Let's do some calculus! (51)

Substituting $u=961x^2$ so that $I=\int^{\infty}_0\frac{\ln(1+3844x^2)}{1+961x^2}\ dx=\frac{1}{62}\int^{\infty}_0\frac{\ln(1+4u)}{\sqrt{u}(u+1)}\ dx.$ Then, we consider the following function $F(a)=\int^{\infty}_0\frac{\ln(1+au)}{\sqrt{u}(u+1)}\ dx$ so that $F'(a)=\int^{\infty}_0\frac{\sqrt{u}}{(u+1)(au+1)}\ du,$ which can be obtained by applying Differentiation under the integral sign. Integrating, we see that $F'(a)=\frac{\pi}{a+\sqrt{a}}\Rightarrow F(a)=\int^a_0\frac{\pi}{a+\sqrt{a}}\ da=2\pi\ln(\sqrt{a}+1).$ Substituting back $a=4,$ we have $I=\frac{\pi}{31}\ln 3.$