$\int_0^1 \frac{\arcsin x\arccos x}{x}dx$ 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.
 A: Let $J(a)=\int_0^\infty \frac{\tan^{-1}(ax)\ln x}{1+x^2}$
\begin{align}
J’(a)=& \int_0^\infty \frac{x\ln x}{(1+x^2)(1+a^2x^2)} \overset{x\to \frac1{ax}}{dx}\\
= & 
 -\frac1{2}\int_0^\infty \frac{x\ln a}{(1+x^2)(1+{a^2}x^2)} {dx}
= \frac{\ln^2 a}{2(1-a^2)}
\end{align}
which leads to
\begin{align}\int_0^\infty \frac{\arctan x\ln x}{1+x^2}dx
=\int_0^1J’(a)da= \frac12 \int_0^1 \frac{\ln^2 a}{1-a^2}da
=\frac78\zeta(3)
\end{align}
Besides, the integral is also related to
$$\int_0^1 \frac{\sin^{-1} x\cos^{-1} x}{x}dx
=\frac14\int_0^\infty \frac{x^2}{\sinh x}dx
=\frac18\int_0^\pi\frac{x(\pi-x)}{\sin x}dx=\frac78\zeta(3)$$
A: Use the relation: $$\arcsin(x)+\arccos(x)=\frac{\pi}2$$
So,
$$J=\frac{\pi}2\int_0^1  \frac{\arcsin(x)}{x}dx-\int_0^1\frac{\arcsin^2(x)}{x} dx=\frac{\pi}2I_1-I_2$$
Let $t=\arcsin(x)$
$$I_1=\int_0^{\pi/2} \frac{t}{\tan(t)} dt,~~~~I_2=\int_0^{\pi/2} \frac{t^2}{\tan(t)}dt$$
For each integral use series:
$$\frac{1}{\tan(t)}=\frac{1}{t}+\sum_{n=1}^\infty \frac{2t}{t^2-n^2\pi^2}$$
Integrate term by term and it is done.
A: Just for the fun (it is to long for a comment)
Using @Quanto's solution
$$I=\frac18\int_0^\pi\frac{x(\pi-x)}{\sin (x)}dx$$ and the $\large 1,400^+$ years old approximation
$$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$
$$I \simeq \frac{1}{128}\int_0^\pi \left(5 \pi ^2-4 (\pi -x) x\right)\,dx=\frac{13 \pi ^3}{384}$$ is in a relative error of $0.2$%
A: $$
\begin{align*}
\int_0^1 \frac{\arcsin(x)\arccos(x)}{x}\,dx& = \int_0^1 \frac{\arcsin(x)\left(\frac{\pi}{2}-\arcsin(x)\right)}{x}\,dx\\
&=\frac{\pi}{2}\int_0^1 \frac{\arcsin(x)}{x}\,dx-\int_0^1 \frac{\arcsin^2(x)}{x}\,dx\\
&=\frac{\pi}{2}\int_0^{\pi/2} \frac{x}{\sin(x)}\cos(x)\,dx-\int_0^{\pi/2} \frac{x^2}{\sin(x)}\cos(x)\,dx\\
&=\frac{\pi}{2}\int_0^{\pi/2} x\cot(x)\,dx-\int_0^{\pi/2} x^2\cot(x)\,dx\\
&=\frac{\pi}{2} J - K\\
&=\frac{\pi^2\ln(2)}{4}-\frac{\pi^2\ln(2)}{4}+\frac{7\zeta(3)}{8}\\
&=\frac{7\zeta(3)}{8} \qquad \blacksquare
\end{align*}
$$
$$
\begin{align*}
J&=\int_0^{\pi/2} x\cot(x)\,dx\\
&=x\ln(\sin(x))\Big|_0^{\pi/2}-\int_0^{\pi/2}\ln(\sin(x))\,dx\\
&=\ln(2)\int_0^{\pi/2} \,dx+\sum_{k=1}^\infty \frac{1}{k} \int_0^{\pi/2}\cos(2 k x)\,dx \\
&=\frac{\pi\ln(2) }{2}
\end{align*}
$$
$$
\begin{aligned}
K&=\int_0^{\pi/2} x^2\cot(x)\,dx\\
&=x^2\ln(\sin(x))\Big|_0^{\pi/2}-2\int_0^{\pi/2}x\ln(\sin(x))\,dx\\
&=-2\int_0^{\pi/2}x \ln\left(\sin(x)\right)\,dx\\
&=-2\left(-\ln(2)\int_0^{\pi/2}x \,dx-\sum_{k=1}^\infty \frac{1}{k} \int_0^{\pi/2}x \cos(2 k x)\,dx\right)\\
&=-2\left(-\frac{\pi^2\ln(2)}{8}-\sum_{k=1}^\infty \frac{1}{k} \left(\frac{x \sin(2 k x)}{2k}\Big|_0^{\pi/2} -\frac{1}{2k}\int_0^{\pi/2}\sin(2 k x)\,dx\right)\right)\\
&=-2\left(-\frac{\pi^2\ln(2)}{8}-\frac{1}{2}\sum_{k=1}^\infty \frac{1}{k^2} \left(\frac{\cos(2 k x)}{2k}\Big|_0^{\pi/2}\right)\right)\\
