## University Calculus: Early Transcendentals (3rd Edition)

$\sin^{-1} x-\sqrt {1-x^2}+C$
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, $=\sin^{-1} x-\sqrt {1-x^2}+C$