Find : $\int_0^{b}\frac{\ln x}{x^2-1}dx$ where $b>1$

Question

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 !

Answer 1

$ \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} $