# Evaluating the definite integral $\int_0^{\infty} \frac{(\ln x)^2}{x^2 + 1} dx$ [duplicate]

I have two integral questions listed below:

$$\int_0^{\infty} \frac{\ln x}{x^2 + 1} dx \qquad (1)$$

$$\int_0^{\infty} \frac{(\ln x)^2}{x^2 + 1} dx \qquad (2)$$

The first one, I've solved it, by separating the integral into integrals over $0$ to $1$ and over $1$ to $\infty$; with substitution, it becomes zero.

But the second one wasn't that easy. Can someone give me help?

Wolfram alpha says the answer is $$\frac{\pi^3}{8}.$$