$\int_0^\infty \frac{\ln\left(1+x-\sqrt{2x}\right)}{1+x^2}\,dx$ Trying to compute Integral $\int_{0}^{\frac{\pi}4} \ln(\sin{x}+\cos{x}+\sqrt{\sin{2x}})dx$
I was facing:
\begin{align}J=\int_0^\infty \frac{\ln\left(1+x-\sqrt{2x}\right)}{1+x^2}\,dx\end{align}
I want to prove that $\displaystyle J=0$, or equivalently, that,
 \begin{align}\int_0^1 \frac{\ln\left(1+x-\sqrt{2x}\right)}{1+x^2}\,dx=-\dfrac{1}{2}\text{G}\end{align}
$\text{G}$ being the Catalan constant.
Read carefully please. 
I know, using so-called Feynman's trick, how to prove this.
I would like to obtain a proof, using only integration by parts and change of variable in simple integrals (that is, no multiple integrals)
I don't know, if, under these restrictions, such computation is possible.
NB:
You're probably wondering what is the link between :
$\int_{0}^{\frac{\pi}4} \ln(\sin{x}+\cos{x}+\sqrt{\sin{2x}})dx$ and $J$.
The link is, for $x\in\left[0;\frac{\pi}{2}\right]$, $(\sin{x}+\cos{x}+\sqrt{\sin{2x}})(\sin{x}+\cos{x}-\sqrt{\sin{2x}})=1$ and $\sin(2x)=2\sin x\cos x$
 A: Sketch of a possible proof:
By roots of unity and  partial fractions one may write
$$I=\int_0^1  \frac{\log{(1+x-\sqrt{2x})}}{1+x^2} dx = 2\int_0^1 \frac{x}{1+x^4}
\log{(1+x^2-\sqrt{2}{x} )} $$
$$ = 2\int_0^1 \frac{dx \quad x}{(x+w)(x-w)(1+1/w)(x-1/w)}\log{((x-w)(x-1/w))} \quad, \quad w=e^{i\pi/4}$$
$$ = \frac{w^2}{(w^2+1)(w^2-1)}\int_0^1 \big(\log{(x-w)} + \log{(x+w)} \big)\Big(\frac{1}{x+w}+\frac{1}{x-w} - \frac{1}{x+1/w} - \frac{1}{x-1/w}\Big) $$
The factor in front of the integral is $-i/2.$ Integrals of the form 
$$ \int dx \frac{ \log{(x-a)} }{x-b} = \log{(x-a)}\log{(1+(x-a)/(b-a))} + \text{Li}_2((x-a)/(b-a)) $$
and when $b=a,$ the integral is a square of a logarithm, that is, there is no dilogarithm involved.  There's a lot of symmetry involved, so maybe the mess can be simplified.  There are also dilog identities which can be invoked.  I suspect this to work because it is known that $\text{Li}_2(i) = i G - \pi^2/48.$  Using dilog identities is probably not an easy way to get the answer.
A: We can avoid complicated integrations, at the price of doing a little complex analysis.
Consider
$$F(z)=\int^\infty_0\frac{\ln(\sqrt t-z)}{t^2+1}dt\qquad z\in\mathbb C\setminus\mathbb R^+$$
where the principal logarithm is used.
One can show without difficulties that $F$ is holomorphic on its domain.

1. Proof of  $F(0)=0$
$$\begin{align}
F(0)
&=\frac12\int^\infty_0\frac{\ln t}{t^2+1}dt \\
&=\frac12\int^1_0\frac{\ln t}{t^2+1}dt+\frac12\int^\infty_1\frac{\ln t}{t^2+1}dt \\
&=\frac12\int^1_0\frac{\ln t}{t^2+1}dt+\frac12\int^1_0\frac{-\ln t}{t^2+1}dt \qquad (1)\\
&=0
\end{align}
$$
$(1)$: the substitution $t\mapsto\frac1t$ is applied on the second integral.

