Here is an explicit way to proceed, we are calculating the last integral in the OP, denoted here by $J$:
$$
\begin{aligned}
J
&=
\int_0^{1/2}\frac 4u\;\arcsin^2 u\; du\qquad\text{(Substitution: $u=\sin x$)}\\
&=
\int_0^a\frac 4{\sin x}\;x^2\; \cos x\; dx\qquad\text{(where $a=\arcsin (1/2)=\pi/6$)}\\
&=
4
\int_0^a\frac x{\tan x}\;x\; dx\\
&=
4
\int_0^a\left(\frac {2ix}{e^{2ix}-1}+ix\right)\;x\; dx\\
&=
4\Re
\int_0^a\frac {2ix}{e^{2ix}-1}\;x\; dx\\
&=
-\Re
\int_0^a\frac {2ix}{e^{2ix}-1}\;2ix\; d(2ix)\\
&=
-\Re
\int_0^{2ia}\frac {X^2}{e^X-1}\;dX\\
&=
-\Re
\left[
\
\color{gray}{
-\frac 13X^3}
+ X^2\log(1-e^X)
+ 2X\operatorname{Li}_2(e^X)
- 2\operatorname{Li}_3(e^X)
\
\right]_0^{2ia}
\\
&=
-\Re
\left[
\
-a^2\log(1-e^{2ia})
+ 4ia\operatorname{Li}_2(e^{2ia})
- 2\operatorname{Li}_3(e^{2ia})
+ 2\operatorname{Li}_3(e^0)
\
\right]
\\
&=
-\Re
\left[
\
0
+ 4ia\operatorname{Li}_2(e^{2ia})
- 2\operatorname{Li}_3(e^{2ia})
+ 2\zeta(3)
\
\right]
\\
&=
-\Re
\left[
\
4ia\operatorname{Li}_2(b)
- 2\operatorname{Li}_3(b)
+ 2\zeta(3)
\
\right]
\ .
\end{aligned}
$$
The logarithmic term disappeared because
$$
\Re \log(1-e^{2ia})
=
\log |1-e^{2ia}|
=
\log |1-e^{2i\pi/6}|
=
\log |e^{-2i\pi/6}|
=\log 1
=0\ .
$$
The value of $b$ is explicitly
$$
b = e^{2ia}
=e^{2i\pi/6}
=e^{i\pi/3}
\ ,
$$
so $b$ is the above primitive root of unity of order six.
One can show the relation:
$$
\Re
\operatorname{Li}_3(b)
=\frac 13\operatorname{Li}_3(1)
=\frac 13\zeta(3)\ .
$$
This is seen for instance after a series expansion:
$$
\begin{aligned}
&2\Re\operatorname{Li}_3(b)
=
\operatorname{Li}_3(b) + \operatorname{Li}_3(\bar b)
\\
&=
\frac 1{1^3}\underbrace{(b+\bar b)}_{=1}
+
\frac 1{2^3}\underbrace{(b^2+\bar b^2)}_{=-1}
+
\frac 1{3^3}\underbrace{(b^3+\bar b^3)}_{=-2}
+
\frac 1{4^3}\underbrace{(b^4+\bar b^4)}_{=-1}
+
\frac 1{5^3}\underbrace{(b^5+\bar b^5)}_{=1}
+
\frac 1{6^3}\underbrace{(b^6+\bar b^6)}_{=2}
+
\dots\text{ with $6$-periodic repetitions}
\\
&=
\frac 1{1^3}(1)
+
\frac 1{2^3}(1-\color{blue}{2})
+
\frac 1{3^3}(1-\color{darkgreen}{3})
+
\frac 1{4^3}(1-\color{blue}{2})
+
\frac 1{5^3}(1)
+
\frac 1{6^3}(1-\color{blue}{2}-\color{darkgreen}{3}+\color{red}{6})
+
\dots\text{ with $6$-periodic repetitions}
\\
&=
\zeta(3)\left(
1
-
\frac{\color{blue}{2}}{2^3}
-
\frac{\color{darkgreen}{3}}{3^3}
+
\frac{\color{red}{6}}{6^3}
\right)
\\
&=
\zeta(3)
\left(
1
-
\frac{\color{blue}{2}}{2^3}
\right)
\left(
1-
\frac1{\color{darkgreen}{3}}{3^3}
\right)
=\zeta(3)\cdot\frac 34\cdot\frac 89
=\zeta(3)\cdot\frac 23\ .
