<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] 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,
$$\boxed{I=-\dfrac{\pi^2}{8}}$$
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