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:
$$\int\limits_{0}^{1} \frac{(x^2+4)\sin^{-1}x}{x^4-12x^2+16} \, dx $$
 A: Since:
$$ \frac{z+4}{z^2-12 z+16} = \frac{\bar{\varphi}}{z-(2\bar\varphi)^2}+\frac{\varphi}{z-(2\varphi)^2}\tag{1}$$
the integral can be computed by exploiting the relations between the dilogarithm and the golden ratio, namely Landen's identity. A first step of partial fraction decomposition followed by integration by parts leads to:
$$ I = \frac{\pi\log 5}{8}+\frac{1}{4}\int_{0}^{\pi/2}\log\left(\frac{4-2\sin(t)-\sin^2(t)}{4+2\sin(t)-\sin^2(t)}\right)\,dt \tag{2}$$
and if we replace $t$ with $\frac{\pi}{2}-2\arctan u$ we get:
$$ I = \frac{\pi\log 5}{8}+\frac{1}{2}\int_{0}^{1}\log\left(\frac{1+10v^2+5v^4}{5+10v^2+v^4}\right)\frac{dv}{1+v^2}\tag{3}$$
where the last integral is a dilogarithmic integral, namely the imaginary part of
$$\text{Li}_2\left[i\left(1-\sqrt{5}+\sqrt{5-2 \sqrt{5}}\right)\right]+\text{Li}_2\left[i\left(1+\sqrt{5}-\sqrt{5+2 \sqrt{5}}\right)\right]\tag{4}$$
that is a special value of the Clausen function $\text{Cl}_2$. It is also useful to notice that:
$$ \frac{1+10v^2+5v^4}{5+10v^2+v^4}=\frac{v}{\tanh(5\text{arctanh}(v))}. \tag{5}$$
A: 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$\int_{\frac{\pi}{20}}^{\frac{3\pi}{20}}\ln \tan a\ da =-\frac25G$
