Convergent integral $\int_0^{+\infty}\frac{\ln (1+x^2)}{1+x^2}dx=\pi \ln(2)$

Question

I've been stuck for a while now, as I use the suggested formula with its respective theorems, but I arrive at a divergent integral.

Prove that $$\int_0^{+\infty}\dfrac{\ln (1+x^2)}{1+x^2}dx=\pi \ln(2)$$ Tip: Use Leibniz's formula with $$F(\lambda)=\int_0^{+\infty}\dfrac{\ln (1+\lambda x^2)}{1+x^2}dx, \lambda \geq 0$$

Answer 1

Note that $F(0)=0$ and that\begin{align}F'(\lambda)&=\int_0^{+\infty}\frac\partial{\partial\lambda}\frac{\ln(1+\lambda x^2)}{1+x^2}\,\mathrm dx\\&=\int_0^{+\infty}\frac1{\lambda-1}\frac1{x^2+1}-\frac1{\lambda-1}\frac1{\lambda x^2+1}\,\mathrm dx\\&=\frac\pi{2\sqrt{\lambda}\left(\sqrt\lambda+1\right)}\end{align}Therefore $F(\lambda)=\pi\log\left(\sqrt\lambda+1\right)$ and, in particular, $F(1)=\pi\log2$.