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$$I=\int_0^1\ln (\sqrt{1-x}+\sqrt{x}+\sqrt{1+x})dx$$ First I tried like this: $$I=\int_0^1\ln (\sqrt{x}+\sqrt{1+x})dx+\int_0^1\ln (\sqrt{1-x^2}-\sqrt{x-x^2}+1)dx$$

I got a stuck,Then I tried using wolframalpha ,but the result is too complicated. how to we evaluating this integral, I don't know how to do it, and any help is welcome .

Thank you very much for your answer

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  • $\begingroup$ The answer is $$\frac{3}{5}\ln2+\frac{4}{5}\ln{(1+\sqrt2)}+\frac{\pi}{10}-\frac{1}{2}$$ $\endgroup$ – user178256 Sep 8 '17 at 9:46
  • $\begingroup$ @user178256,Thanks you for your comment,where did you get the result? $\endgroup$ – JamesJ Sep 8 '17 at 9:53
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$\begin{align}I=\int_0^1\ln (\sqrt{1-x}+\sqrt{x}+\sqrt{1+x})dx\end{align}$

Perform the change of variable $\displaystyle y=\sqrt{\frac{1-x}{1+x}}$,

$\begin{align}I=4\int_0^1 \frac{x\ln\left(\sqrt{1-x^2}+\sqrt{2}x+\sqrt{2}\right)}{(1+x^2)^2}\,dx-2\int_0^1 \frac{x\ln(1+x^2)}{(1+x^2)^2}\,dx\end{align}$

In the first integral perform the change of variable $x=\sin \theta$,

In the second integral perform the change of variable $y=x^2$,

$\begin{align}I&=4\int_0^{\frac{\pi}{2}} \frac{\sin\theta\cos\theta\ln\left(\cos \theta+\sqrt{2}\sin \theta+\sqrt{2}\right)}{(1+\sin^2\theta)^2}\,d\theta-\int_0^1\frac{\ln(1+x)}{(1+x)^2}\,dx\\ &=4\int_0^{\frac{\pi}{2}} \frac{\sin\theta\cos\theta\ln\left(\cos \theta+\sqrt{2}\sin \theta+\sqrt{2}\right)}{(1+\sin^2\theta)^2}\,d\theta+\left[\frac{\ln(1+x)}{1+x}+\frac{1}{1+x}\right]_0^1\\ &=4\int_0^{\frac{\pi}{2}} \frac{\sin\theta\cos\theta\ln\left(\cos \theta+\sqrt{2}\sin \theta+\sqrt{2}\right)}{(1+\sin^2\theta)^2}\,d\theta+\frac{1}{2}\ln 2-\frac{1}{2} \end{align}$

Perform the change of variable $\displaystyle y=\frac{\pi}{2}-x$,

$\begin{align}I&=4\int_0^{\frac{\pi}{2}} \frac{\sin\theta\cos\theta\ln\left(\sin \theta+\sqrt{2}\cos \theta+\sqrt{2}\right)}{(1+\cos^2\theta)^2}\,d\theta+\frac{1}{2}\ln 2-\frac{1}{2}\\ &=4\int_0^{\frac{\pi}{2}} \frac{\sin\theta\cos\theta\ln\left(2\cos\left(\frac{\theta}{2}\right)\left(\sin\left(\frac{\theta}{2}\right)+\sqrt{2}\cos\left(\frac{\theta}{2}\right)\right)\right)}{(1+\cos^2\theta)^2}\,d\theta+\frac{1}{2}\ln 2-\frac{1}{2}\\ &=4\ln 2\int_0^{\frac{\pi}{2}} \frac{\sin\theta\cos\theta}{(1+\cos^2\theta)^2}\,d\theta+8\int_0^{\frac{\pi}{2}}\frac{\sin\theta\cos\theta\ln\left(\cos\left(\frac{\theta}{2}\right)\right)}{(1+\cos^2\theta)^2}\,d\theta+\\ &4\int_0^{\frac{\pi}{2}}\frac{\sin\theta\cos\theta\ln\left(\sqrt{2}+\tan\left(\frac{\theta}{2}\right)\right)}{(1+\cos^2\theta)^2}\,d\theta+\frac{1}{2}\ln 2-\frac{1}{2} \end{align}$

$\begin{align}A&:=\int_0^{\frac{\pi}{2}} \frac{\sin\theta\cos\theta}{(1+\cos^2\theta)^2}\,d\theta\\ &=\left[\frac{1}{\cos(2x)+3}\right]_0^{\frac{\pi}{2}}\\ &=\frac{1}{2}-\frac{1}{4}\\ &=\boxed{\frac{1}{4}} \end{align}$

$\begin{align}B&:=\int_0^{\frac{\pi}{2}}\frac{\sin\theta\cos\theta\ln\left(\cos\left(\frac{\theta}{2}\right)\right)}{(1+\cos^2\theta)^2}\,d\theta\\ &=-\frac{1}{2}\int_0^{\frac{\pi}{2}}\frac{2\tan\left(\frac{\theta}{2}\right)\left(1-\tan^2\left(\frac{\theta}{2}\right)\right)\ln\left(1+\tan^2\left(\frac{\theta}{2}\right)\right)}{\left(1+\tan^2\left(\frac{\theta}{2}\right)\right)^2\left(1+\left(\frac{1-\tan^2\left(\frac{\theta}{2}\right)}{1+\tan^2\left(\frac{\theta}{2}\right)}\right)^2\right)^2}\,d\theta \end{align}$

Perform the change of variable $\displaystyle x=\tan\left(\frac{\theta}{2}\right)$,

