Please help me evaluate the following integral analytically $$\int_{0}^{5} \left(\arccos\left(\frac{2}{x+2}\right)\right)^2 \, dx
$$
I tried evaluating this integral using traditional techniques such as integration by parts,  u-substitution and trigonometric identities, however,
I was unsuccessful. If this integral is analytically derivable, please specify the technique or analysis framework that could yield results, including multivariable calculus, complex analysis or tricks (such as Feynman's trick). I would prefer a closed-form solution to this integral if possible, for the positive real numbers.
Thanks in advance!
 A: Starting with
$$I = \int \left[ \arccos{ \left( \frac{2}{x+2} \right)} \right]^2  \, dx,$$
change variables to
$$y = \arccos{ \left( \frac{2}{x+2} \right)}$$
to get
$$I = 2 \int y^2 \frac{\sin{y} \, dy}{\cos^2{y}},$$
which can be integrated by parts:
$$I = 2 \int y^2 \, d\left( \frac{1}{\cos{y}} \right) = \frac{2y^2}{\cos{y}} - 4J$$
where
$$J = \int \frac{y\,dy}{\cos{y}}.$$
To integrate $J$, express it as
$$J = 2 \int \frac{y\,dy}{e^{iy}+e^{-iy}}$$
and change variables to
$$z = e^{iy}$$
to get
$$J = -2 \int \log{z}\,\frac{dz}{z^2 + 1}.$$
This can then also be integrated by parts:
$$J = -2 \int (\log{z}) \, d(\arctan(z)) = -2 (\log{z}) \arctan{z} + 2K$$
where
$$K = \int \frac{\arctan{z}}{z}\,dz.$$
Now use the identity
$$\arctan{z} = -\frac{i}{2}\log\frac{1+iz}{1-iz}$$
to write
$$K = -\frac{i}{2} \left[ \int \frac{\log{(1+iz)}}{z} \, dz - \int \frac{\log{(1-iz)}}{z} \right]$$
and recognize these integrals as dilogarithms:
$$K = -\frac{i}{2} \left[ \text{Li}_2(iz) - \text{Li}_2(-iz) \right].$$
Assembling these results,
$$I = \frac{2y^2}{\cos{y}} + 4y \left[ \log{(1+ie^{iy})} - \log{(1-ie^{iy})} \right] + 4i \left[ \text{Li}_2(ie^{iy}) - \text{Li}_2(-ie^{iy}) \right]$$
where
$$y = \arccos{ \left( \frac{2}{x+2} \right)}.$$
Plotting $I$ between $x = 0$ and $x = 5$ reveals that it is in fact real and well-behaved. The value at $x = 0$ is $-7.32772$ and the value at $x = 5$ is $-1.77109$, giving $5.55664$ for the definite integral, in agreement with numeric evaluation of the original integral you asked about.
A: $$I=\int \left(\arccos\left(\frac{2}{x+2}\right)\right)^2 \, dx$$
Let $t=\frac{2}{x+2}$
$$I=2\int \frac{ \big[\cosh ^{-1}(t)\big]^2}{t^2}\,dt$$
$$I=-\frac 2t \Bigg[\cosh ^{-1}(t) \left(\cosh ^{-1}(t)+4 t \cot ^{-1}\left(e^{\cosh
   ^{-1}(t)}\right)\right)-2 i t \left(\text{Li}_2\left(i e^{-\cosh
   ^{-1}(t)}\right)-\text{Li}_2\left(-i e^{-\cosh ^{-1}(t)}\right)\right)\Bigg]$$  No way to avoid the polylogarithms.
If you want a series solution, write
$$\big[\cosh ^{-1}(t)\big]^2=\sum_{n=0}^\infty a_n\, t^n$$ where
$$a_n=\frac {(n-2)^2} {n(n-1)}a_{n-2} \quad\quad  \text{with}\quad\quad  a_0=-\frac{\pi ^2}{4}\quad \quad a_1=\pi\quad\quad a_2=1$$ Integrating termwise
$$I=\sum_{n=0}^\infty \frac {a_n}{n-1}\, t^{n-1}$$
Using the given bounds
$$J=\int_0^5 \left(\arccos\left(\frac{2}{x+2}\right)\right)^2 \, dx=-5 a_0-2 a_1 \log \left(\frac{7}{2}\right)-\frac{10 }{7}a_2+$$ $$
2\sum_{n=3}^\infty \frac {a_n}{n-1}\,\left(1-\left(\frac{2}{7}\right)^{n-1}\right)$$ Computing the partial sum after the expansion to $O(t^{k+1})$
$$\left(
\begin{array}{cc}
k & \text{partial sum} \\
 10 & 5.56163 \\
 20 & 5.55774 \\
 30 & 5.55707 \\
 40 & 5.55685 \\
 50 & 5.55676 \\
\cdots & \cdots \\
\infty &5.55664
\end{array}
\right)$$
A: If you make the substitution $\cos\theta=\frac{2}{x+2}$ then you get
$$ dx=2\sec\theta\tan\theta\,d\theta $$
and the integral becomes
$$ \int_0^{\arccos(2/7)}2\theta^2\sec\theta\tan\theta\,d\theta$$
I suspect Feynman's trick will work on this version since Wolfram gives an antiderivative. But it is not pretty:


