<s>[INST] Find the Value of Integral Find the Value of $$\begin{align}I=\int_{0}^{1}\frac{\ln(x)\,dx}{1-x^2}\end{align}$$
I have tried like this: We have $$\begin{align}2I=\int_{0}^{1}\frac{\ln(x^2)\,dx}{1-x^2}=\int_{0}^{1}\frac{\ln(1-(1-x^2))\,dx}{1-x^2}\end{align}$$ So
$$2I=\begin{align}\int_{0}^{1}\frac{-(1-x^2)-\frac{(1-x^2)^2}{2}-\frac{(1-x^2)^3}{3}-\cdots }{1-x^2}\end{align}=$$
I need help from here..
 [/INST] Let
$$ I(\alpha)=\int_{0}^{1}\frac{x^{\alpha}dx}{1-x^2}. $$
Then $I'(0)=I$. Now
\begin{eqnarray*}
I(\alpha)=\lim_{a\to 1^-}\int_{0}^{a}\sum_{n=0}^\infty x^{\alpha+2n}dx=\lim_{a\to 1^-}\sum_{n=0}^\infty \frac{1}{\alpha+2n+1}a^{\alpha+2n+1}
\end{eqnarray*}
and hence
\begin{eqnarray*}
I'(\alpha)&=&\lim_{a\to 1^-}\sum_{n=0}^\infty \left(\frac{-1}{(\alpha+2n+1)^2}a^{\alpha+2n+1}+\frac{1}{\alpha+2n+1}a^{\alpha+2n+1}\ln a\right)\\
&=&-\sum_{n=0}^\infty\frac{1}{(\alpha+2n+1)^2}
\end{eqnarray*}
Thus
$$ I=I'(0)=-\sum_{n=0}^\infty\frac{1}{(2n+1)^2}=-\frac{\pi^2}{8}. $$
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