<s>[INST] Seemingly Simple Integral $\int_0^1\frac{x^2\ln x}{\sqrt{1-x^2}}dx$. Evaluate $$\int_0^1 f(x) dx$$ where
$$f(x) = \frac{x^2\ln x}{\sqrt{1-x^2}}$$
I started off with the substitution $x=\sin y$, which resulted in the integrand reducing to
$$\sin^2y\cdot \ln (\sin y) dy$$
Then I used the property of definite integrals that
$$\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$$
Then too it wasn't getting simplified.
I tried $e^z=\sin x$, but this gave no headway because after a while I reached a complete full-stop. How should I go about this?
 [/INST] $$I(\alpha)=\int_0^1\dfrac{x^\alpha\ln x}{\sqrt{1-x^2}}\ dx$$
Using Beta function we have
$$\int_0^1\dfrac{x^\alpha}{\sqrt{1-x^2}}\ dx=\dfrac12{\bf B}\left(\frac{\alpha+1}{2},\frac{1}{2}\right)$$
then with Digamma function $\psi$
\begin{align}
I(\alpha)
= \dfrac{d}{d\alpha}\int_0^1\dfrac{x^\alpha}{\sqrt{1-x^2}}\ dx \\
= \dfrac12{\bf B}\left(\frac{\alpha+1}{2},\frac{1}{2}\right)\left(\psi(\frac{\alpha+1}{2})-\psi(\frac{\alpha+2}{2})\right)
\end{align}
now let $\alpha=2$,
$$\psi(\frac{3}{2})-\psi(\frac{4}{2})=2\sum_{n=0}^\infty\left(\dfrac{1}{2n+4}-\dfrac{1}{2n+3}\right)=1-\ln4$$
therefore
$$I(2)=\color{blue}{\dfrac{\pi}{8}(1-\ln4)}$$
</s>