Answer
$$\int \frac{3x^3-9}{x^2}dx
=\frac{3}{2}x^2+\frac{9}{x}+c.$$
Work Step by Step
We have
$$\int \frac{3x^3-9}{x^2}dx=\int 3x-9x^{-2}dx=\frac{3}{2}x^2+9x^{-1}+c\\
=\frac{3}{2}x^2+\frac{9}{x}+c.$$
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