<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] 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
$$ \begin{align} \int_{0}^{1} \frac{\ln (x)}{1-x^{2}} \ dx &= \frac{1}{2} \Big( \text{Li}_{2}(-1) - \text{Li}_{2}(1) \Big) \\ &= \frac{1}{2} \Big( - \frac{\zeta(2)}{2} - \zeta(2)  \Big) \\ &= \frac{1}{2} \left(- \frac{\pi^{2}}{12}-\frac{\pi^{2}}{6} \right) \\ &= -\frac{\pi^{2}}{8} . \end{align}$$
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