<s>[INST] Evaluation of $ \int_{0}^{1}\ln\left(\sqrt{1+x}-\sqrt{1-x}\right)dx$ 
Evaluation of $\displaystyle \int_{0}^{1}\ln\left(\sqrt{1+x}+\sqrt{1-x}\right)dx$

$\bf{My\; Try::}$ Let $$I = \frac{1}{2}\int_{0}^{1}\ln\left(\sqrt{1+x}+\sqrt{1-x}\right)^2dx = \frac{1}{2}\int_{0}^{1}\ln\left(2+2\sqrt{1-x^2}\right)dx$$
So $$I =\frac{1}{2}\ln(2)+\frac{1}{2}\int_{0}^{1}\ln\left(1+\sqrt{1-x^2}\right)\cdot 1dx$$
So $$I = \frac{\ln 2}{2}+\frac{1}{2}\left[x\cdot \ln(1+\sqrt{1-x^2})\right]_{0}^{1}+\frac{1}{2}\int_{0}^{1}\left(\frac{x^2}{1+\sqrt{1-x^2}}\right)\cdot\frac{1}{\sqrt{1-x^2}}dx$$
Now Put $x=\sin \phi\;,$ Then $dx = \cos \phi d\phi$ and changing limits, We get
So $$I = \frac{\ln 2}{2}+\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\left(1-\cos \phi\right) d\phi = \frac{\ln 2}{2}+\frac{\pi}{4}-\frac{1}{2}$$
Can we solve it any other way, If yes Then plz explain here
Thanks 
 [/INST] Integrating by parts,
$$\int\ln(\sqrt{1+x}+\sqrt{1-x})\ dx$$
$$=\ln(\sqrt{1+x}+\sqrt{1-x})\int dx-\int\left(\dfrac{d[\ln(\sqrt{1+x}+\sqrt{1-x}])}{dx}\cdot\int dx\right)dx$$
$$=x\ln(\sqrt{1+x}+\sqrt{1-x})-\int\dfrac{2x}{\sqrt{1-x^2}}dx$$
Hope you can take it from here!
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