\end{aligned}
$$
In a similar manner we can (try to) look into the dilogarithmic part.
$$
\begin{aligned}
&2\Re i\operatorname{Li}_2(b)
=\Re i(
\operatorname{Li}_2(b) - \operatorname{Li}_2(\bar b))
\\
&=
\frac i{1^2}\underbrace{(b-\bar b)}_{=i\sqrt 3}
+
\frac i{2^2}\underbrace{(b^2-\bar b^2)}_{=i\sqrt 3}
+
\frac i{3^2}\underbrace{(b^3-\bar b^3)}_{=0}
+
\frac i{4^2}\underbrace{(b^4-\bar b^4)}_{=-i\sqrt 3}
+
\frac i{5^2}\underbrace{(b^5-\bar b^5)}_{=-i\sqrt 3}
+
\frac i{6^2}\underbrace{(b^6-\bar b^6)}_{=0}
+
\dots\text{ with $6$-periodic repetitions}
\\
&=
\sqrt3\left(
\frac1{1^2}
+
\frac 1{2^2}
-
\frac1{4^2}
-
\frac 1{5^2}
+
\dots
\right)\text{ with $6$-periodic repetitions}
\\
&=
\frac {\sqrt3}{18}
\left(
\psi'\left(\frac 16\right)
+
\psi'\left(\frac 13\right)
\right)
-\frac {4 {\sqrt3}}{27}\pi^2
\ .
\end{aligned}
$$
We have a thus some formula for the expression involving corresponding
special values $L(2,\chi)$ of the Dirichlet $L$-function computed in $2$ w.r.t. some characters $\chi$ with periodicity modulo six,
and in general we can expect some $\psi$-values as in:
https://mathworld.wolfram.com/DirichletL-Series.html
We can do slightly better. Using $b^4=-b$ and the dilogarithmic relation
$\operatorname{Li}_2(x)+\operatorname{Li}_2(-x)=\frac 12\operatorname{Li}_2(x^2)$,
we obtain:
$\operatorname{Li}_2(b)+\operatorname{Li}_2(b^4)=\frac 12\operatorname{Li}_2(b^2)$.
Here, $b^2$ and $b^4$ are cubic primitive units, so the periodicity is reduced.
Writing series for instance, we get $\operatorname{Li}_2(b^4)=
-\operatorname{Li}_2(b^2)$. This gives:
$$
\begin{aligned}
\Im \operatorname{Li}_2(b)
&=\frac 32
\Im \operatorname{Li}_2(b^2)
\\
&=
\frac {3\sqrt 3}4
\left(
\frac1{1^2}
-
\frac 1{2^2}
+
\frac1{4^2}
-
\frac 1{5^2}
+
\dots
\right)\text{ with $3$-periodic repetitions}
\\
&=
\frac {3\sqrt 3}4
\left(
\frac 29\psi'\left(\frac 13\right)
-\frac4{27}\pi^2
\right)\ .
\end{aligned}
$$
Here, for instance, the sum of the inverses of $1^2$, $4^2$, $7^2$, ... is $\frac 19\psi'\left(\frac 13
\right)$.
Putting all together:
$$
\color{blue}{
\begin{aligned}
J
&=
\frac 23\pi\cdot \Im \operatorname{Li}_2(b)
- \frac 43\zeta(3)
\\
&=
\pi\cdot \Im \operatorname{Li}_2(b^2)
- \frac 43\zeta(3)
\ .
\end{aligned}
}
$$
(And the dilogarithmic values have expressions in terms of $\psi'$ computed in $1/3$.)
Numerical check using pari/gp:
? \p 50
? J = intnum( u=0, 1/2, 4/u*asin(u)^2);
? b = exp(I*Pi/3);
? J
%158 = 0.52294619213333510849118518352730354016304459174398
? 2/3*Pi*imag(dilog(b)) - 4/3*zeta(3)
%159 = 0.52294619213333510849118518352730354016304459174398
? Pi*imag(dilog(b^2)) - 4/3*zeta(3)
%160 = 0.52294619213333510849118518352730354016304459174398
Numerical check using sage:
sage: var('x,u,k');
sage: J = integral( 4/u * arcsin(u)^2, u, 0, 1/2, hold=True )
sage: J.n()
0.5229461921333352
sage: b = (1 + i*sqrt(3))/2
sage: ( 2/3 * pi * imag(dilog(b)) - 4/3 * zeta(3) ).n( digits=50 )
0.52294619213333510849118518352730354016304459174398
sage: ( pi * imag(dilog(b^2)) - 4/3 * zeta(3) ).n( digits=50 )
0.52294619213333510849118518352730354016304459174398
sage: imag( dilog(b) ).n( digits=50 )
1.0149416064096536250212025542745202859416893075303
sage: imag( 3/2 * dilog(b^2) ).n( digits=50 )
1.0149416064096536250212025542745202859416893075303
sage: ( sum( 1/(3*k+1)^2 - 1/(3*k+2)^2, k, 0, oo) * 3*sqrt(3)/4 ).n( digits=50 )
1.0149416064096536250212025542745202859416893075303
sage: sum( 1/(3*k+1)^2 - 1/(3*k+2)^2, k, 0, oo)
-4/27*pi^2 + 2/9*psi(1, 1/3)
sage: var('X');
sage: integral( X^2 / (exp(X) - 1), X)
-1/3*X^3 + X^2*log(-e^X + 1) + 2*X*dilog(e^X) - 2*polylog(3, e^X)
(Please omit this coding section, if this feels misplaced, it was done only for my calm sleep, so that i can easily double check the computations next day with better eyes.)
0.522946192133335
. If we plug this into an inverse symbolic calculator the only symbolic representation it knows about is exactly the sum you're trying to compute. $\endgroup$