Answer
$\displaystyle \ln|x-1|-\frac{1}{2}\ln(x^{2}+9)-\frac{1}{3} \arctan(\frac{x}{3})+C$
Work Step by Step
$ I=\displaystyle \int\frac{10}{(x-1)(x^{2}+9)}dx$
$\displaystyle \frac{10}{(x-1)(x^{2}+9)}=\frac{A}{x-1}+\frac{Bx+C}{x^{2}+9}$
$10=A(x^{2}+9)+(Bx+C)(x-1)$
$10=Ax^{2}+9A+Bx^{2}-Bx+Cx-C$
$10=x^{2}(A+B)+x(-B+C)+(9A-C)$
$\left\{\begin{array}{lllll}
A & +B & & =0 & \Rightarrow A=-B\\
& -B & +C & =0 & \Rightarrow C=B=-A\\
9A & & -C & =10 &
\end{array}\right.$
Eq 3$\Rightarrow$9A+A=10 $\Rightarrow A=1,B=C=-1$
$\displaystyle \frac{10}{(x-1)(x^{2}+9)}=\frac{1}{x-1}+\frac{-x-1}{x^{2}+9}=\frac{1}{x-1}-\frac{x+1}{x^{2}+9}$
$=\displaystyle \frac{1}{x-1}-\frac{1}{2}\cdot\frac{2x}{x^{2}+9}-\frac{1}{x^{2}+3^2}$
$I=\displaystyle \ln|x-1|-\frac{1}{2}\ln(x^{2}+9)-\frac{1}{3} \arctan(\frac{x}{3})+C$
