Let:
\alpha = (2t-1)^{-2}
\Rightarrow p = x(\alpha - x)
I = \int_\limits{0}^\alpha \ln (x) \log_p(e)dx = \int_\limits{0}^\alpha \cfrac{\ln(x)}{\ln(p)}dx
I = \int_\limits{0}^\alpha \cfrac{\ln(x)}{\ln(x)+ \ln(\alpha -x )}dx
Applying property of Definite Integral
I = \int_\limits{0}^\alpha \cfrac{\ln(\alpha - x )}{\ln(x)+ \ln(\alpha -x)} dx
\Rightarrow 2I = \int_\limits{0}^\alpha dx = \alpha
\Rightarrow I = \cfrac{\alpha}{2}
\sum_\limits{t=1}^{\infty}(\int_\limits{0}^\alpha \ln (x) \log_p(e)dx )= \sum_\limits{t=1}^{\infty}\cfrac{\alpha}{2}
\Rightarrow \cfrac{1}{2}\sum_\limits{t=1}^{\infty}\cfrac{1}{(2t-1)^2} = \cfrac{1}{2}\left( \cfrac{1}{1}+ \cfrac{1}{3^2}+ \cfrac{1}{5^2}+ ... \right)= \cfrac{1}{2} \left( \cfrac{\pi ^2}{8}\right) = \cfrac{\pi ^2}{16}
@Tushar_Rathore
