<s>[INST] Another improper integral Show that :
$$\int_0^1\frac{(\sin ^{-1}x)^2}{x}\text{d}x=\frac{\pi ^2\ln 2}{4}-\frac78\zeta(3)$$
This integral is in "irresistible integrals" on page 122. I can't prove this one.
 [/INST] This is not an answer, but you might be interested in knowing why this can work. I thought of writing this as a comment, but it's quite big. 


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*Note that $\displaystyle \ln(1+x) = x-\frac{x^2}{2} + \frac{x^3}{3} + \cdots$

*So $\ln(2) = \displaystyle 1-\frac{1}{2} + \frac{1}{3} + \cdots$
Now I tried the standard integration by parts method. Put $x =\sin\theta$.  Then the integral becomes $$I =\int_{0}^{\pi/2} \theta^{2} \cdot \cot\theta \ d\theta$$ Using the standard ILATE rule take $dv=\cot\theta \ d\theta$ so that you have $v=\ln(\sin\theta)$. Take $u=x^{2} \implies du = 2\theta \ d\theta$. So we have $$I = \underbrace{\bigl( u\cdot v\bigr)_{0}^{\pi/2}} - 2\int_{0}^{\pi/2} \theta \cdot \ln(\sin\theta)\ d\theta .$$
The underbrace term $\to 0$ so we are left with  $$-2\int_{0}^{\pi/2} \theta \cdot \ln(\sin\theta) \ d\theta$$
Now
\begin{align*}
\int_{0}^{\pi/2} \theta \cdot \ln(\sin\theta) \ d\theta &=\int_{0}^{\pi/2} \theta \cdot \ln\bigl(1+ (\sin\theta-1)\bigr)\ d\theta \\ &= \int_{0}^{\pi/2} \theta \cdot \biggl[ (\sin\theta-1) - \frac{(\sin\theta -1)^{2}}{2}+\frac{(\sin\theta-1)^{3} }{3}- \cdots \biggr] \ d\theta
\end{align*}
Now collecting the $\theta$ involving terms we see that its: 
\begin{align*}
\int_{0}^{\pi/2} \biggl( -\theta + \frac{\theta}{2} -\frac{\theta}{3} + \cdots \biggr) \ d\theta &=-\biggl(1-\frac{1}{2}+\frac{1}{3} - \cdots\biggr)\cdot \int_{0}^{\pi/2} \theta \ d\theta \\ &= -\ln(2) \cdot \frac{\pi^{2}}{8}
\end{align*}
Note that we have to multiply by $-2$ to complete the solution. 
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