Improper integral convergence question Prove that the following integral converges:

We divided the integral to 2 integrals (one from 0 to 1/2 and the other from 1/2 to 1).
We managed to prove that the integral from 1/2 to 1 converges but we're struggling with the other one.
Thanks
 A: We have $$\int_{0}^{1}\frac{\log\left(x\right)}{1-x^{2}}dx=\sum_{k\geq0}\int_{0}^{1}\log\left(x\right)x^{2k}dx
 $$ and by parts $$\int_{0}^{1}\log\left(x\right)x^{2k}dx=\left.\frac{x^{2k+1}\log\left(x\right)}{2k+1}\right|_{0}^{1}-\frac{1}{2k+1}\int_{0}^{1}x^{2k}dx=-\frac{1}{\left(2k+1\right)^{2}}
 $$ hence $$\int_{0}^{1}\frac{\log\left(x\right)}{1-x^{2}}dx=-\sum_{k\geq0}\frac{1}{\left(2k+1\right)^{2}}=-\sum_{k\geq1}\frac{1}{k^{2}}+\sum_{k\geq1}\frac{1}{\left(2k\right)^{2}}=-\frac{\pi^{2}}{6}+\frac{\pi^{2}}{24}=-\frac{\pi^{2}}{8}.
 $$
A: We look at the interval $(0,1/2]$. In order to deal with positive quantities, we will show instead that $\int_0^{1/2}\frac{-\ln x}{1-x^2}\,dx$ exists. Note that in our interval $1-x^2\ge 3/4$. So it is enough to prove that $\int_0^{1/2} \frac{-\ln x}{3/4}\,dx$ exists, or equivalently that $\int_0^{1/2}(-\ln x)\,dx$ exists.
This can be viewed as a standard fact. To be formal, we calculate $\int_{\epsilon}^{1/2}(-\ln x)\,dx$ and find the limit as $\epsilon$ approaches $0$ from the right.
An antiderivative of $-\ln x$ is $x-x\ln x$ (integration by parts). Now plug in and find the limit as $\epsilon\to 0^+$. We use the fact that $t\ln t$ approaches $0$ as $t$ approaches $0$ from the right.
