Question

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

Answer 1

$$\begin{gathered} \int \left(\frac{x^3-1}{x^2}\right)dx \\ =\int \left(\frac{x^3}{x^2}-\frac{1}{x^2}\right)dx \\ =\int \left(x-x^{-2}\right)dx \\ =\frac{x^2}{2}-\frac{x^{-2+1}}{-2+1}+C \\ =\frac{x^2}{2}+\frac{1}{x}+C \end{gathered}$$