Calculate $\int_0^1\ln{(\sqrt{1+x}+\sqrt{1-x})}dx$ I am looking for the fastest possible way to calculate:
$$\int_0^1\ln{(\sqrt{1+x}+\sqrt{1-x})}dx$$
The integral appeared on an integration bee where only a few minutes are given to calculate it, so please do not hesitate to use shortcuts in your solution, even if they involve advanced mathematics. Thank you in advance.
 A: Note 
$$ (\sqrt{1+x}+\sqrt{1-x})^2=2(1+\sqrt{1-x^2})$$
and so
\begin{eqnarray}
&&\int_0^1\ln(\sqrt{1+x}+\sqrt{1-x})dx\\
&=&\frac{1}{2}\int_0^1\ln\left[2\left(1+\sqrt{1-x^2}\right)\right]dx\\
&=&\frac{1}{2}\ln2+\frac12\int_0^1\ln\left(1+\sqrt{1-x^2}\right)dx\\
&=&\frac{1}{2}\ln2+\frac12\int_0^{\pi/2}\ln(1+\cos t)d\sin t\\
&=&\frac{1}{2}\ln2+\frac12\bigg[\ln(1+\cos t)\sin t\bigg|_0^{\pi/2}+\int_0^{\pi/2}\frac{\sin^2t}{1+\cos t}dt\bigg]\\
&=&\frac{1}{2}\ln2+\frac12\int_0^{\pi/2}(1-\cos t)dt\\
&=&\frac{1}{2}\ln2+\frac12\left(\frac{\pi}{2}-1\right)\\
\end{eqnarray}
A: Hint:
$$\int_0^1\ln{(\sqrt{1+x}+\sqrt{1-x})}dx=\int_0^1\ln\sqrt{1+x}\left({1+\dfrac{\sqrt{1-x}}{\sqrt{1+x}}}\right)dx$$
and  after separation for second integral let $x=\cos2t$.
A: $\begin{align}J=\int_0^1\ln{(\sqrt{1+x}+\sqrt{1-x})}dx\end{align}$
Perform the change of variable $\displaystyle y=\frac{1-x}{1+x}$,
$\begin{align}J&=\ln 2\int_0^1 \frac{1}{(1+x)^2}\,dx-\int_0^1 \frac{\ln(1+x)}{(1+x)^2}\,dx+2\int_0^1 \frac{\ln(1+\sqrt{x})}{(1+x)^2}\,dx\\
&=\ln 2\left[\frac{-1}{1+x}\right]_0^1+\left(\left[\frac{\ln(1+x)}{1+x}\right]_0^1-\int_0^1\frac{1}{(1+x)^2}\,dx\right)+2\int_0^1 \frac{\ln(1+\sqrt{x})}{(1+x)^2}\,dx\\
&=\left(-\frac{\ln 2}{2}+\ln 2\right)+\frac{\ln 2}{2}+\left[\frac{1}{1+x}\right]_0^1+2\int_0^1 \frac{\ln(1+\sqrt{x})}{(1+x)^2}\,dx\\
&=\ln 2-\frac{1}{2}+2\int_0^1 \frac{\ln(1+\sqrt{x})}{(1+x)^2}\,dx\\
&=\ln 2-\frac{1}{2}+2\left(\left[-\frac{\ln(1+\sqrt{x})}{1+x}\right]_0^1+\int_0^1\frac{1}{2(1+x)(1+\sqrt{x})\sqrt{x}}\,dx\right)\\
&=-\frac{1}{2}+\int_0^1\frac{1}{(1+x)(1+\sqrt{x})\sqrt{x}}\,dx\\
\end{align}$
Perform the change of variable $\displaystyle y=\sqrt{x}$,
$\begin{align}J&=-\frac{1}{2}+\int_0^1\frac{2}{(1+x^2)(1+x)}\,dx\\
&=-\frac{1}{2}+\int_0^1\left(\frac{1}{1+x}+\frac{1}{1+x^2}-\frac{x}{1+x^2}\right)\,dx\\
&=-\frac{1}{2}+\Big[\ln(1+x)\Big]_0^1+\Big[\arctan x\Big]_0^1-\frac{1}{2}\Big[\ln(1+x^2)\Big]_0^1\\
&=-\frac{1}{2}+\ln 2+\dfrac{\pi}{4}-\frac{1}{2}\ln 2\\
&=\boxed{\dfrac{\pi}{4}+\frac{1}{2}\ln 2-\frac{1}{2}}
\end{align}$
A: First note that $\sqrt a+\sqrt b=\sqrt{a+b+2\sqrt{ab}}$ and then rewrite the integral as
$$\int \ln(\sqrt{1+x}+\sqrt{1-x})dx=\int \ln \left(\left(2+2\sqrt{1-x^2}\right)^{1/2}\right)dx$$
$$=\frac{1}{2}\int \ln\left(2+2\sqrt{1-x^2}\right)dx$$
Substitute $x=\sin t \implies dx=\cos t dt$
$$=\frac{1}{2}\int \ln(2+2\cos t)\cos t dt$$
This can easily be integrated by parts:
$$=\frac{1}{2}\ln(2+2\cos t)\sin t-\int\frac{-\sin^2 t}{2+2\cos t}dt$$
$$=\frac{1}{2}\ln(2+2\cos t)\sin t-\int \frac{(\cos t+1)(\cos t-1)}{2(1+\cos t)}dt$$
$$=\frac{1}{2}\ln(2+2\cos t)\sin t-\frac{1}{2}\int(\cos t-1)dt$$
$$=\frac{1}{2}\ln(2+2\cos t)\sin t+\frac{t-\sin t}{2}+C$$
When evaluating the original integral, remember to change integration limits to $0\to\pi/2$.
