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?
 A: We first deal with the improper fraction by division and then resolve it into partial fractions, $$
\begin{array}{l} \displaystyle 
\frac{x^{3}+5 x^{2}+3 x+6}{x(2 x+3)}=\frac{1}{2} x+\frac{7}{4}+\frac{-
\frac{9}{4} x+6}{2 x^{2}+3 x} \\ 
\displaystyle \frac{-\frac{9}{4} x+6}{x(2 x+3)} \equiv \frac{A}{x}+\frac{B}{2 x+3} \\
 \displaystyle -\frac{9}{4} x+6 \equiv A(2 x+3)+B x
\end{array}
$$
Putting $x=0$ yields $$6=3 A \Rightarrow A=2$$
Putting $ \displaystyle x=-\frac{3}{2}$ yields $$ \displaystyle \quad \frac{27}{8}+6=-\frac{3}{2} B \Rightarrow B=-\frac{25}{4}$$
Now we can conclude that
$$
\begin{aligned}
I &=\int\left[\frac{1}{2} x+\frac{7}{4}+\frac{1}{2 x}-\frac{25}{4(2 x+3)}\right] d x \\
&=\frac{x^{2}}{4}+\frac{7}{4} x+\frac{1}{2} \ln |x|-\frac{25}{8} \ln |2 x+3|+C
\end{aligned}
$$
A: 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}
