Question

Calculate the following integrals.
${\int }_1^2\frac{dx}{x^3}.$

Answer 1

${\int }_1^2\frac{dx}{x^3}$

$={\left[\frac{-1}{2x^2}\right]}_1^2$

$=\frac{-1}{8}+\frac{1}{2}$

$=\frac{3}{8}$