Answer
$$\int \sqrt{\frac{1+x}{1-x}}dx=arcsin\,x-\sqrt{1-x^{2}}+C$$
Work Step by Step
$$\int \sqrt{\frac{1+x}{1-x}}dx=\int \sqrt{\frac{(1+x)^{2}}{1-x^{2}}}dx$$
$$let\ x=sint,\,dx=cost\,dt$$
$$\int \sqrt{\frac{(1+x)^{2}}{1-x^{2}}}dx=\int \sqrt{\frac{(1+sint)^{2}}{1-sint^{2}}}cost\,dt$$
$$=\int (1+sint)\,dt=t-cost+C$$
$$=arcsin\,x-\sqrt{1-x^{2}}+C$$
