<s>[INST] 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.
 [/INST] Hyperbolic substitution is very convenient for this integral. We can substitute $x+1=\cosh t$
$$I=\int_0^1 \ln{\left(\frac{x+1}{2x^2}\right)}\frac{dx}{\sqrt{x^2+2x}}=\int_0^{\ln{(2+\sqrt 3)}}\ln\left(\frac{\cosh t}{2(\cosh t-1)^2}\right)dt$$$$=\int_0^{\ln{(2+\sqrt 3)}}\ln\left(\frac{e^{-t}(1+e^{-2t})}{(1-e^{-t})^4}\right)dt$$
When we split the integral we get
$$I=-\frac{\ln^2(2+\sqrt 3)}{2}+\int_0^{\ln{(2+\sqrt 3)}}\ln (1+e^{-2t})dt-4\int_0^{\ln{(2+\sqrt 3)}}\ln(1-e^{-t})dt$$
For the first integral we can apply the substitution $u=-e^{-2t}$, and for the second integral we can use $v=e^{-t}$
This will allow us to use the dilogarithm.
$$I=-\frac{\ln^2(2+\sqrt 3)}{2}+\frac{1}{2}\int_{-1}^{-7+4\sqrt3}\frac{-\ln (1-u)}{u}du-4\int_1^{2-\sqrt 3}\frac{-\ln(1-v)}{v}dv$$
Using $\int \frac{-\ln(1-x)}{x}dx=\operatorname{Li}_2(x)+C$ we get
$$I=\frac{17\pi^2}{24}-\frac{\ln^2(2+\sqrt3)}{2}+\frac{1}{2}\operatorname{Li}_2(-7+4\sqrt3)-4\operatorname{Li}_2(2-\sqrt3)$$
ATTEMPT 2
Refer to C. Leibovici's answer, apply partial fractions and integration by parts
$$I=8\int_0^{\frac{1}{\sqrt3}}\frac{\operatorname{artanh} x}{x(1-x^4)}dx=8\int_0^{\frac{1}{\sqrt3}}\frac{\operatorname{artanh}x}{x}dx+2\int_0^{\frac{1}{\sqrt3}}\frac{4x^3\operatorname{artanh}x}{1-x^4}dx$$$$=8X_2\left(\frac{1}{\sqrt3}\right)+\ln(2-\sqrt3)\ln\left(\frac{8}{9}\right)+2\int_0^\frac{1}{\sqrt3} \frac{\ln(1-x^4)}{1-x^2}dx$$
$X_2(t)$ is the Legendre chi function
Using $x^4-1=(x+1)(x-1)(x+i)(x-i)$ in conjunction with logarithm properties and partial fractions, we can decompose the integral into 8 integrals of the form $\int \frac{\ln(x+a)}{x+b}dx$. 2 of them, $\int \frac{\ln(1+x)}{1+x}dx$ and $\int \frac{\ln(1+x)}{1+x}dx$ can be evaluated without dilogarithms. The other 6 each give us 2 dilogarithmic terms. This gives us 12 dilogarithms to simplify in total, and it might work
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