Question

Evaluate the given integral.
$\int \frac{x^3-1}{x^2}dx$

Answer 1

$I=\int \frac{x^3-1}{x^2}dx$

$=\int \frac{x^3}{x^2}dx-\int \frac{1}{x^2}dx$

$=\int xdx-\int x^{-2}dx$

We know that $\int x^{n}dx=\frac{x^{n+1}}{n+1}+c$

$=\frac{x^{1+1}}{1+1}-\frac{x^{-2+1}}{-2+1}+c$ ..........where $c$ is the constant of integration

$=\frac{x^2}{2}-\frac{x^{-1}}{-1}+c$

$=\frac{x^2}{2}+\frac{1}{x}+c$