<s>[INST] $\int_0^1 \frac{\arcsin x\arccos x}{x}dx$ Someone on Youtube posted a video solving this integral.
I can't find on Math.stack.exchange this integral using search engine https://approach0.xyz
It is related to $\displaystyle \int_0^\infty \frac{\arctan x\ln x}{1+x^2}dx$
Following is a solution that is not requiring the use of series:
\begin{align}J&=\int_0^1 \frac{\arcsin x\arccos x}{x}dx\\
&\overset{\text{IBP}}=\underbrace{\Big[\arcsin x\arccos x\ln x\Big]_0^1}_{=0}-\underbrace{\int_0^1 \frac{\arccos x\ln x}{\sqrt{1-x^2}}dx}_{x=\cos t }+\underbrace{\int_0^1 \frac{\arcsin x\ln x}{\sqrt{1-x^2}}dx}_{x=\sin t}\\
&=\int_0^{\frac{\pi}{2}} t\ln(\tan t)dt\\
&\overset{u=\tan t}=\int_0^\infty \frac{\arctan u\ln u}{1+u^2}du\\
&\overset{\text{IBP}}=\underbrace{\left[\arctan u\left(\int_0^u \frac{\ln t}{1+t^2}dt\right)\right]_0^\infty}_{=0}-\int_0^\infty \frac{1}{1+u^2}\left(\underbrace{\int_0^u \frac{\ln t}{1+t^2}dt}_{y(t)=\frac{t}{u}}\right)du\\
&=-\int_0^\infty \left(\int_0^1 \frac{u\ln(uy)}{(1+u^2)(1+u^2y^2)}dy\right)du\\
&=-\int_0^\infty \left(\int_0^1 \frac{u\ln u}{(1+u^2)(1+u^2y^2)}dy\right)du-\int_0^1 \left(\int_0^\infty \frac{u\ln y}{(1+u^2)(1+u^2y^2)}du\right)dy\\
&=-\int_0^\infty \left[\frac{\arctan(uy)}{1+u^2}\right]_{y=0}^{y=1}\ln udu-\frac{1}{2}\int_0^1 \left[\frac{\ln\left(\frac{1+u^2}{1+u^2y^2}\right)}{1-y^2}\right]_{u=0}^{u=\infty}\ln ydy\\
&=-J+\int_0^1 \frac{\ln^2 y}{1-y^2}dy\\
&=\frac{1}{2}\int_0^1 \frac{\ln^2 y}{1-y}dy-\frac{1}{2}\underbrace{\int_0^1 \frac{y\ln^2 y}{1-y^2}dy}_{z=y^2}\\
&=\frac{7}{16}\int_0^1 \frac{\ln^2 y}{1-y}dy\\
&=\frac{7}{16}\times 2\zeta(3)=\boxed{\frac{7}{8}\zeta(3)}
\end{align}
NB:I assume $\displaystyle \int_0^1 \frac{\ln^2 y}{1-y}dy=2\zeta(3)$
Feel free to post your solution.
 [/INST] $$
\begin{align*}
\int_0^1 \frac{\arcsin(x)\arccos(x)}{x}\,dx& = \int_0^1 \frac{\arcsin(x)\left(\frac{\pi}{2}-\arcsin(x)\right)}{x}\,dx\\
&=\frac{\pi}{2}\int_0^1 \frac{\arcsin(x)}{x}\,dx-\int_0^1 \frac{\arcsin^2(x)}{x}\,dx\\
&=\frac{\pi}{2}\int_0^{\pi/2} \frac{x}{\sin(x)}\cos(x)\,dx-\int_0^{\pi/2} \frac{x^2}{\sin(x)}\cos(x)\,dx\\
&=\frac{\pi}{2}\int_0^{\pi/2} x\cot(x)\,dx-\int_0^{\pi/2} x^2\cot(x)\,dx\\
&=\frac{\pi}{2} J - K\\
&=\frac{\pi^2\ln(2)}{4}-\frac{\pi^2\ln(2)}{4}+\frac{7\zeta(3)}{8}\\
&=\frac{7\zeta(3)}{8} \qquad \blacksquare
\end{align*}
$$
$$
\begin{align*}
J&=\int_0^{\pi/2} x\cot(x)\,dx\\
&=x\ln(\sin(x))\Big|_0^{\pi/2}-\int_0^{\pi/2}\ln(\sin(x))\,dx\\
&=\ln(2)\int_0^{\pi/2} \,dx+\sum_{k=1}^\infty \frac{1}{k} \int_0^{\pi/2}\cos(2 k x)\,dx \\
&=\frac{\pi\ln(2) }{2}
\end{align*}
$$
$$
\begin{aligned}
K&=\int_0^{\pi/2} x^2\cot(x)\,dx\\
&=x^2\ln(\sin(x))\Big|_0^{\pi/2}-2\int_0^{\pi/2}x\ln(\sin(x))\,dx\\
&=-2\int_0^{\pi/2}x \ln\left(\sin(x)\right)\,dx\\
&=-2\left(-\ln(2)\int_0^{\pi/2}x \,dx-\sum_{k=1}^\infty \frac{1}{k} \int_0^{\pi/2}x \cos(2 k x)\,dx\right)\\
&=-2\left(-\frac{\pi^2\ln(2)}{8}-\sum_{k=1}^\infty \frac{1}{k} \left(\frac{x \sin(2 k x)}{2k}\Big|_0^{\pi/2} -\frac{1}{2k}\int_0^{\pi/2}\sin(2 k x)\,dx\right)\right)\\
&=-2\left(-\frac{\pi^2\ln(2)}{8}-\frac{1}{2}\sum_{k=1}^\infty \frac{1}{k^2} \left(\frac{\cos(2 k x)}{2k}\Big|_0^{\pi/2}\right)\right)\\
&=-2\left(-\frac{\pi^2\ln(2)}{8}-\frac{1}{4}\sum_{k=1}^\infty \frac{1}{k^3} \left((-1)^k-1\right)\right)\\
&=-2\left(-\frac{\pi^2\ln(2)}{8}+\frac{1}{4}\zeta(3)+\frac14\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^3}\right)\\
&=-2\left(-\frac{\pi^2\ln(2)}{8}+\frac{\zeta(3)}{4}+\frac{\eta(3)}{4}\right)\\
&=-2\left(-\frac{\pi^2\ln(2)}{8}+\frac{\zeta(3)}{4}+\frac{3\zeta(3)}{16}\right)\\
&=-2\left(-\frac{\pi^2\ln(2)}{8}+\frac{7\zeta(3)}{16}\right) \\
&=\frac{\pi^2\ln(2)}{4}-\frac{7\zeta(3)}{8}
\end{aligned}
$$
Where we used
$$ \arcsin(x)+\arccos(x)=\frac{\pi}{2}$$
$$\eta(s)=(1-2^{1-s})\zeta(s)$$
$$\ln\left(\sin(x)\right)=-\ln(2)-\sum_{k=1}^\infty \frac{\cos(2 k x)}{k}$$
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