Answer
$\ln|x-1|-\frac{1}{2}\ln|x^2+9|-\frac{1}{3}\tan^{-1}\left(\frac{x}{3}\right)+C$
Work Step by Step
$\int\frac{10}{(x-1)(x^{2}+9)}dx$ = $\int\left(\frac{A}{x-1}+\frac{Bx+C}{x^2+9}\right)dx$
$10$ = $A(x^2+9)+(Bx+C)(x-1)$
$10$ = $(Ax^2+9A)+(Bx^2-Bx+Cx-C)$
$0$ = $A+B$
$0$ = $-B+C$
$10$ = $9A-C$
$A$ = $1$, $B$ = $-1$, $C$ = $-1$
So
$\int\frac{10}{(x-1)(x^{2}+9)}dx$ = $\int\left(\frac{1}{x-1}-\frac{x+1}{x^2+9}\right)dx$ = $\int\left[\frac{1}{x-1}-\frac{x}{x^2+9}-\frac{1}{x^2+9}\right]dx$ = $\ln|x-1|-\frac{1}{2}\ln|x^2+9|-\frac{1}{3}\tan^{-1}\left(\frac{x}{3}\right)+C$