Evaluating definite integral I'm having a hard time understanding how to compute this integral.
$$\int_1^4\frac{3x^3-2x^2+4}{x^2}\,\mathrm dx$$
The steps I do is $\dfrac{3x^4}{4} - \dfrac{2x^3}{3} + 4x$ but I don't know how to integrate the $x^2$ in the integral. I know it's suppose to be $\dfrac{x^3}3$.
Is this how the answer is supposed to look like  $$\left.\frac{\dfrac{3x^4}{4} - \dfrac{2x^3}{3} + 4x}{\dfrac{x^3}{3}}\right|_1^4?$$
The answer to this equation is $\displaystyle{39\over2}$ and I don't know how they got that answer.

$$
\int_1^4 \frac{3x^3 - 2x^2 + 4}{x^2} dx = \int_1^4 (\frac{3x^3}{x^2} - \frac{2x^2}{x^2} + \frac{4}{x^2})dx = \int_1^4 (3x - 2 + 4x^-2)dx 
$$
$$ = \frac{3}{2}x^2 - 2x - \frac{4}{x} 
$$
Now evaluate this function from 1 to 4.  First evaluating the function at 4 and then subtracting off the function evaluated at 1.
$$
(\frac{3}{2}(4^2) - 2(4) - \frac{4}{4}) - (\frac{3}{2}(1^2) - 2(1) - \frac{4}{1}) = \frac{39}{2}
$$