2. An identity for $\text{Im } F$
By the substitution $t\mapsto\frac1t$,
$$\begin{align}
F(z)
&=\int^\infty_0\frac{\ln\left(\frac1{\sqrt t}-z\right)}{t^2+1}dt \\
&=\int^\infty_0\frac{\ln\left[\frac{-z}{\sqrt t}(\sqrt t-\frac1z)\right]}{t^2+1}dt \\
&=\int^\infty_0\frac{\ln(-z)}{t^2+1}dt-\frac12\int^\infty_0\frac{\ln(t)}{t^2+1}dt+F\left(\frac1z\right) \\
&=\frac{\pi}{2}\ln(-z)+F\left(\frac1z\right)
\end{align}
$$
If $|z|=1$, $\frac1z=\bar z$. Also, from the integral representation of $F$, $F(\bar z)=\overline{F(z)}$.
Hence, for $z\ne 1$ on the unit circle,
$$F(z)=\frac{\pi}{2}\ln\left|-z\right|+\frac{\pi}{2}\arg(-z)+\overline{F(z)}=\frac{\pi}{2}\arg(-z)+\overline{F(z)}$$
$$F(z)-\overline{F(z)}=\frac{\pi}{2}\arg(-z)\implies \text{Im }F(z)=\frac{\pi}{4}\arg(-z)$$

3. Proof of $F(z)=\frac{\pi}{4}\ln(-z)$
The result $\text{Im }F(z)=\frac{\pi}{4}\arg(-z)$ is already a big hint at $F(z)=\frac{\pi}{4}\ln(-z)$. Indeed, $\text{Im}\big(F(z)-\frac{\pi}{4}\ln(-z)\big)=0$ in $\mathbb C\setminus\mathbb R^+$, thus the desired equality follows by Identity Theorem.

4. Evaluation of the integral
By noticing $$1+x-\sqrt{2x}=(\sqrt x-e^{i\pi/4})(\sqrt x-e^{-i\pi/4})$$
we have
$$\begin{align}
J&=\int_0^\infty \frac{\ln\left(1+x-\sqrt{2x}\right)}{1+x^2}\,dx \\
&=F(e^{i\pi/4})+F(e^{-i\pi/4}) \\
&=\frac{\pi}{4}\left[\ln(-e^{i\pi/4})+\ln(-e^{-i\pi/4})\right] \\
&=\frac{\pi}{4}\left[\ln(e^{-3\pi i/4})+\ln(e^{3\pi i/4})\right] \\
&=\frac{\pi}{4}\left(-\frac{3\pi i}{4}+\frac{3\pi i}{4}\right) \\
&\color{red}{=0}\qquad\blacksquare
\end{align}
$$
A: Let $
 K=\int_0^\infty \frac{\ln(1+x+\sqrt{2x})}{x^2+1}dx$
and note that $J+K=\int_0^\infty \frac{\ln(x^2+1)}{x^2+1}\ {dx}$
\begin{align}
J-K=&\int_{0}^\infty \frac{\ln({1+x-\sqrt{2x}})-\ln({1+x+\sqrt{2x}})}{x^2+1}\ \overset{x=t^2}{dx}\\
=&\int_{-\infty}^\infty \frac{t\ln({1+t^2-\sqrt{2}t})-t\ln({1+t^2+\sqrt{2}t})}{t^4+1}\ dt\\
=& \ \frac1{\sqrt2}\int_{-\infty}^\infty \overset{x=\sqrt2t-1}{\frac{\ln(t^2-\sqrt2t+1)}{t^2-\sqrt2t+1}}+ \overset{x=\sqrt2t+1}{ \frac{\ln(t^2+\sqrt2t+1)}{t^2+\sqrt2t+1}}-\overset{x={(t-t^{-1})}/{\sqrt2}}{\frac{(1+t^2)\ln(t^4+1)}{t^4+1}}\ dt\\
=& \ 2\int_{-\infty}^\infty \frac{\ln \frac{x^2+1}2}{x^2+1}dx
 - \int_{-\infty}^\infty \frac{\ln (2(x^2+1))}{x^2+1}dx
 =2 \int_{0}^\infty \frac{\ln \frac{x^2+1}8}{x^2+1}dx
\end{align}
As a result
\begin{align}
J =\frac12 (J+K) + \frac12 (J-K)
 =\ \frac32 \int_0^\infty  \frac{\ln \frac{x^2+1}4}{x^2+1}dx=0
\end{align}
A: I will focus on the integral $\int_0^1 \frac{\ln(1+x-\sqrt{2x})}{1+x^2}dx$.
I am not sure, if it is straightforward enough, but I used only simple substitutions and some of your hints. First split up the integral:
$$\int_0^1 \frac{\ln(1+x-\sqrt{2x})}{1+x^2}dx=\int_0^1 \frac{\ln(\sqrt{x}(1/\sqrt{x}+\sqrt{x}-\sqrt{2})}{1+x^2}dx=$$
$$=\frac{1}{2}\int_0^1\frac{\ln x}{1+x^2}dx + \int_0^1\frac{\ln(\sqrt{x}-\sqrt{2}+1/\sqrt{x})}{1+x^2}dx$$
And investigate the two integrals separately.


