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

You can split the integral into two parts
$$\int \frac{(\sqrt{1+x}+\sqrt{1-x})}{2x} \, dx=\frac{1}{2}\left[\int \frac{\sqrt{1+x}}{x}\,dx+\int \frac{\sqrt{1-x}}{x}\,dx\right].$$
Solve these separately as follows:
\begin{align*}
\int \frac{\sqrt{1+x}}{x}\,dx & =\int \frac{t^2}{(t^2-1)}\,dt && (\text{ let } 1+x=t^2) \\
& =\int \frac{t^2-1+1}{(t^2-1)}\,dt\\ 
& =\int 1 \, dt+\int \frac{1}{(t^2-1)}\,dt\\ 
& =t+\frac{1}{2}\left[\int \frac{1}{(t-1)}\,dt-\int \frac{1}{(t+1)}\,dt\right]\\
&=t+\ln\frac{|t-1|}{|t+1|}+c\\
&=\sqrt{1+x}+\ln\frac{|\sqrt{1+x}-1|}{|\sqrt{1+x}+1|}+c\\
\end{align*}
Observe that the second part is pretty much the same. If you use $x=-u$, then
$$\int \frac{\sqrt{1-x}}{x}\, dx=\int \frac{\sqrt{1+u}}{u}\, du.$$
So you can write the answer without any further computation.
$$\int \frac{\sqrt{1-x}}{x}\, dx=\sqrt{1\color{red}{-x}}+\ln\frac{|\sqrt{1\color{red}{-x}}-1|}{|\sqrt{1\color{red}{-x}}+1|}+c$$
