Finding $\int^{\infty}_{0}\frac{\ln^2(x)}{(1-x^2)^2} dx$ 
Calculate  $$\int^{\infty}_{0}\frac{\ln^2(x)}{(1-x^2)^2}dx$$

I have tried to put $\displaystyle x=\frac{1}{t}$ and $\displaystyle dx=-\frac{1}{t^2}dt$
$$ \int^{\infty}_{0}\frac{t^2\ln^2(t)}{(t^2-1)^2}dt$$
$$\frac{1}{2}\int^{\infty}_{0}t\ln^2(t)\frac{2t}{(t^2-1)^2}dt$$
$$ \frac{1}{2}\bigg[-t\ln^2(t)\frac{1}{t^2-1}+\int^{\infty}_{0}\frac{\ln^2(t)}{t^2-1}+2\int^{\infty}_{0}\frac{\ln(t)}{t^2-1}dt\bigg]$$
How can I solve it?

You're definetly on the right track with that substitution of $x=\frac1t$ 
Basically we have: $$I=\int^{\infty}_{0}\frac{\ln^2(x)}{(1-x^2)^2}dx=\int_0^\infty \frac{x^2\ln^2 x}{(1-x^2)^2}dx$$
Now what if we add them up?
$$2I=\int_0^\infty \ln^2 x \frac{1+x^2}{(1-x^2)^2}dx$$
If you don't know how to deal easily with the integral  $$\int \frac{1+x^2}{(1-x^2)^2}dx=\frac{x}{1-x^2}+C$$
I recommend you to take a look here.
Anyway we have, integrating by parts: 
$$2I= \underbrace{\frac{x}{1-x^2}\ln^2x \bigg|_0^\infty}_{=0} +2\underbrace{\int_0^\infty \frac{\ln x}{x^2-1}dx}_{\large =\frac{\pi^2}{4}}$$
$$\Rightarrow 2I= 2\cdot \frac{\pi^2}{4} \Rightarrow I=\frac{\pi^2}{4}$$
For the last integral see here for example.
