Integral?

$\text{}\left[\text{Int}(a)=\int_0^t \frac{x^a}{1-x^2} \, dx=\int_0^t \left(\sum _{n=0}^{\infty } x^{a+2 n}\right) \, dx=\sum _{n=0}^{\infty } \int_0^t x^{a+2 n} \, dx=\sum _{n=0}^{\infty } \frac{t^{a+2 n+1}}{a+2 n+1}\right]$

$\text{}\left[\text{Int}'(a)=\int_0^t \frac{x^a \log (x)}{1-x^2} \, dx=\sum _{n=0}^{\infty } \left(\frac{\log (t) t^{a+2 n+1}}{a+2 n+1}-\frac{t^{a+2 n+1}}{(a+2 n+1)^2}\right)\right]$

$\text{Int}'(0)=\int_0^t \frac{\log (x)}{1-x^2} \, dx= \left(\sum _{n=0}^{\infty } \frac{t^{2 n+1} (2 n \log (t)+\log (t)-1)}{(2 n+1)^2}\right)$

$\int_0^1 \frac{\log (x)}{1-x^2} \, dx =\sum _{n=0}^{\infty } -\frac{1}{(2 n+1)^2}=-\frac{\pi ^2}{8}$