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I would like to evaluate: $$ \int {\frac{1}{2+\sqrt{1-x}+\sqrt{1+x}}}\mathrm dx $$

$$ \frac{1}{2+\sqrt{1-x}+\sqrt{1+x}}=\frac{\sqrt{1-x}+\sqrt{1+x}-2}{2(\sqrt{1-x^2}-1)} $$

The substitution $ x \rightarrow \sin(x) $ or $ \cos(x) $ can only simplify the denominator, and $ x \rightarrow \sqrt{1+x}$ or $ \sqrt{1-x} $ is also useless... Can you help me find a useful substitution?

$$ x=\cos(2t) $$ $$ \int {\frac{1}{2+\sqrt{1-x}+\sqrt{1+x}}}\mathrm dx=-\int {\frac{\sqrt{2}\sin(t)\cos(t)}{\sqrt{2}+\sin(t)+\cos(t)}}\mathrm dt $$

$$ u=\tan(t/2) $$

$$ -4\sqrt{2}\int \frac{u(1-u^2)}{(1+u^2)^2((\sqrt{2}-1)u^2+2u+1+\sqrt{2})}\mathrm du $$

But now it looks even more complicated... ?

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    Use $1-\cos(x) = 2\sin^2(\frac{x}{2})$ and $1+\cos(x) = 2 \cos^2(\frac{x}{2})$.2011-09-24
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    I tried $u = \sqrt{1-x}$, $u^2 = 1-x$, $2u\;du = -dx$. That reduced it to an expression in which only one radical appeared: $\sqrt{2-u^2}$. Then I tried $v=\sqrt{2-u^2}$, and that transformed it to exactly the same expression with $v$ in place of $u$. I'm not sure I've seen exactly that happen before, although I wouldn't be surprised if I have.2011-09-24
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    It seems that the integral is not really simplified after using $x=\cos(2t)$ and $ u=\tan(t/2) $, as I wrote it above (if there is no mistake in my calculus)... What can I do?2011-09-24
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    I see you added my answer into your question. You say it "looks more complicated", but it fits right into the standard algorithm involving partial fractions: $\frac{\text{numerator}}{(1+u^2)^2((\sqrt{2}-1)u^2+2u+1+\sqrt{2})} = \frac{Au+B}{1+u^2} + \frac{Cu+D}{(1+u^2)^2}+\frac{Eu+F}{(\sqrt{2}-1)u^2+2u+1+\sqrt{2}}$.2011-09-24
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    ...AND: $(\sqrt{2}-1)u^2 + 2u + (\sqrt{2}+1)$ is a perfect square, since it's $\Big( \sqrt{\sqrt{2}-1}\; u + \sqrt{\sqrt{2}+1}\Big)^2$.2011-09-24
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    I see... Thanks!2011-09-24
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    I suppose I should add that since that last one is a perfect square, the partial fraction should be $\frac{E}{\sqrt{\sqrt{2}-1}\;u + \sqrt{\sqrt{2} + 1}} + \frac{F}{(\sqrt{\sqrt{2}-1}\;u + \sqrt{\sqrt{2} + 1})^2}$.2011-09-24

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