Answer
$\sin^{-1} x-\sqrt {1-x^2}+C $
Work Step by Step
Since, we have $\int \sqrt{\dfrac{x+1}{1-x}} dx$
Now, the given integral can be written as:
$\int \sqrt{\dfrac{x+1}{1-x}} dx =\int \dfrac{x}{\sqrt{1-x^2}} dx +\int \dfrac{1}{\sqrt{1-x^2}} dx $
$=-\sqrt {1-x^2}+\sin^{-1} x+C$
or, $=\sin^{-1} x-\sqrt {1-x^2}+C $
