How to solve the partial fraction decomposition $\frac{x^3+5x^2+3x+6}{2x^2+3x}$. I have the following integral:

$$\int\frac{x^3+5x^2+3x+6}{2x^2+3x}dx$$

I'm trying to use partial fraction decomposition but I'm getting stuck at the following formula:
$$\int\frac{(x+6)(1+5x+x^2)}{x(2x+3)}-\frac{x+27}{2x+3}dx$$
I can't necessarily guarantee that this problem has a nice solution. But is there a way to circumvent the fact that $(1+5x+x^2)$ has a somewhat messy root? (I'm getting $\frac{5+-\sqrt{21}}{2}$). Or do I have to use that value?

I will let you work out the detail, but the solution should be: 
$ \cfrac{x^3 + 5x^2 + 3x + 6}{2x^2 + 3x} =\cfrac{7}{4} + \cfrac{2}{x} + \cfrac{x}{2} - \cfrac{25}{4(2x+3)} $
Now, you can evaluate the integral: 
\begin{equation} \int  \cfrac{x^3 + 5x^2 + 3x + 6}{2x^2 + 3x} dx  = \int \cfrac{7}{4} dx + \int \cfrac{2}{x} dx + \int \cfrac{x}{2} dx - \int \cfrac{25}{4(2x+3)} dx \end{equation}
\begin{equation} = \cfrac{7}{4}x + 2\ln(x) + \frac{x^2}{4} -  \cfrac{25}{8}\ln(2x + 3) \end{equation}