&=-2\left(-\frac{\pi^2\ln(2)}{8}-\frac{1}{4}\sum_{k=1}^\infty \frac{1}{k^3} \left((-1)^k-1\right)\right)\\
&=-2\left(-\frac{\pi^2\ln(2)}{8}+\frac{1}{4}\zeta(3)+\frac14\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^3}\right)\\
&=-2\left(-\frac{\pi^2\ln(2)}{8}+\frac{\zeta(3)}{4}+\frac{\eta(3)}{4}\right)\\
&=-2\left(-\frac{\pi^2\ln(2)}{8}+\frac{\zeta(3)}{4}+\frac{3\zeta(3)}{16}\right)\\
&=-2\left(-\frac{\pi^2\ln(2)}{8}+\frac{7\zeta(3)}{16}\right) \\
&=\frac{\pi^2\ln(2)}{4}-\frac{7\zeta(3)}{8}
\end{aligned}
$$
Where we used
$$ \arcsin(x)+\arccos(x)=\frac{\pi}{2}$$
$$\eta(s)=(1-2^{1-s})\zeta(s)$$
$$\ln\left(\sin(x)\right)=-\ln(2)-\sum_{k=1}^\infty \frac{\cos(2 k x)}{k}$$
A: Letting $y=\arcsin x$ transforms our integral
$$
\begin{aligned}
I &=\int_{0}^{\frac{\pi}{2}} \frac{y\left(\frac{\pi}{2}-y\right) \cos y d y}{\sin y} \\
&=\frac{\pi}{2} \underbrace{\int_{0}^{\frac{\pi}{2}} \frac{y \cos y}{\sin y} d y}_{J}-\underbrace{\int_{0}^{\frac{\pi}{2}} \frac{y^{2} \cos y}{\sin y} d y}_{K}
\end{aligned}
$$
$$
\begin{aligned}
J &=\int_{0}^{\frac{\pi}{2}} y d(\ln \sin y)=-\int_{0}^{\frac{\pi}{2}} \ln (\sin y) d y =\frac{\pi}{2} \ln 2
\end{aligned}
$$
where the last result comes from my post 1.
$$
\begin{aligned}
K=\int_{0}^{\frac{\pi}{2}} y^{2} d(\ln (\sin y)) \stackrel{IBP}{=}  2 \int_{0}^{\frac{\pi}{2}} y \ln (\sin y) d y
\end{aligned}
$$
By the result that  $\int_{0}^{\frac{\pi}{2}} y \ln (\sin y) d y=\frac{\pi^2}{8}\ln 2 -\frac{7}{16}\zeta(3)$ shown below as Footnote, we can conclude that
$$
\boxed{I=\frac{\pi}{2}\left(\frac{\pi}{2} \ln 2\right)-\left(\frac{\pi^{2}}{4} \ln 2-\frac{7}{8} \zeta(3)\right)=\frac{7}{8} \zeta(3)}
$$

Footnote:
By the Fourier series of $\ln(\sin x)$,
$$\ln (\sin x)+\ln 2=-\sum_{k=1}^{\infty} \frac{\cos (2 k x)}{k}$$
Multiplying the equation by $x$ followed by integration from $0$ to $\infty$, we have
$$
\begin{aligned}
\int_{0}^{\frac{\pi}{2}} x \ln (\sin x) d x+\int_{0}^{\frac{\pi}{2}} x\ln 2 d x&=-\sum_{k=1}^{\infty} \int_{0}^{\frac{\pi}{2}} \frac{x \cos (2 k x)}{k} d x\\
\int_{0}^{\frac{\pi}{2}} x \ln (\sin x) d x+\left[\frac{x^{2}}{2} \ln 2\right]_{0}^{\frac{\pi}{2}}&=-\sum_{k=1}^{\infty} \frac{1}{2 k^{2}} \int_{0}^{\frac{\pi}{2}} x d(\sin 2 k x)\\
\int_{0}^{\frac{\pi}{2}} x \ln (\sin x) d x&=-\frac{\pi^{2}}{8} \ln 2-\frac{1}{2} \sum_{k=1}^{\infty} \frac{1}{k^{2}}\left[\frac{\cos 2(x)}{2 k}\right]_{0}^{\frac{\pi}{2}} \\
&=-\frac{\pi^{2}}{8} \ln 2-\frac{1}{4} \sum_{k=1}^{\infty} \frac{(-1)^{k}-1}{k^{3}}\\
&=\frac{\pi^{2}}{8} \ln 2-\frac{1}{4}\left(\sum_{k=1}^{\infty} \frac{2}{(2 k+1)^{3}}\right)\\
&=\frac{\pi^{2}}{8} \ln 2-\frac{1}{2}\left[\sum_{k=1}^{\infty} \frac{1}{k^{3}}-\sum_{k=1}^{\infty} \frac{1}{(2 k)^{3}}\right]\\
&=\frac{\pi^{2}}{8} \ln 2-\frac{7}{16}\zeta(3) \blacksquare
\end{aligned}
$$
