TITLE: How to evaluate $\int_0^1\ln (\sqrt{1-x}+\sqrt{x}+\sqrt{1+x})dx$
QUESTION [6 upvotes]: Evaluate
$$I=\int_0^1\ln (\sqrt{1-x}+\sqrt{x}+\sqrt{1+x})dx$$
First, I tried like this
$$I=\int_0^1\ln (\sqrt{x}+\sqrt{1+x})dx+\int_0^1\ln (\sqrt{1-x^2}-\sqrt{x-x^2}+1)dx$$
and I got a stuck. Then I tried using wolframalpha. But, the result is too complicated. How to we evaluate this integral. I don't know how to do it. Any help is welcome .
Thank you very much for your answer

REPLY [5 votes]: Integrate by parts
\begin{align}
I=&\int_0^1\ln (\sqrt{1-x}+\sqrt{x}+\sqrt{1+x})dx\\
=& \ln(1+\sqrt2) -\int_0^1 x\ d[\ln (\sqrt{1-x}+\sqrt{x}+\sqrt{1+x})]\\
 =& \ln(1+\sqrt2) +\frac12 \int_0^1\left(\frac{\sqrt{1-x}+\sqrt{1+x}} {\sqrt{1-x^2}(\sqrt{1-x}+\sqrt{x}+\sqrt{1+x}) }-1\right)dx\\
 =& \ln(1+\sqrt2) - \frac12 +\frac12J\tag1
\end{align}
where, with $x= \frac{2t^2}{1+t^4}$ and partial decomposition
\begin{align}
J& =\int_0^1 \frac{\sqrt{1-x}+\sqrt{1+x}} {\sqrt{1-x^2}(\sqrt{1-x}+\sqrt{x}+\sqrt{1+x}) }dx\\
&= \int_0^1\frac{8t}{(2+\sqrt2 t)(1+t^4)}dt
=\frac\pi5+\frac65\ln2-\frac25\ln(1+\sqrt2)
\end{align}
Plug into (1) to obtain
$$I=-\frac12 +\frac\pi{10}+\frac35\ln2+\frac45\ln(1+\sqrt2)
$$