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
 A: 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.
$$\int \frac{\sqrt{1-x}}{x}\, dx=\sqrt{1\color{red}{-x}}+\ln\frac{|\sqrt{1\color{red}{-x}}-1|}{|\sqrt{1\color{red}{-x}}+1|}+c$$
A: $$\int \frac{dx}{\sqrt{1+x}-\sqrt{1-x}}=\int \frac{dx}{\sqrt{(\sqrt{1+x}-\sqrt{1-x})^2}}$$
$$=\int \frac{dx}{\sqrt{2-2\sqrt{1-x^2}}}$$
Let $x=\sin\theta\implies dx=\cos\theta d\theta$
$$=\int \frac{\cos\theta d\theta}{\sqrt{2-2\cos\theta}}$$
$$=\int \frac{\cos\theta d\theta}{\sqrt{4\sin^2\frac{\theta}{2}}}\quad \quad \left(\because \cos\theta=1-2\sin^2\frac{\theta}{2}\right)$$
$$=\int \frac{\left(1-2\sin^2\frac{\theta}{2}\right)d\theta}{2\sin\frac{\theta}{2}}$$
$$=\int \left(\frac12\csc\frac{\theta}{2}-\sin\frac{\theta}{2}\right)\ d\theta$$
$$=\ln \left|\tan\frac{\theta}{4}\right|+2\cos\frac{\theta}{2}+C$$
A: With the change of variable $x=\sin 2t$, we have
$$\sqrt{1+x}-\sqrt{1-x}=\sqrt{\cos^2t+2\cos t\sin t+\sin^2t}-\sqrt{\cos^2t-2\cos t\sin t+\sin^2t}=2\sin t.$$
Then
$$\int\frac{dx}{\sqrt{1+x}-\sqrt{1-x}}=\int\frac{2\cos2t}{2\sin t}dt=\int\left(\frac1{\sin t}-2\sin t\right)dt
\\=\text{arcoth}(\cos t)+2\cos t+C.$$

From the biquadratic equation
$$4\cos^2t\,(1-\cos^2t)=x^2$$ you draw $\cos t$ as a function of $x$.
A: $\begin{aligned} I &:= \int \frac{d x}{\sqrt{1+x}-\sqrt{1-x}}=\int \frac{\sqrt{1+x}+\sqrt{1-x}}{2 x} d x =\frac{1}{2}\left[ \underbrace{\int \frac{\sqrt{1+x}}{x}-d x}_{J}+\underbrace{\int \frac{\sqrt{1-x}}{x} d x}_{K}\right] \end{aligned}$
$$
\begin{aligned}
J &=\int \frac{\sqrt{1+x}}{x} d x \\
&=\int \frac{1+x}{x \sqrt{1+x}} d x \\
&=\int \frac{1+x}{x} d (\sqrt{1+x}) \\
&=\int\left(\frac{1}{x}+1\right) d(\sqrt{1+x}) \\
&=\int \frac{d (\sqrt{1+x})}{(\sqrt{1+x})^{2}-1}+\sqrt{1+x} \\
&=\frac{1}{2} \ln \left|\frac{\sqrt{1+x}-1}{\sqrt{1+x}+1}\right|+\sqrt{1+x}+c_{1}
\end{aligned}
$$$$
K \stackrel{x\mapsto -x}{=} \sqrt{1-x}+\frac{1}{2} \ln \left|\frac{\sqrt{1-x}-1}{\sqrt{1-x}+1}\right|+c_{2}
$$
Now we can conclude that $$
I=\sqrt{1-x}+\sqrt{1+x}+\frac{1}{2} \ln \left| \frac{(\sqrt{1+x}-1)(\sqrt{1-x}-1)}{(\sqrt{1+x}+1)(\sqrt{1-x}+1)}\right|+C
$$
