Complex Integration $\int^\infty_0\frac{(\ln x)^2}{1+x^2}\,dx$ using theory of Residue and Branch Cut

Question

The question asks to compute $$\displaystyle\int^\infty_0\frac{(\ln x)^2}{1+x^2}\,dx$$ but I have no idea how to do that. I have thought about using the following branch cut but doesn't seem to work. Branch Cut Here

Answer 1

Substitute $y = \ln x$ and get $$\displaystyle\int^\infty_0\dfrac{(\ln x)^2}{1+x^2}\,dx = \displaystyle\int^\infty_{-\infty}\dfrac{y^2}{1+e^{2y}}e^y\,dy = \frac{1}{2}\displaystyle\int^\infty_{-\infty}\dfrac{y^2}{\cosh(y)}\,dy = -\frac{1}{2}g''(0)$$ where $g = \mathscr{F}(\frac{1}{\cosh}) = \pi\operatorname{sech}(\frac{\pi}{2}y)$. So $g''(0) = -\pi(\frac{\pi}{2})^2$, which gives
$$\displaystyle\int^\infty_0\dfrac{(\ln x)^2}{1+x^2}\,dx = -\frac{1}{2}g''(0) = (\frac{\pi}{2})^3 = \frac{\pi^3}{8}$$