Find : $\int_0^{b}\frac{\ln x}{x^2-1}dx$ where $b>1$ Evaluate : 
$I=\int_0^{b}\frac{\ln x}{x^2-1}dx$ where $b>1$
My attempt : 
I use integral by part : $u=\ln x$ $dv=\frac{1}{x^2-1}$
Then : $u'=\frac{1}{x}$ and $v=\frac{\ln \frac{x-1}{x+1}}{2}$ 
So :
$I=[\frac{\ln x\ln\frac{x-1}{x+1}}{2}]_0^{b}$
$+\int_0^{b}\frac{\ln (1+x)-\ln (x-1)}{2x}dx$
My problem here is limit of integral because $b>1$ ?
And I can't complete the last integral !
 A: $
\begin{align}
\int_0^b\dfrac{\ln x}{x^2-1}~dx&=\int_0^1\dfrac{\ln x}{x^2-1}~dx+\int_1^b\dfrac{\ln x}{x^2-1}~dx
\\&=\dfrac{\pi^2}{8}+\int_1^b\frac{\ln x}{x^2\left(1-\dfrac{1}{x^2}\right)}~dx
\\&=\dfrac{\pi^2}{8}+\sum\limits_{n=0}^\infty\int_1^b\frac{\ln x}{x^{2n+2}}~dx
\\&=\dfrac{\pi^2}{8}-\sum\limits_{n=0}^\infty\int_1^b\ln x~d\left(\dfrac{1}{(2n+1)x^{2n+1}}\right)
\\&=\dfrac{\pi^2}{8}-\left[\dfrac{\ln x}{(2n+1)x^{2n+1}}\right]_1^b+\sum\limits_{n=0}^\infty\int_1^b\dfrac{1}{(2n+1)x^{2n+1}}~d(\ln x)
\\&=\dfrac{\pi^2}{8}-\dfrac{\ln b}{(2n+1)b^{2n+1}}+\sum\limits_{n=0}^\infty\int_1^b\dfrac{1}{(2n+1)x^{2n+2}}~dx
\\&=\dfrac{\pi^2}{8}-\dfrac{\ln b}{(2n+1)b^{2n+1}}-\sum\limits_{n=0}^\infty\left[\dfrac{1}{(2n+1)^2x^{2n+1}}\right]_1^b
\\&=\dfrac{\pi^2}{8}-\dfrac{\ln b}{(2n+1)b^{2n+1}}+\sum\limits_{n=0}^\infty\dfrac{1}{(2n+1)^2}-\sum\limits_{n=0}^\infty\dfrac{1}{(2n+1)^2b^{2n+1}}
\\&=\dfrac{\pi^2}{4}-\dfrac{\ln b}{(2n+1)b^{2n+1}}-\sum\limits_{n=0}^\infty\dfrac{1}{(2n+1)^2b^{2n+1}}
\end{align}
$
