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

I am attempting to evaluate the following integral: $$\int_0^\infty \frac{(ln(x))^2 }{1+x^2}dx$$
Using the substitution $x=e^u$ and $dx=e^u du$, I get: $$\int_{-\infty}^\infty \frac{u^2}{e^{-u} + e^u} du$$