Evaluating $\int_0^\infty \frac{\log (1+x)}{1+x^2}dx$ Can this integral be solved with contour integral or by some application of residue theorem?
$$\int_0^\infty \frac{\log (1+x)}{1+x^2}dx = \frac{\pi}{4}\log 2 + \text{Catalan constant}$$
It has two poles at $\pm i$ and  branch point of $-1$ while the integral is to be evaluated from $0\to \infty$. How to get $\text{Catalan Constant}$? Please give some hints.

Following the same approach in this answer, we have
\begin{gather*}
 \int_0^\infty\frac{\ln^a(1+x)}{1+x^2}\mathrm{d}x=\int_0^\infty\frac{\ln^a\left(\frac{1+y}{y}\right)}{1+y^2}\mathrm{d}y\\
 \overset{\frac{y}{1+y}=x}{=}(-1)^a\int_0^1\frac{\ln^a(x)}{x^2+(1-x)^2}\mathrm{d}x\\
\left\{\text{write $\frac{1}{x^2+(1-x)^2}=\mathfrak{J} \frac{1+i}{1-(1+i)x}$}\right\}\\
 =(-1)^a \mathfrak{J} \int_0^1\frac{(1+i)\ln^a(x)}{1-(1+i)x}\mathrm{d}x\\
=a!\ \mathfrak{J}\{\operatorname{Li}_{a+1}(1+i)\} 
 \end{gather*}
where the last step follows from using the integral form of the polylogarithm function:
$$\operatorname{Li}_{a}(z)=\frac{(-1)^{a-1}}{(a-1)!}\int_0^1\frac{z\ln^{a-1}(t)}{1-zt}\mathrm{d}t.$$
