Integration of $\int\log(\sqrt{1-x}+\sqrt{1+x}) \, dx$ Integration of 
$$\int\log\left(\sqrt{1-x}+\sqrt{1+x}\right) \, dx$$
Please help to go through this problem as i have started with putting $x$= $cos2y$.

Consider the integral
\begin{align}
I = \int\log(\sqrt{1-x}+\sqrt{1+x})dx 
\end{align}
It is seen that
\begin{align}
I &= \int \ln\left[ \sqrt{1 -x} \, \left(1 + \sqrt{\frac{1+x}{1-x}} \right) \right] dx \\
&= \frac{1}{2} \int \ln(1-x) \, dx + \int \ln\left(1 + \sqrt{\frac{1+x}{1-x}} \right) dx
\end{align}
Now make the substitution 
\begin{align}
u^{2} = \frac{1+x}{1-x}
\end{align}
to obtain
\begin{align}
I &= \frac{1}{2} [ (x-1) \ln(1-x) - x] + \int \frac{4 \ln(1+u) \, u \, du}{(1+u^{2})^{2}}.
\end{align}
Integration by parts leads to
\begin{align}
I &= \frac{1}{2} [ (x-1) \ln(1-x) - x] + \frac{2 \ln(1+u)}{1+u^{2}} - \int \frac{du}{(1+u)(1+u^{2})} \\
&= \frac{1}{2} [ (x-1) \ln(1-x) - x] + \frac{2 \ln(1+u)}{1+u^{2}} + \frac{1}{4} \ln(1 + u^{2}) - \frac{1}{2} \ln(1+u) - \frac{1}{2} \tan^{-1}(u).
\end{align}
Now reverse substitution yields
\begin{align}
I &= \frac{1}{2} [ (x-1) \ln(1-x) - x] + \left(\frac{1}{2}-x\right) \ln\left(1+ \sqrt{\frac{1 + x}{1-x}} \right) + \frac{1}{4} \ln\left(\frac{2}{1-x} \right) 
- \frac{1}{2} \tan^{-1}\left( \sqrt{\frac{1 + x}{1-x}} \right) \\
&= \frac{\ln(2)}{4} - \frac{x}{2} + \frac{1}{2} \left( x - \frac{3}{2}\right) \ln(1-x) + \left(\frac{1}{2}-x\right) \ln\left(1+ \sqrt{\frac{1 + x}{1-x}} \right) - \frac{1}{2} \tan^{-1}\left( \sqrt{\frac{1 + x}{1-x}} \right)
\end{align}
