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..
 A: 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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\begin{align}
I&=\color{#66f}{\Large\int_{0}^{1}{\ln\pars{x} \over 1 - x^{2}}\,\dd x}
=\half\int_{0}^{1}{\ln\pars{x} \over 1 - x}\,\dd x
+\half\int_{0}^{1}{\ln\pars{x} \over 1 + x}\,\dd x
\\[3mm]&=\half\int_{0}^{1}{\ln\pars{x} \over 1 - x}\,\dd x
-\half\int_{0}^{-1}{\ln\pars{-x} \over 1 - x}\,\dd x
=\half\int_{0}^{1}{\ln\pars{1 - x} \over x}\,\dd x
-\half\int_{0}^{-1}{\ln\pars{1 - x} \over x}\,\dd x
\\[3mm]&=-\half\sum_{n = 1}^{\infty}{1 \over n^{2}}
+\half\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over n^{2}}
=-\sum_{n = 1}^{\infty}{1 \over \pars{2n - 1}^{2}}
=-\sum_{n = 1}^{\infty}{1 \over n^{2}}
+\sum_{n = 1}^{\infty}{1 \over \pars{2n}^{2}}
\\[3mm]&=-\,{3 \over 4}\ \underbrace{\sum_{n = 1}^{\infty}{1 \over n^{2}}}
_{\ds{\color{#c00000}{\zeta\pars{2} = {\pi^{2} \over 6}}}}
=-\,{3 \over 4}\,{\pi^{2} \over 6} = \color{#66f}{\Large -\,{\pi^{2} \over 8}}
\end{align}

since
  \begin{align}
\int_{0}^{a}{\ln\pars{1 - x} \over x}\,\dd x\,
&=
\int_{0}^{a}{1 \over x}\pars{-\sum_{n = 1}^{\infty}{x^{n} \over n}}\,\dd x
=-\sum_{n = 1}^{\infty}{a^{n} \over n^{2}}\,,\qquad
\verts{a}\ \leq\ 1
\end{align}

A: Integrating by parts,
$$ \begin{align}  \int \frac{\ln (x)}{1-x^{2}} \ dx &= \ln(x) \ \text{arctanh} (x) - \int \frac{\text{arctanh}(x)}{x} \ dx \\  &= \ln (x) \  \text{arctanh} (x) - \frac{1}{2} \int \frac{\ln (1+x)}{x} \ dx + \frac{1}{2} \int \frac{\ln (1-x)}{x} \ dx  \\ &= \ln(x) \ \text{arctanh}(x) + \frac{1}{2} \text{Li}_{2}(-x) - \frac{1}{2} \text{Li}_{2}(x) + C  \end{align}$$
where $\text{Li}_{2}(x)$ is the dilogarithm function.
Then
$$ \begin{align} \int_{0}^{1} \frac{\ln (x)}{1-x^{2}} \ dx &= \frac{1}{2} \Big( \text{Li}_{2}(-1) - \text{Li}_{2}(1) \Big) \\ &= \frac{1}{2} \Big( - \frac{\zeta(2)}{2} - \zeta(2)  \Big) \\ &= \frac{1}{2} \left(- \frac{\pi^{2}}{12}-\frac{\pi^{2}}{6} \right) \\ &= -\frac{\pi^{2}}{8} . \end{align}$$
A: Use the series expansion for $\dfrac{1}{1-x^2}$ i.e
$\displaystyle \frac{1}{1-x^2}=\sum_{k=0}^{\infty} x^{2k}\,dx$
Hence,
$$I=\int_0^1 \ln(x)\left(\sum_{k=0}^{\infty} x^{2k}\right)\,dx=\sum_{k=0}^{\infty} \int_0^1 x^{2k}\ln x\,dx=-\sum_{k=0}^{\infty}\frac{1}{(2k+1)^2}$$
where the final integral can be evaluated using the substitution $\ln x=-t$. 
Since,
$$\sum_{k=1}^{\infty} \frac{1}{k^2}=\sum_{k=1}^{\infty} \frac{1}{(2k)^2}+\sum_{k=0}^{\infty} \frac{1}{(2k+1)^2} \Rightarrow \sum_{k=0}^{\infty} \frac{1}{(2k+1)^2}=\frac{\pi^2}{6}-\frac{1}{4}\frac{\pi^2}{6}$$
$$\Rightarrow \sum_{k=0}^{\infty} \frac{1}{(2k+1)^2}=\frac{3\pi^2}{24}=\frac{\pi^2}{8}$$
Hence,
$$\boxed{I=-\dfrac{\pi^2}{8}}$$
