Question:
Evaluate: $\int \frac{x^{3}-1}{x^{2}} d x$
Solution:
Given, $\int \frac{x^{2}-1}{x^{2}} d x$
$=\int \frac{x^{3}}{x^{2}}-\frac{1}{x^{2}} d x$
$=\int x-\frac{1}{x^{2}} d x$
$\left[\right.$ since, $\left.\int x^{n} d x=\frac{x^{n+1}}{n+1}\right]$
$=\frac{x^{2}}{2}-\frac{x^{-2+1}}{-2+1}+c$
$=\frac{x^{2}}{2}-\frac{x^{-1}}{-1}+c$
$=\frac{x^{2}}{2}+\frac{1}{x}+c$