Answer
$\frac{2}{3}(2-\frac{1}{x})^{3/2}+C$
Work Step by Step
Let u=2-$\frac{1}{x}=>du=\frac{1}{x^2}dx$
=$\int \frac{1}{x^2}\sqrt{2-\frac{1}{x}}dx$
=$\int \sqrt{u}du$
=$\int u^{1/2}du$
=$\frac{2}{3}u^{3/2}+C$
=$\frac{2}{3}(2-\frac{1}{x})^{3/2}+C$
