How to evaluate $\int \frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}} dx$? I am trying to evaluate
$$\int \frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}} dx \quad (1)$$

The typical way to confront this kind of integrals are the conjugates i.e:
$$\int \frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}} dx  = $$
$$ \int \left(\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}\right)\left(\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}\right) dx = $$
$$\int \left(\frac{(\sqrt{1+x})^2-(\sqrt{1-x})^2}{(\sqrt{1+x})^2-(\sqrt{1-x})^2}\right)\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)dx = $$
$$\int 1*\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)dx $$
That's a dead end.

I also tried other conjugate approaches (only the numerator, only the denominator etc) with no better luck.
Any ideas?
 A: Let $\sqrt{\frac{1+x}{1-x}}=t.$
Thus, $$x=\frac{t^2-1}{t^2+1},$$ $$dx=\frac{2t(t^2+1)-(t^2-1)2t}{(t^2+1)^2}dt=\frac{4tdt}{(t^2+1)^2}$$ and we need to calculate $$\int\frac{4t(t+1)}{(t-1)(t^2+1)^2}dt.$$
Now, easy to show that:
$$\int\frac{4t(t+1)}{(t-1)(t^2+1)^2}dt=\frac{2t}{t^2+1}+\ln\frac{(t-1)^2}{t^2+1}+C.$$
Can you end it now?
A: Multiplying by the conjugate is not a dead end.  I'm not sure why you multiplied by the conjugate of the numerator and denominator, you can easily evaluate this integral just by multiplying by the conjugate of the denominator:
$$I=\int \left(\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}\right)\left(\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right) dx$$
$$I=\int \frac{{(1+x)}+(1-x)+2\sqrt{1-x^2}}{2x} dx$$
$$I=\ln{|x|}+\int \frac{\sqrt{1-x^2}}{x} \; dx$$
Let $x=\sin{\theta}$ to evaluate the integral:
$$I=\ln{|x|}+\int \frac{\cos^2{\theta}}{\sin{\theta}} \; d\theta $$
$$I=\ln{|x|}+\int \csc{\theta} \; d\theta - \int \sin{\theta} \; d\theta$$
$$I=\ln{|x|}-\ln{\big | \csc{\theta}+\cot{\theta}\big |}+ \cos{\theta}+C$$
Substitute $\theta$ back for $x$:
$$I=2\ln{|x|}-\ln{\big | \sqrt{1-x^2}+1\big |} + \sqrt{1-x^2}+C$$
A: $$I=\int \frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}} \,dx $$
Rationalize:
$$\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}=\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}\times\frac{-\sqrt{1+x}-\sqrt{1-x}}{-\sqrt{1+x}-\sqrt{1-x}}$$
$$=\frac{\sqrt{(1+x)(1-x)}+1}{x}$$
$$=\frac{\sqrt{1-x^2}}{x}+\frac 1x$$
Therefore
$$I=\int\left(\frac{\sqrt{1-x^2}}{x}+\frac 1x\right)\,dx=\int\frac{\sqrt{1-x^2}}{x}\,dx+\ln|x|+C_1$$
Now, integrate the second integral by parts:
$$\int \frac{\sqrt{1-x^2}}{x^2}dx=\sqrt{1 - x^2}\left( -\frac{1}{x}\right) +  \int \frac{x}{\sqrt{1-x^2}}\left( -\frac{1}{x}\right)\,dx$$
$$=-\frac{\sqrt{1 - x^2}}{x} -  \int \frac{1}{\sqrt{1-x^2}}\,dx$$
$$= -\frac{\sqrt{1 - x^2}}{x} -  \arcsin(x) +C_2$$
Thus, letting $C=C_1+C_2$ gives
$$I=-\frac{\sqrt{1 - x^2}}{x} -  \arcsin(x)+\ln|x|+C$$
A: Substitute $x$ with $\cos{2\theta}$.
The above integral will simplify to:
$-2\int \frac{\cos\theta + \sin\theta}{\cos\theta - \sin\theta} \sin{2\theta} \mathrm d\theta $.
Then you can use integration by parts.
Note that: $-\mathrm d(\cos\theta - \sin\theta) = (\cos\theta + \sin\theta) \mathrm d\theta$.
