Answer
$\int\frac{x^3 + 3x-4}{x^2} dx=\frac{1}{2}x^2 + 3ln|x| +\frac{4}{x}+C $
Work Step by Step
$\int\frac{x^3 + 3x-4}{x^2} dx= \int x + \frac{3}{x} -\frac{4}{x^2} dx = \int x + \frac{3}{x}-4x^{-2}dx = \frac{1}{2}x^2 + 3ln|x| - 4\times\frac{1}{-1}x^{-1}+C=\frac{1}{2}x^2 + 3ln|x| +\frac{4}{x}+C $
