Integral of a logarithmic function What is the value of integral   $$\int_0^1 \log[(1+x)^{1/2} +(1-x)^{1/2}]dx$$ ? I'm applying integration by parts but couldn't find any substitution after couple of steps
 A: Let $$I = \int^{1}_{0}\ln\bigg(\sqrt{1+x}+\sqrt{1-x}\bigg)\cdot 1dx$$
Integration by parts
$$I = \log\bigg(\sqrt{1+x}+\sqrt{1-x}\bigg)\cdot x\bigg|_{0}^{1}+\frac{1}{2}\int^{1}_{0}\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\cdot \frac{1}{\left(\sqrt{1+x}\cdot \sqrt{1-x}\right)}xdx$$
$$I = \frac{\ln2}{2}+\frac{1}{2}\int^{1}_{0}\frac{x^2}{(1+\sqrt{1-x^2})\cdot \sqrt{1-x^2}}dx$$
Put $x= \sin \theta$ Then  $dx = \cos \theta d\theta$
$$I = \frac{\ln2}{2}+\frac{1}{2}\int^{\frac{\pi}{2}}_{0}\frac{\sin^2 \theta\cdot \cos \theta}{(1+\cos \theta)\cdot \cos \theta}d\theta$$
$$I = \frac{\ln2}{2}+\frac{1}{2}\int^{\frac{\pi}{2}}_{0}(1-\cos \theta)d\theta =\frac{\ln2}{2}+\frac{\pi}{4}-\frac{1}{2}.$$
A: If $x\in(0,1)$ we have $\sqrt{1-x}+\sqrt{1+x} = \sqrt{2+2\sqrt{1-x^2}}$, hence
$$ \int_{0}^{1}\log\left(\sqrt{1+x}+\sqrt{1-x}\right)\,dx = \frac{1}{2}\left(\log 2+\int_{0}^{1}\log\left(1+\sqrt{1-x^2}\right)\,dx\right).$$
On the other hand,
$$\begin{eqnarray*} \int_{0}^{1}\log\left(1+\sqrt{1-x^2}\right)\,dx &=& \int_{0}^{\pi/2}\cos(\theta)\log(1+\cos\theta)\,d\theta\\(IBP)\quad &=&\int_{0}^{\pi/2}\frac{\sin^2\theta}{1+\cos\theta}\,d\theta\\&=&\int_{0}^{\pi/2}\left(1-\cos\theta\right)\,d\theta=\frac{\pi-2}{2}\end{eqnarray*} $$
hence:
$$ \int_{0}^{1}\log\left(\sqrt{1+x}+\sqrt{1-x}\right)\,dx = \color{red}{\frac{\pi-2+2\log 2}{4}}.$$
