Do you really need "contour" here?

Relevant wiki: Differentiation Under the Integral Sign

Let $F(a)= \displaystyle\int_0^\infty \dfrac {\ln(1+a^2x^2)}{1+x^2} \mathrm{d}x~~,\small{F(0)=0}$

$F'(a)= 2a\int_0^\infty \dfrac {x^2}{(1+a^2x^2)(1+x^2)} \mathrm{d}x$

$=\dfrac{2a}{a^2-1}\int_0^\infty \left[\dfrac {1}{1+x^2}-\dfrac{1}{1+a^2x^2}\right] \mathrm{d}x$

$=\dfrac{2a}{a^2-1}\left[\tan^{-1} x-\dfrac 1a\tan^{-1} ax\right]_0^{\infty}$

$=\dfrac{2a}{(a-1)(a+1)}\left[\dfrac{\pi}2-\dfrac{\pi}{2a}\right]$

$=\dfrac{\pi}{a+1}\implies F(a)=\int\dfrac{\pi\mathrm{d}a}{a+1}$

$\implies F(a)=\pi\ln(a+1)~~~~\small{(F(0)=0)}$

In question we need to evaluate $F(1)$ which comes out to be $\pi\ln(1+1)\approx \boxed{2.178}$ .

Substitute x =tanx in the integral

to get the integral from 0 to pi/2

-2ln(cosx)

Now its a pretty standard integral from 0 to pi/2 of ln(cosx) = -pi/2*log2