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.
 A: 
Generally$$\displaystyle\int\frac{p(x)}{g(x)}\,dx\neq\frac{\int p(x)\,dx}{\int g(x)\,dx}$$

That means You generally cannot Integrate numerator and denominator seperate to geet correct answer
We can separate the terms and then integrate
$$I=\int_1^4 \frac{3x^3-2x^2+4}{x^2}\,dx=\int_1^4 3x-2+\frac{4}{x^2}\, dx=\int_1^4 3x\,dx-\int_1^4 2\,dx+\int_1^4 \frac{4}{x^2}\,dx$$
$$I=\left[\frac{3x^2}{2}\right]_{1}^{4}-\left[2x\right]_{1}^{4}-\left[\frac{4}{x}\right]_{1}^{4}=\frac{39}{2}$$
A: Hint: $$\frac{3x^3-2x^2+4}{x^2} = 3x-2+\frac{4}{x^2}$$
A: Hint: $${3x^3-2x^2+4\over x^2}=3x-2+4/x^2,\quad \int x^n=\begin{cases}x^{n+1} /(n+1)\quad n\ne-1\\\ln x,\quad n=-1\end{cases}$$
A: $$
\int_1^4 \frac{3x^3 - 2x^2 + 4}{x^2} dx = \int_1^4 (\frac{3x^3}{x^2} - \frac{2x^2}{x^2} + \frac{4}{x^2})dx = \int_1^4 (3x - 2 + 4x^-2)dx 
$$
$$ = \frac{3}{2}x^2 - 2x - \frac{4}{x} 
$$
Now evaluate this function from 1 to 4.  First evaluating the function at 4 and then subtracting off the function evaluated at 1.
$$
(\frac{3}{2}(4^2) - 2(4) - \frac{4}{4}) - (\frac{3}{2}(1^2) - 2(1) - \frac{4}{1}) = \frac{39}{2}
$$
