Question

Find the following integral: $\int \bigg( \sqrt x - \frac{1}{\sqrt x}\bigg)^2dx$

Answer 1

$\int \bigg( \sqrt x - \frac{1}{\sqrt x}\bigg)^2dx$

= $\int \bigg( x+\frac{1}{x}-2\bigg)dx$

= $\int xdx + \int \frac{1}{x} dx-2 \int 1.dx$

= $\frac{x^2}{2}+\log\mid x \mid -2x+C$