integrate ln(x^2-1)
Do exactly what galactus said but be careful like mr fantastic said!
$\displaystyle ln(x^2- 1)$ is an even function so you really only need to integrate for x> 1, where galactus' method has no problem, and then use the same integral for x< -1.
Or you can go with integration by parts
$\displaystyle \int{\ln(x^2-1)}$
sub. $\displaystyle \ln(x^2-1)=u \Rightarrow \frac{2x dx}{x^2-1}=du; dv = dx \Rightarrow v = x;$
so we have
$\displaystyle x\ln(x^2-1)-2\int{\frac{x^2}{x^2-1}dx} $
$\displaystyle = x\ln(x^2-1)-2\int{\frac{x^2-{\color{red}1}+{\color{red} 1}}{x^2-1}dx}$
$\displaystyle =x\ln(x^2-1)-2\int{\left (1+\frac{1}{x^2-1}\right )dx} =...$
correct me anyone if I did anything wrong.