<s>[INST] A variation of Ahmed's integral $\int_{0}^{1} \frac{(x^2+4)\sin^{-1}x}{x^4-12x^2+16} \, dx $ Given that the closed form exist, evaluate the following Integral:
$$\int\limits_{0}^{1} \frac{(x^2+4)\sin^{-1}x}{x^4-12x^2+16} \, dx $$
 [/INST] Let
$J(a)=\int_0^\infty\frac1{1+t^2} \ln \frac{t^2+2t\sin 2a+1}{t^2-2t\sin 2a+1}dt
$
and establish $J’(a)=8a\csc(2a)$. Note\begin{align}
&I=\int_{0}^{1} \frac{(x^2+4)\sin^{-1}x}{x^4-12x^2+16} \, dx 
=\frac14\int_{0}^{1}\sin^{-1}x \ d\left(\ln \frac{4+2x-x^2}{4-2x-x^2} \right)
\end{align}
Integrate by parts and then substitute  $x=\frac{2t}{1+t^2}$ to obtain
\begin{align}
I= &\ \frac\pi8\ln5-\frac14(J(\frac{3\pi}{20})-J(\frac{\pi}{20}))
=\frac\pi8\ln5 -\int_{\frac{\pi}{20}}^{\frac{3\pi}{20}}2a\csc(2a)da\\
\overset{ibp}
=&\ \frac\pi8\ln5 - a \ln\tan a\bigg|_{\frac{\pi}{20}}^{\frac{3\pi}{20}}
+\int_{\frac{\pi}{20}}^{\frac{3\pi}{20}}\ln \tan a\ da\\
 =&\ \frac\pi8\ln5 - \frac{3\pi}{20}\ln \tan \frac{3\pi}{20}
+\frac{\pi}{20}\ln \tan \frac{\pi}{20}-\frac25 G\\
\end{align}
where$\int_{\frac{\pi}{20}}^{\frac{3\pi}{20}}\ln \tan a\ da =-\frac25G$
</s>