integration of $\int \frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}$

Question

$$\int \frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}$$

$$\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}=\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}*\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}=\frac{{1+x}+2\sqrt{({1+x})*({1-x})}+1-x}{{1+x}-{1+x}}=\frac{{2}+2\sqrt{1-x^2}}{2x}=\frac{2(1+\sqrt{1-x^2})}{2x}=\frac{(1+\sqrt{1-x^2})}{x}$$

$$\int \frac{(1+\sqrt{1-x^2})}{x}=\int \frac{1}{x}+\int\frac{(\sqrt{1-x^2})}{x}=\ln|x|+\int\frac{(\sqrt{1-x^2})}{x}$$

$x=\sin t$

$dx=\cos t \; dt$

$$\int\frac{(\sqrt{1-\sin^2t})}{\sin t}\cos t\;dt=\int \frac{\cos^2t}{\sin t}\;dt =\int \frac{1-\sin^2t}{\sin t}\; dt=\int \frac{1}{\sin t}\;dt -\int \sin t \; dt$$ $$=\ln\left(\tan\frac{t}{2}\right)+\cos t+c$$

How do I substitute t back to x?

Answer 1

Since $x= \sin t$, you have that $t=\arcsin x$

Answer 2

Since $x=sin(t)$ then $cos(t)= \sqrt{1-x^2}$

To deal with the $tan(t/2)$ term use the half angle formulae to put this in terms of $cos(t)$ and $sin(t)$