right. \\
& \left\{\begin{array}{l}
A_{1}+B=3 \\
-A_{1}+A_{2}-2 B+C=0 \\
9 A_{1}+B-2 C=1 \\
9 B-4 C=23
\end{array}\right.
\end{aligned}
$$

Add the third equation multiplied by -2 to the fourth equation:

$$
\left\{\begin{array}{l}
A_{1}+B=3 \\
-A_{1}+A_{2}-2 B+C=0 \\
9 A_{1}+B-2 C=1 \\
-18 A_{1}+7 B=21
\end{array}\right.
$$

Add the first equation multiplied by 18 to the fourth equation:

$$
\begin{aligned}
& \left\{\begin{array}{l}
A_{1}+B=3 \\
-A_{1}+A_{2}-2 B+C=0 \\
9 A_{1}+B-2 C=1 \\
25 B=75
\end{array}\right. \\
& \left\{\begin{array}{l}
A_{1}=0 \\
-A_{1}+A_{2}+C=6 \\
9 A_{1}-2 C=-2 \\
B=3
\end{array}\right. \\
& \left\{\begin{array}{l}
A_{1}=0 \\
A_{2}+C=6 \\
-2 C=-2 \\
B=3
\end{array}\right. \\
& \left\{\begin{array}{l}
A_{1}=0 \\
A_{2}=5 \\
C=1 \\
B=3
\end{array}\right. \\
& \frac{3 x^{3}+x+46}{(x-1)^{2}\left(x^{2}+9\right)}=\frac{5}{(x-1)^{2}}+\frac{3 x+1}{x^{2}+9}
\end{aligned}
$$

Then:

$$
\begin{aligned}
& \int \frac{3 x^{3}+x+46}{(x-1)^{2}\left(x^{2}+9\right)} d x=\int\left(\frac{5}{(x-1)^{2}}+\frac{3 x+1}{x^{2}+9}\right) d x= \\
& =\int \frac{5}{(x-1)^{2}} d x+\int \frac{3 x+1}{x^{2}+9} d x=-\frac{5}{x-1}+\int \frac{3 x}{x^{2}+9} d x+\int \frac{1}{x^{2}+9} d x= \\
& =-\frac{5}{x-1}+\frac{3}{2} \cdot \int \frac{d\left(x^{2}+9\right)}{x^{2}+9}+\frac{1}{3} \cdot \operatorname{arctg}\left(\frac{x}{3}\right)=-\frac{5}{x-1}+\frac{3}{2} \cdot \ln \left(x^{2}+9\right)+\frac{1}{3} \cdot \operatorname{arctg}\left(\frac{x}{3}\right)+C
\end{aligned}
$$

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 $\% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+7-14$ »

Categories: Kuznetsov's Integral Problems 7 | Integrals

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## Problem Kuznetsov Integrals $7-15$

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