I have recently come across the following integral: $$\int_{0}^{1} \frac{\ln x}{1-x^{2}}\,dx$$
I was perplexed but then the thought of using the beta functions struck me. Consider $$B(x, y)=2\int_{0}^{\frac{\pi}{2}}\sin^{2x-1}\theta\,\cos^{2y-1}\theta\,d\theta$$ $$\frac{\partial}{\partial x}B(x, y)=2\int_{0}^{\frac{\pi}{2}}2\sin^{2x-1}\theta\,\ln\sin\theta\,\cos^{2y-1}\theta\,d\theta$$ $$\frac{\partial}{\partial x}\frac{\Gamma{(x)}\Gamma{(y)}}{\Gamma{(x+y)}}=\Gamma{(y)}\frac{\psi{(x)}\,\Gamma(x+y)-\Gamma{(x)}\,\psi{(x+y)}}{(\Gamma{(x+y)})^2}=4\int_{0}^{\frac{\pi}{2}}2\sin^{2x-1}\theta\,\ln\sin\theta\,\cos^{2y-1}\theta\,d\theta$$
Put $x=\frac{1}{2}$ and let $y\to0$, $$4\int_{0}^{\frac{\pi}{2}}2\sin^{2x-1}\theta\,\ln\sin\theta\,\cos^{2y-1}\theta\,d\theta=\lim_{y \to 0} \Gamma(y) \frac{\psi(\frac{1}{2})\,\Gamma(\frac{1}{2}+y)-\Gamma(\frac{1}{2})\psi(\frac{1}{2}+y)}{(\Gamma(\frac{1}{2}+y))^{2}}$$
I believe substituting $\sin x$ into the above will yield the desired integral. However, I am not sure how to progress further, nor do I know how to derive the answer, which is $-\frac{\pi^2}{8}$. Is there a solution using the beta, gamma and digamma functions? Otherwise, is there an easier solution?
As @ Daniel Fischer commented, start with $$\frac 1 {1-t}=\sum_{n=0}^\infty t^n$$ Make $t=x^2$ and use $$\int_{0}^{1} \frac{\log (x)}{1-x^{2}}\,dx=\sum_{n=0}^\infty \int_0^1 x^{2n} \log(x)\,dx$$ One integration by parts to get $$\int_0^1 x^{2n} \log(x)\,dx=-\frac{1}{(2 n+1)^2}$$ making $$\int_{0}^{1} \frac{\log (x)}{1-x^{2}}\,dx=-\sum_{n=0}^\infty \frac{1}{(2 n+1)^2}$$