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.
 A: 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
A: Let
$$x=y^2,\quad y^2+1=t,\tag1$$
then
$$I=\int\limits_0^1 \dfrac{\ln\left(\dfrac{x+1}{2x^2}\right)}{\sqrt{x^2+2x}}dx
=\int\limits_0^1 \dfrac{\ln\left(\dfrac{y^2+1}{2y^4}\right)}{\sqrt{y^4+2y^2}}\, 2y\,\text dy
=\int\limits_1^2 \dfrac{\ln\left(\dfrac{t}{2(t-1)^2}\right)}{\sqrt{t^2-1}}\,\text dt,$$
$$I=I_1-2I_2-I_3,$$
where
\begin{cases}
I_1=\int\limits_1^2 \dfrac{\ln t\,\text dt}{\sqrt{t^2-1}}
= \dfrac{\pi^2}{24} -\dfrac12 \operatorname{Li_2}\left(\dfrac{2 - \sqrt3}4\right) - \ln^2 2-\dfrac14\arccos^2 2\\
I_2=\int\limits_1^2 \dfrac{\ln(t-1)\,\text dt}{\sqrt{t^2-1}}
=-\dfrac{\pi^2}3+2\operatorname{Li}_2(2-\sqrt3) +6\operatorname{arcsinh}^2\dfrac1{\sqrt2}
-4 \operatorname{arcsinh}\dfrac 1{\sqrt2}\,\log(1+\sqrt3)\\
I_3=\int\limits_1^2 \dfrac{\ln 2\text dt}{\sqrt{t^2-1}} = \dfrac12\ln2 \ln\left(4\sqrt{3}+7\right) = \ln2 \ln\left(2+\sqrt{3}\right)\\
\arccos 2 = i\ln(2+\sqrt3),
\end{cases}
Therefore,
$$I=\dfrac{17\pi^2}{24}-\dfrac12 \operatorname{Li_2}\left(\dfrac{2 - \sqrt3}4\right)-4\operatorname{Li}_2(2-\sqrt3)-\ln^2 2 -\ln\left(2+\sqrt{3}\right)\ln 2$$
$$+\dfrac1{4}\ln^2(2+\sqrt3) 
-12 \operatorname{arcsinh}^2 \dfrac1{\sqrt2}+8\operatorname{arcsinh}\dfrac 1{\sqrt2}\,\ln(1+\sqrt3),$$
$$I\approx4.9348022005446793094$$
(see also Wolfram Alpha calculations),
in accordance with numeric calculations.
A: 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 $\frac{5 \pi ^2}{24}$
A: Let the given integral be $I$. After the substitution $x+1=\sec t $, integrtation by parts and some tricks in the integrand, we have
$$I=\int_0^{\frac{\pi}{3}}\frac{\sin t+\tan t}{1-\cos t}\ln\left(\frac{\cos t}{1-\sin t}\right)dt.$$
After the $z=\tan\left(\frac{t}{2}\right)$ substitution,
$$I=\int_0^{\frac{1}{\large\sqrt{3}}}\frac{4\ln\left(\frac{1+z}{1-z}\right)}{z(1-z^2)(1+z^2)}dz.$$
I think this expression is obtained by Claude Leibovici and FDP.
user12030145 mentioned about this contour integral. Let $C_1=\{z=-\frac{1}{\sqrt{3}}+iy| y:\infty\rightarrow 0\}$, $C_2=\{z=\frac{1}{\sqrt{3}}+iy| 0<y<\infty\}$ and $f(z)=\frac{4\ln\left(\frac{1+z}{1-z}\right)}{z(1-z^2)(1+z^2)}$. Then, branch cut of the logarithm is outside and the only singularity inside and on the contour $C_1+[-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}]+C_2$ is $z=i$ with residue $Res_{z=i}f(z)=\pi^2$. $z=0$ is fake singularity. Hence, since $f(z)$ is even function, the integral on $C_1+C_2$ is equal to the integral on the line $\{\frac{1}{\sqrt{3}}+iy|y\in\Bbb{R}\}$ and by residue theorem we have
$$2I+\int_{\frac{1}{\sqrt{3}}-i\infty}^{\frac{1}{\sqrt{3}}+i\infty}f(z)dz=\pi^2$$
So, the problem reduces to showing that $\int_{\frac{1}{\sqrt{3}}-i\infty}^{\frac{1}{\sqrt{3}}+i\infty}f(z)dz=0.$ Nothing new so far, since I think user12030145 conjectured about such a thing. I think the real part of this integral is zero because of the Maclaurin series of $f(z)$, so it has to be zero.
A: We want to evaluate
$$
I=\int^{1}_{0}\frac{\log\left((x+1)/x^2\right)}{\sqrt{x^2+2x}}dx.
$$
We can easily write $I$ in the form
$$
I=2\int^{1}_{1/2}\frac{\log(\sqrt{t}/(1-t))}{t\sqrt{1-t^2}}dt=4\int^{1}_{1/\sqrt{2}}\log\left(\frac{t}{1-t^2}\right)\frac{dt}{t\sqrt{1-t^4}}=
$$
$$
=\tanh^{(-1)}\left(\frac{\sqrt{3}}{2}\right)\log 2-\int^{1}_{1/2}\frac{t+1}{t-1}t^{-1}\tanh^{(-1)}\left(\sqrt{1-t^2}\right)dt.
$$
Set $t=i\tan t'$, then $I$ becomes
$$
I=\tanh^{(-1)}\left(\frac{\sqrt{3}}{2}\right)\log 2-
$$
$$
-\int^{-i\infty}_{-i\tanh^{(-1)}(1/2)}\left(-2i-\cot t'+\tan t'\right)\tanh^{(-1)}(\sec t')dt'=
$$
$$
\frac{1}{2}i\pi \log(2+\sqrt{3})-\int^{-i\infty}_{-i\tanh^{(-1)}\left(\frac{1}{2}\right)}\left(-2it'-\log\left(\frac{1}{2}\sin(2t')\right)\right)\csc(t')dt'=
$$
$$
\frac{1}{2}i\pi \log(2+\sqrt{3})+2i\int^{-i\infty}_{-i\tanh^{(-1)}\left(\frac{1}{2}\right)}t'\csc(t')dt'+
$$
$$
+\int^{-i\infty}_{-i\tanh^{(-1)}\left(\frac{1}{2}\right)}\log\left(\frac{1}{2}\sin(2t')\right)\csc(t')dt'.
$$
But
$$
\int t\csc tdt=-2t\tanh^{(-1)}(e^{it})-2iLi_2(e^{it})+\frac{1}{2}iLi_2(e^{2it})
$$
and
$$
\int\log\left(\sin(t)\right)\csc(t)dt=
$$
$$
=[\log(\sin t)+\log(1-i\tan(t/2))+\log(1+i\tan(t/2))-
$$
$$
-1/2\log(\tan(t/2))]\log(\tan(t/2))+Li_2(-i\tan(t/2))+Li_2(i\tan(t/2)).
$$
Hence using the above we get the evaluation. (Note that $\int \frac{1}{\sqrt{t^2+2t}}dt$ can evaluated easily).
