Answer
$$-\frac{1}{5} \ln |x-1|-\frac{1}{x-1}+\frac{1}{10} \ln \left|x^{2}+9\right|-\frac{4}{15} \tan ^{-1}\left(\frac{x}{3}\right)+C$$
Work Step by Step
Given $$\int \frac{10 d x}{(x-1)^{2}\left(x^{2}+9\right)} $$
Since \begin{aligned}
\frac{10}{(x-1)^{2}\left(x^{2}+9\right)}&=\frac{A}{x-1}+\frac{B}{(x-1)^{2}}+\frac{C x+D}{x^{2}+9}\\
&= \frac{A(x-1)\left(x^{2}+9\right)+B\left(x^{2}+9\right)+(C x+D)(x-1)^{2}}{(x-1)^{2}\left(x^{2}+9\right)}\\
10&=A(x-1)\left(x^{2}+9\right)+B\left(x^{2}+9\right)+(C x+D)(x-1)^{2}
\end{aligned}
Then$$
\text{at } x =1 \ \ \ \ \to B=1 $$
By comparing the coefficients, we get
\begin{aligned}
A+C &=0 \ \ \ \ \text{ coefficient of } x^3\\
1-A-2 C+D &=0\ \ \text{ coefficient of } x^2\\
9 A+C-2 D &=0\ \ \text{ coefficient of } x \\
9-9 A+D &=10\ \ \ \ \text{ coefficient of } x^0\\
\end{aligned}
Then
$$A=\frac{-1}{5},\ \ \ \ C=\frac{1}{5},\ \ \ \ D=\frac{-1}{5} $$
Hence
\begin{aligned}
\int \frac{10 d x}{(x-1)^{2}\left(x^{2}+9\right)} &=-\frac{1}{5} \int \frac{d x}{x-1}+\int \frac{d x}{(x-1)^{2}}+\frac{1}{5} \int \frac{x d x}{x^{2}+9}-\frac{4}{5} \int \frac{d x}{x^{2}+9} \\
&=-\frac{1}{5} \ln |x-1|-\frac{1}{x-1}+\frac{1}{10} \ln \left|x^{2}+9\right|-\frac{4}{15} \tan ^{-1}\left(\frac{x}{3}\right)+C
\end{aligned}
