The value of $\int\left(x+\frac{1}{x}\right)^{3 / 2}\left(\frac{x^2-1}{x^2}\right) d x$, is

$\frac{2}{3}\left(x+\frac{1}{x}\right)^{3 / 2}+C$ $\frac{2}{5}\left(x+\frac{1}{x}\right)^{5 / 2}+C$ $2\left(x+\frac{1}{x}\right)^{1 / 2}+C$ none of these

$\frac{2}{5}\left(x+\frac{1}{x}\right)^{5 / 2}+C$

We have, $I=\int\left(x+\frac{1}{x}\right)^{3 / 2}\left(\frac{x^2-1}{x^2}\right) d x$ $\Rightarrow I=\int\left(x+\frac{1}{x}\right)^{3 / 2} d\left(x+\frac{1}{x}\right)=\frac{2}{5}\left(x+\frac{1}{x}\right)^{5 / 2}+C$