Inspired By Kishore S Shenoy

Let us consider the function $\displaystyle J(a) = \int_{0}^{\infty} \frac{x^a}{(1+x^2)^2 } dx$ .

We need to obtain $\displaystyle J''(a)]_{a=2} = \int_0^\infty \left( \dfrac{x \ln x} {1+x^2} \right)^2 \, dx$

Substitute $x^2=u$ and the integral turns,

$\displaystyle J(a) = \frac{1}{2} \int_{0}^{\infty} \frac{u^{\frac{a-1}{2}}}{(1+u)^2} du$

Using definition of Beta Function , $\displaystyle \beta(a,b)=\int_{0}^{\infty} \frac{u^{a-1}}{(1+u)^{a+b}}du$ ,

$\displaystyle J(a) = \frac{1}{2} \beta(\frac{a+1}{2},\frac{3-a}{2})$

Differentiating twice with respect to $a$ we get $\displaystyle J''(a) = \frac{1}{8}\beta(\frac{a+1}{2},\frac{3-a}{2}) [ (\psi_{(0)}(\frac{a+1}{2})-\psi_{(0)}(\frac{3-a}{2}))^2 + \psi_{(1)}(\frac{a+1}{2})+\psi_{(1)}(\frac{3-a}{2})]$

Putting $a=2$ and substituting values for $\displaystyle \psi_{(0)}(\frac{3}{2})=-\gamma -2ln2 +2,\psi_{(0)}(\frac{1}{2})=-\gamma-2ln2,\psi_{(1)}(\frac{3}{2}) = \frac{\pi^2}{2} -4 , \psi_{(1)}(\frac{1}{2}) = \frac{\pi^2}{2}$ we get ,

$\large J''(2) = \frac{\pi^3}{16}\implies \boxed{A+B+C=20}$