<s>[INST] $\int_0^\infty \frac{\ln\left(1+x-\sqrt{2x}\right)}{1+x^2}\,dx$ 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 $\sin(2x)=2\sin x\cos x$
 [/INST] 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}
</s>