Integrate $\int \frac{x\ln (x)}{(x^2-1)^{3/2}} dx$ 
Integrate $$\int \frac{x\ln (x)}{(x^2-1)^{3/2}} dx$$.

My Try:
$$\int \frac{x\ln (x)}{x^3(1-1/x^2)^{3/2}} dx=\int \frac{\frac{1}{x^2}\ln (x)}{(1-\frac{1}{x^2})^{3/2}} dx.$$
Putting $x=\frac{1}{t}$,
I get
$$\int \frac{t^2\ln (\frac{1}{t})}{(1-t^2)^{3/2}(-t^2)} dt=\int \frac{\ln ({t})}{(1-t^2)^{3/2}} dt.$$
After this what to do?

Check that
$$\int\frac1{t^{3/2}}dx=-2t^{-1/2}+C\implies \int\frac{f'(x)\,dx}{f(x)^{3/2}} =-2f(x)^{-1/2}+C$$
and now just observe that $\;2x=(x^2-1)'\;$ , so by parts:$${}$$
$$\begin{cases}&u=\log x,&u'=\cfrac1x\\{}\\&v'=\cfrac x{(x-1)^{3/2}},&v=-\cfrac1{\sqrt{x^2-1}}\end{cases}\;\;\implies$$$${}$$
$$\int\frac{x\log x}{(x^2-1)^{3/2}}dx=-\frac{\log x}{(x^2-1)^{3/2}}+\int\frac{dx}{x\sqrt{x^2-1}}$$
and remember (or prove...)
$$\int\frac{dx}{x\sqrt{x^2-1}}=-\arctan\frac1{\sqrt{x^2-1}}+C$$
