Question
Find the integral: $\int \frac{x^{3}-1}{x^{2}} d x$

Answer

We have
$\int \frac{x^{3}-1}{x^{2}} d x=\int x d x-\int x^{-2} d x$
= $\left(\frac{x^{1+1}}{1+1}+C_{1}\right)-\left(\frac{x^{-2+1}}{-2+1}+C_{2}\right); C_1, C_2$ are constants of integration
= $\frac{x^{2}}{2}+C_{1}-\frac{x^{-1}}{-1}-C_{2}$ = $\frac{x^{2}}{2}+\frac{1}{x}+C_{1}-C_{2}$
= $\frac{x^{2}}{2}+\frac{1}{x}+C$ where $C = C_1 - C_2$ is another constant of integration

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