$\displaystyle \displaystyle\frac{15}{x^2-1}=\frac{A}{x-1}+\frac{B}{x+1}=\frac{x+1}{x+1}\ \frac{A}{x-1}+\frac{x-1}{x-1}\ \frac{B}{x+1}$
$\displaystyle =\displaystyle\frac{A(x+1)+B(x-1)}{(x-1)(x+1)}=\frac{(A+B)x+(A-B)}{x^2-1}$
Therefore $\displaystyle 15=15+(0)x=(A-B)+(A+B)x$
which gives us 2 simultaneous equations for A and B
$\displaystyle A+B=0\Rightarrow\ A=-B$
$\displaystyle A-B=15\Rightarrow\ A+A=15$
Get the idea?
factor the constant 15 out from the integral ...
$\displaystyle \displaystyle \frac{1}{x^2-1} = \frac{A}{x-1} + \frac{B}{x+1}$
$\displaystyle \displaystyle \frac{1}{x^2-1} = \frac{A(x+1)}{(x-1)(x+1)} + \frac{B(x-1)}{(x-1)(x+1)}$
$\displaystyle 1 = A(x+1) + B(x-1)$
let $\displaystyle x = 1$ ... $\displaystyle \displaystyle A = \frac{1}{2}$
let $\displaystyle x = -1$ ... $\displaystyle \displaystyle B = -\frac{1}{2}$
$\displaystyle \displaystyle \frac{1}{x^2-1} = \frac{1}{2(x-1)} - \frac{1}{2(x+1)}$
$\displaystyle \displaystyle \frac{15}{2} \int \frac{1}{x-1} - \frac{1}{x+1} \, dx
$
now integrate.