A calculus problem by Hamza A

Rewriting using partial fractions as $\displaystyle 2\int_{0}^{1}\frac{\ln{x}}{2(x-1)}dx+\frac{-\ln{x}}{2(x+1)}dx$

$\displaystyle\int_{0}^{1}\frac{\ln{x}}{(x+1)}dx = [ \ln(x) \ln(x+1)]^1_0 - \int_{0}^{1}\frac{\ln(x+1)}{x}dx = 0 - \frac{\pi^2}{12}$

$\displaystyle\int_{0}^{1}\frac{\ln{x}}{(x-1)}dx = \frac{\pi^2}{6}$ , both by expanding and integrating the power series for $\ln(x)$ about $x=1$

$\displaystyle\frac{\pi^2}{6}+\frac{\pi^2}{12} = \boxed{\frac{\pi^2}{4}}$

$I=-\int_0^1 \dfrac{\ln(x^2)}{1-x^2} dx= -\int_0^1 \sum_{n=0}^\infty x^{2n} \ln{x^2}dx=-\sum_{n=0}^\infty\int_0^1 x^{2n} \ln{x^2}dx\\ =-\sum_{n=0}^\infty\dfrac{d}{dn} \int_0^1 x^{2n}dx=\sum_{n=0}^\infty-\dfrac{d}{dn} \dfrac{1}{2n+1}=\sum_{n=0}^\infty \dfrac{2}{(2n+1)^2}\\=\sum_{n=1}^\infty \dfrac{2}{n^2}-\sum_{n=1}^\infty \dfrac{2}{(2n)^2}=\dfrac{2\pi^2}{6}-\dfrac{\pi^2}{12}=\dfrac{\pi^2}{4}$

Since the integral is from $0$ to $1$ , we must have $x≤1$ . After using partial fractions, the integral splits into two integrals,each of which could be written as an infinite geometric series and which when calculated, give $\zeta(2)$ and $\eta(2)$ respectively. So the answer is $\zeta(2)$ + $\eta(2)$ = $\zeta(2)(1+1-2^{-1})$ = $\frac{\pi^2}{4}$ . Hence answer is $2*4$ = $\boxed{8}$

(I'm not actually showing that the two integrals obtained after using partial fractions evaluate to $\zeta(2)$ and $\eta(2)$ because there are many steps involved (including LHopital's rule). But the answer was indeed having this surprisingly closed expression in terms of zeta and eta). :)