The value of $\int\left(x+\frac{1}{x}\right)^{3 / 2}\left(\frac{x^2-1}{x^2}\right) d x$, is
Answer & explanation
Correct answer: option 2
We have,
$I=\int\left(x+\frac{1}{x}\right)^{3 / 2}\left(\frac{x^2-1}{x^2}\right) d x$
$\Rightarrow I=\int\left(x+\frac{1}{x}\right)^{3 / 2} d\left(x+\frac{1}{x}\right)=\frac{2}{5}\left(x+\frac{1}{x}\right)^{5 / 2}+C$