$\begin{align}B&=-2\int_0^{1}\frac{x(1-x^2)\ln(1+x^2)}{(1+x^2)^3\left(1+\left(\frac{1-x^2}{1+x^2}\right)^2\right)^2}\,dx \end{align}$

Perform the change of variable $y=x^2$,

$\begin{align}B&=-\int_0^{1}\frac{(1-x)\ln(1+x)}{(1+x)^3\left(1+\left(\frac{1-x}{1+x}\right)^2\right)^2}\,dx\\ &=-\frac{1}{4}\int_0^{1}\frac{(1-x^2)\ln(1+x)}{(1+x^2)^2}\,dx\\ &=-\frac{1}{4}\left(\left[\frac{x}{1+x^2}\ln(1+x)\right]_0^1-\int_0^1\frac{x}{(1+x)(1+x^2)}\,dx\right)\\ &=-\frac{1}{8}\ln 2+\frac{1}{4}\int_0^1\frac{x}{(1+x)(1+x^2)}\,dx\\ &=-\frac{1}{8}\ln 2+\frac{1}{8}\int_0^1\left(\frac{x}{1+x^2}+\frac{1}{1+x^2}-\frac{1}{1+x}\right)\,dx\\ &=-\frac{1}{8}\ln 2+\frac{1}{16}\ln^2 2+\frac{1}{32}\pi-\frac{1}{8}\ln 2\\ &=\boxed{\frac{1}{32}\pi-\frac{3}{16}\ln 2} \end{align}$

$\begin{align}C&= \int_0^{\frac{\pi}{2}}\frac{\sin\theta\cos\theta\ln\left(\sqrt{2}+\tan\left(\frac{\theta}{2}\right)\right)}{(1+\cos^2\theta)^2}\,d\theta\\ &=\int_0^{\frac{\pi}{2}}\frac{2\tan\left(\frac{\theta}{2}\right)\left(1-\tan^2\left(\frac{\theta}{2}\right)\right)\ln\left(\sqrt{2}+\tan\left(\frac{\theta}{2}\right)\right)}{\left(1+\tan^2\left(\frac{\theta}{2}\right)\right)^2\left(1+\left(\frac{1-\tan^2\left(\frac{\theta}{2}\right)}{1+\tan^2\left(\frac{\theta}{2}\right)}\right)^2\right)^2}\,d\theta \end{align}$

Perform the change of variable $\displaystyle x=\tan\left(\frac{\theta}{2}\right)$ and observe that, $\ln\left(\sqrt{2}-1\right)=-\ln\left(\sqrt{2}+1\right)$,

$\begin{align}C&=4\int_0^1 \frac{x(1-x^2)\ln\left(\sqrt{2}+x\right)}{(1+x^2)^3\left(1+\left(\frac{1-x^2}{1+x^2}\right)^2\right)^2}\,dx\\ &=\int_0^1\frac{x(1-x^4)\ln\left(\sqrt{2}+x\right)}{(1+x^4)^2}\,dx\\ &=\frac{1}{2}\left[\frac{x^2\ln\left(\sqrt{2}+x\right)}{1+x^4}\right]_0^1-\frac{1}{2}\int_0^1 \frac{x^2}{(1+x^4)\left(\sqrt{2}+x\right)}\,dx\\ &=\frac{1}{4}\ln\left(1+\sqrt{2}\right)-\frac{1}{2}\int_0^1 \frac{x^2}{\left(x^2+\sqrt{2}x+1\right)\left(x^2-\sqrt{2}x+1\right)\left(\sqrt{2}+x\right)}\,dx\\ &=\frac{1}{4}\ln\left(1+\sqrt{2}\right)+\frac{1}{4\sqrt{2}}\int_0^1 \frac{1}{x^2+\sqrt{2}x+1}\,dx+\frac{1}{4}\int_0^1 \frac{x}{x^2+\sqrt{2}x+1}\,dx-\\ &\frac{1}{20\sqrt{2}}\int_0^1 \frac{1}{x^2-\sqrt{2}x+1}\,dx-\frac{1}{20}\int_0^1 \frac{x}{x^2-\sqrt{2}x+1}\,dx-\frac{1}{5}\int_0^1 \frac{1}{x+\sqrt{2}}\,dx\\ &=\frac{1}{4}\ln\left(1+\sqrt{2}\right)+\frac{1}{32}\pi+\left(\frac{1}{16}\ln 2+\frac{1}{8}\ln\left(1+\sqrt{2}\right)-\frac{1}{32}\pi\right)-\frac{3}{160}\pi+\\ &\left(\frac{1}{40}\ln\left(1+\sqrt{2}\right)-\frac{1}{80}\ln 2-\frac{3}{160}\pi\right)+\left(\frac{1}{10}\ln 2-\frac{1}{5}\ln\left(1+\sqrt{2}\right)\right)\\ &=\boxed{\frac{1}{5}\ln\left(1+\sqrt{2}\right)-\frac{3}{80}\pi+\frac{3}{20}\ln 2} \end{align}$

Therefore,

$\displaystyle \boxed{I=\frac{4}{5}\ln\left(1+\sqrt{2}\right)+\frac{3}{5}\ln 2+\frac{1}{10}\pi-\frac{1}{2}}$

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    $\begingroup$ * bows to you * $\endgroup$ – valer Sep 9 '17 at 22:01
  • $\begingroup$ toute mon admiration pour cette démonstration (+1) $\endgroup$ – user178256 Sep 10 '17 at 9:45
  • $\begingroup$ Merci beaucoup/Thanks alot $\endgroup$ – FDP Sep 10 '17 at 12:59

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