Question

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

Answer 1

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

$=\int (\sqrt{x}{)}^2+{\left(\frac{1}{\sqrt{x}}\right)}^2-2\sqrt{x}\cdot \frac{1}{\sqrt{x}}dx$

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

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

$=\frac{x^2}{2}+ln|x|-2x+c$.