let us suppose that we are asked to integrate following rational function
Integral
$\int \frac{1} {(x-4)^2-15} dx$
for integrate this function, let us denote $u=x-4$, from where we will get
$du=dx$
so original function will be same as $\int \frac{1} {u^2-15} du$
now we can write
$ \frac{1} {u^2-15}=\frac {A}{u-\sqrt{15}}+\frac {B}{u+\sqrt{15}} $
from where we will get
$ \frac{1} {u^2-15}=\frac {0.5}{u-\sqrt{15}}-\frac {0.5}{u+\sqrt{15}} $
so that integral from each part will be
$ 0.5*log(u-\sqrt{15})-0.5*log(u+\sqrt{15})$
after inserting of $u=x-4$ we will get
$ 0.5*log(x-4-\sqrt{15})-0.5*log(x+\sqrt{15}-4)+C$
i think everything is correct right? am i missing anything ?
EDDIT :
aaaa i get point what i am missing, there must be $\sqrt{15}$ in denominator, so that $ \frac{0.5}{\sqrt{15}}*log(x-4-\sqrt{15})- \frac{0.5}{\sqrt{15}}*log(x+\sqrt{15}-4)+C$
because after equality of
$ \frac{1} {u^2-15}=\frac {A}{u-\sqrt{15}}+\frac {B}{u+\sqrt{15}} $
we will get
$u*(A+B)=0 $
and from second equality
$\sqrt{15}*(A-B)=1$
from where
$A-B=\frac{1} {\sqrt{15}}$