How to Evaluate the Integral? $\int_{0}^{1}\frac{\ln\left( \frac{x+1}{2x^2} \right)}{\sqrt{x^2+2x}}dx=\frac{\pi^2}{2}$ I am trying to find a closed form for
$$
\int_{0}^{1}\ln\left(\frac{x + 1}{2x^{2}}\right)
{{\rm d}x \over \,\sqrt{\,{x^{2} + 2x}\,}\,}.
$$
I have done trig substitution and it results in
$$
\int_{0}^{1}\ln\left(\frac{x + 1}{2x^{2}}\right)
{{\rm d}x \over \,\sqrt{\,{x^{2} + 2x}\,}\,} =
\int_{0}^{\pi/3}\sec\left(\theta\right)
\ln\left(\frac{\sec\left(\theta\right)}
{2\left[\sec\left(\theta\right) - 1\right]^{\,2}} \right){\rm d}\theta
$$
which doesn't help.
By part integration with
$\displaystyle u = \ln\left(\frac{x + 1}{2x^{2}} \right)$, $\displaystyle\,\,{\rm d}v=\frac{\displaystyle\,\,{\rm d}x}{\,\sqrt{\,{x^{2} + 2x}\,}\,}$ also makes it more complicated.
I appreciate any help on this problem.

You can "simplify" the problem using a first integration by parts to get rid of the logarithm
$$u=\log \left(\frac{x+1}{2 x^2}\right)\quad \implies \quad du=-\frac{x+2}{x^2+x}$$
$$dv=\frac{1}{\sqrt{x^2+2 x}}\quad \implies \quad v=2 \tanh ^{-1}\left(\sqrt{\frac{x}{x+2}}\right)$$
Using the bounds $u\,v=0$ and we are left with
$$I=2\int_0^1\frac{(x+2) }{x^2+x}\tanh ^{-1}\left(\sqrt{\frac{x}{x+2}}\right)dx$$
Now
$$\sqrt{\frac{x}{x+2}}=t \implies x=\frac{2 t^2}{1-t^2}\implies dx=\frac{4 t}{\left(1-t^2\right)^2}$$
$$I=8\int_0^{\frac{1}{\sqrt{3}}}\frac{\tanh ^{-1}(t)}{t-t^5}\,dt$$ Now, using partial fraction decomposition
$$\frac{1}{t-t^5}=\frac{1}{t(1-t^2)(1+t^2)}=-\frac{t}{2 \left(t^2+1\right)}-\frac{1}{4 (t-1)}-\frac{1}{4
   (t+1)}+\frac{1}{t}$$ and now would arrive a bunch of polylogarithm functions.
The simplest is
$$\int \frac{\tanh ^{-1}(t)}{t}\,dt=\frac{1}{2} (\text{Li}_2(t)-\text{Li}_2(-t))$$
Fortunately, between the given bounds everything simplify a lot.
I let you the pleasure of computing the pieces.
Edit
If we write
$$\frac{\tanh ^{-1}(t)}{t-t^5}=\frac{\tanh ^{-1}(t)}{t}+\sum_{n=1}^\infty t^{4n-1}\,\tanh ^{-1}(t)$$ we have
$$I=8\int_0^{\frac{1}{\sqrt{3}}}\frac{\tanh ^{-1}(t)}{t-t^5}\,dt=4
   \left(\text{Li}_2\left(\frac{1}{\sqrt{3}}\right)-\text{Li}_2\left
   (-\frac{1}{\sqrt{3}}\right)\right)+$$
$$\sum_{n=1}^\infty\frac{9^{-n} \log \left(2+\sqrt{3}\right)-B_{\frac{1}{3}}\left(2 n+\frac{1}{2},0\right)}{n}$$ that is to say
$$I=4
   \left(\text{Li}_2\left(\frac{1}{\sqrt{3}}\right)-\text{Li}_2\left
   (-\frac{1}{\sqrt{3}}\right)\right)+\log \left(\frac{9}{8}\right) \log \left(2+\sqrt{3}\right)-\sum_{n=1}^\infty \frac{B_{\frac{1}{3}}\left(2 n+\frac{1}{2},0\right)}{n}$$  Numerically, the sum of the first and second terms is $4.96991$ and the summation is only $0.03511$
Edit
There is something very strange : two different $CAS$ give as result
$$I =\frac{17 \pi ^2}{24}-\frac{1}{2} \left(8
   \text{Li}_2\left(2-\sqrt{3}\right)-\text{Li}_2\left(-7+4
   \sqrt{3}\right)+\log ^2\left(2-\sqrt{3}\right)\right)$$ without any further simplification while Wolfram Alpha gives $1.$ when it is written as
$$\int_0^1  \frac2{ \pi ^2} \frac{\log \left(\frac{x+1}{2 x^2}\right)}{\sqrt{x^2+2 x}}\,dx$$ Without the factor, it just return a decimal value.
Using RootApproximant[%] also behaves differently with or without the factor inside the integrand.
$$\frac{1}{2} \left(8
   \text{Li}_2\left(2-\sqrt{3}\right)-\text{Li}_2\left(-7+4
   \sqrt{3}\right)+\log ^2\left(2-\sqrt{3}\right)\right)=2.056167583560283045590519$$ Passed to the $ISC$, it is returned as $\frac{5 \pi ^2}{24}$