*

*The first integral is by definition of Catalan constant equal to $-\frac{1}{2}G$.

*The second integral can be broken up further:
Substitute $$\arctan x = u$$ we get
$$\int_0^{\pi/4}\ln (\sqrt{\tan u}+\sqrt{cotan u}-\sqrt{2})du=$$
$$ = \int_0^{\pi/4}\ln\left(\sqrt{\frac{\sin u}{\cos u}} + \sqrt{\frac{\cos u}{\sin u}} - \sqrt{2}\right)du$$
$$ = \int_0^{\pi/4}\ln\frac{\sin u + \cos u - \sqrt{\sin 2u}}{\sqrt{\sin u\cos u}}du$$ 
Next trick is to use the identity (which you mention), i.e., 
$$(\sin u + \cos u - \sqrt{\sin 2u})(\sin u + \cos u + \sqrt{\sin 2u}) = 1.$$
$$\int_0^{\pi/4}\ln\frac{\sin u + \cos u - \sqrt{\sin 2u}}{\sqrt{\sin u\cos u}}du = -\int_0^{\pi/4}\ln\left(\sqrt{\cos u \sin u}(\sin u + \cos u + \sqrt{\sin 2u})\right)du = $$
$$ = -\frac{1}{2}\int_0^{\pi/4}\ln(\sin u\cos u)du - \int_0^{\pi/4}\ln(\sin u + \cos u + \sqrt{\sin 2u})du.$$
The first integral:
$$\int_0^{\pi/4}\ln\left(\sin u \cos u\right)du = \int_0^{\pi/4}\ln\sin u du+\int_0^{\pi/4}\ln \cos u du = \int_0^{\pi/4}\ln\sin u du+\int_0^{\pi/4}\ln \sin (\pi/2 - u)du = \int_0^{\pi/4}\ln\sin u du+\int_{\pi/4}^{\pi/2}\ln\sin u du = \int_0^{\pi/2}\ln\sin u du.$$
It is easy to show (e.g. computing-the-integral-of-log-sin-x) that $\int_0^{\pi/2}\ln\sin u du = -\frac{\pi}{2}\ln 2$.
The second part of the integral is mentioned in your link above (integral-int-0-frac-pi4-ln-sinx-cosx-sqrt-sin2xdx) and is equal to:
$$\int_0^{\pi/4}\ln(\sin u + \cos u + \sqrt{\sin 2u})du = \frac{\pi}{4}\ln2$$
Putting everything together leads to (as expected):
$$\int_0^1 \frac{\ln(1+x-\sqrt{2x})}{1+x^2}dx=$$
$$=\frac{1}{2}\int_0^1\frac{\ln x}{1+x^2}dx + \int_0^1\frac{\ln(\sqrt{x}-\sqrt{2}+1/\sqrt{x})}{1+x^2}dx = -\frac{1}{2}G + \frac{1}{2}\frac{\pi}{2}\ln 2 - \frac{\pi}{4}\ln 2 = -\frac{1}{2}G.\ \square$$
