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?
 A: One may first integrate by parts,
$$
\int \frac{x\ln (x)}{(x^2-1)^{3/2}} \:dx=-\frac{\ln (x)}{(x^2-1)^{1/2}} +\int \frac1{x(x^2-1)^{1/2}} \:dx
$$ then using the change of variable $u=\dfrac1{(x^2-1)^{1/2}}$, one gets
$$
\int \frac1{x(x^2-1)^{1/2}} \:dx=-\int \frac1{1+u^2} \:du=-\arctan \left(\dfrac1{(x^2-1)^{1/2}}\right),
$$ finally

$$
\int \frac{x\ln (x)}{(x^2-1)^{3/2}} \:dx=-\frac{\ln (x)}{(x^2-1)^{1/2}}-\arctan \left(\dfrac1{(x^2-1)^{1/2}}\right)+C.
$$

A: 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...)
$$\int\frac{dx}{x\sqrt{x^2-1}}=-\arctan\frac1{\sqrt{x^2-1}}+C$$
A: Using the substitution $x=\sec(\theta)$, we get
$$
\begin{align}
\int\frac{x\log(x)}{\left(x^2-1\right)^{3/2}}\,\mathrm{d}x
&=-\int\log(x)\,\mathrm{d}\left(x^2-1\right)^{-1/2}\\
&=-\frac{\log(x)}{\sqrt{x^2-1}}+\int\frac{\mathrm{d}x}{x\sqrt{x^2-1}}\\
&=-\frac{\log(x)}{\sqrt{x^2-1}}+\int\mathrm{d}\theta\\
&=-\frac{\log(x)}{\sqrt{x^2-1}}+\sec^{-1}(x)+C
\end{align}
$$
A: Integrate by parts taking into account that
$$
\frac{d}{dt}\left[\frac{t}{\sqrt{1-t^2}}\right]=\frac{1}{(1-t^2)^{3/2}}.
$$
You will get
$$
\int\ln td\frac{t}{\sqrt{1-t^2}}=\frac{t\ln t}{\sqrt{1-t^2}}-\int\frac{1}{\sqrt{1-t^2}}dt=\frac{t\ln t}{\sqrt{1-t^2}}-\arcsin t.
$$ 
