Question: $\int\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^{2} d x$
Solution:
$\int\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^{2} d x$
$=\int\left(x+\frac{1}{x}-2\right) d x$
$=\int x d x+\int \frac{1}{x} d x-2 \int 1 \cdot d x$
$=\frac{x^{2}}{2}+\log |x|-2 x+\mathrm{C}$
