<s>[INST] How to solve $\int \frac{dx}{\sqrt{1+x}-\sqrt{1-x}}$? 
How can i evaluate the following integral $$\int \frac{dx}{\sqrt{1+x}-\sqrt{1-x}}=?$$

This is taken from a definite integral where $x$ varies from $0$ to $1$.
My attempt:
multiplied by conjugate
$$\int \frac{dx}{\sqrt{1+x}-\sqrt{1-x}}=\int \frac{(\sqrt{1+x}+\sqrt{1-x})dx}{(\sqrt{1+x}-\sqrt{1-x})(\sqrt{1+x}+\sqrt{1-x})}$$
$$=\int \frac{(\sqrt{1+x}+\sqrt{1-x})dx}{1+x-1+x}$$
$$=\int \frac{(\sqrt{1+x}+\sqrt{1-x})dx}{2x}$$

*

*if i use $x=\sin^2\theta$
$$\int \frac{(\sqrt{1+\sin^2\theta}+\cos\theta)}{2\sin^2\theta}\sin2\theta\ d\theta=\int (\sqrt{1+\sin^2\theta}+\cos\theta)\cot\theta d\theta$$

*if i use $x=\tan^2\theta$
$$\int \frac{(\sec\theta-\sqrt{1-\tan^2\theta})}{2\tan^2\theta}2\tan\theta\sec^2\theta d\theta\ d\theta=\int \frac{(\sec\theta-\sqrt{1-\tan^2\theta})}{\sin\theta\cos\theta} d\theta$$
Should I use substitution $x=\sin^2\theta$ or $x=\tan^2\theta$?. I can't decide which substitution will work further. Please help me solve this integration.
Thanks
 [/INST] $$\int \frac{dx}{\sqrt{1+x}-\sqrt{1-x}}=\int \frac{dx}{\sqrt{(\sqrt{1+x}-\sqrt{1-x})^2}}$$
$$=\int \frac{dx}{\sqrt{2-2\sqrt{1-x^2}}}$$
Let $x=\sin\theta\implies dx=\cos\theta d\theta$
$$=\int \frac{\cos\theta d\theta}{\sqrt{2-2\cos\theta}}$$
$$=\int \frac{\cos\theta d\theta}{\sqrt{4\sin^2\frac{\theta}{2}}}\quad \quad \left(\because \cos\theta=1-2\sin^2\frac{\theta}{2}\right)$$
$$=\int \frac{\left(1-2\sin^2\frac{\theta}{2}\right)d\theta}{2\sin\frac{\theta}{2}}$$
$$=\int \left(\frac12\csc\frac{\theta}{2}-\sin\frac{\theta}{2}\right)\ d\theta$$
$$=\ln \left|\tan\frac{\theta}{4}\right|+2\cos\frac{\theta}{2}+C$$
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