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?

$$I=\int \frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}} \,dx $$
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
$$I=-\frac{\sqrt{1 - x^2}}{x} -  \arcsin(x)+\ln|x|+C$$
