How to integrate $$\int_{2}^{\infty} \frac{\pi(x) \ln^2(x^{\sqrt{x}}) \cdot (x^2 + 1)}{(x^2 - 1)^3} \,dx \quad?$$
Wolfram gives the numerical value
$$\int_{2}^{\infty} \frac{\pi x (1 + x^2) \log^2(x^{\sqrt{x}})}{(-1 + x^2)^3} \, dx = 1.46036 $$ My idea
Let $$I = \int_{2}^{\infty} \frac{\pi(x) \ln^2(x^{\sqrt{x}}) (x^2 + 1)}{(x^2 - 1)^3} \, dx$$
The $\ln^2(x^{\sqrt{x}})$ looks kinda annoying , so let's manipulate it so looks more familiar using the property of logarithms we get:
$$\ln^2(x^{\sqrt{x}}) = (\sqrt{x} \ln(x))^2 = x \ln^2(x) \implies I = \int_{2}^{\infty} \frac{\pi(x) x \ln^2(x) (x^2 + 1)}{(x^2 - 1)^3} \, dx$$
Also, notice that $x^2 + 1 = x^2 - 1 + 2$, so we can express $I$ as:
$$I = \int_{2}^{\infty} \left(\frac{\pi(x) x \ln^2(x)}{(x^2 - 1)^2} + \frac{2\pi(x) x \ln^2(x)}{(x^2 - 1)^3}\right) \, dx$$
I will assume that you do mean $\log^2(x)$ instead of $\log(x)$ as you change this halfway through. We can write your integral from $0$ instead of $2$ as $\pi(x)=0$ for $x<2$ and the zero at $x=1$ is of high enough order that the pole there from the denominator becomes removable.
Recall that for $\Re(s)>1$, $$\frac{1}{s}\log\zeta(s)=\int_{0}^{\infty}\frac{\pi(x)}{x(x^s-1)}\,\mathrm{d}x.$$ Taking the second derivative in $s$, we have $$\frac{d^2}{ds^2}\left(\frac{1}{s}\log\zeta(s)\right)=\int_{0}^{\infty}\frac{\pi(x)x^{s-1}(x^s+1)\log^2(x)}{(x^s-1)^3}\,\mathrm{d}x.$$ Taking $s\mapsto 2$, we recover your integral immediately, giving $$I=\frac{1}{4}\log\zeta(2)-\frac{\zeta’(2)}{2\zeta(2)}-\frac{1}{2}\left(\frac{\zeta’(2)}{\zeta(2)}\right)^2+\frac{\zeta’’(2)}{2\zeta(2)}.$$
Using $\zeta(2)=\frac{\pi^2}{6}$ and $\zeta’(2)=\frac{\pi^2}{6}\left(\gamma+\log(2\pi)-12\log(A)\right)$, where $A$ is the Glaisher-Kinkelin constant, we can give the integral as $$I=\frac{1}{4}\log\left(\frac{\pi^2}{6}\right)-\frac{1}{2}\left(\gamma+\log(2\pi)-12\log(A)\right)\left(1+\gamma+\log(2\pi)-12\log(A)\right)+\frac{3}{\pi^2}\zeta’’(2).$$