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

Question

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?

Answer 1

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 $$ \begin{aligned} I &=\int\left[\frac{1}{2} x+\frac{7}{4}+\frac{1}{2 x}-\frac{25}{4(2 x+3)}\right] d x \\ &=\frac{x^{2}}{4}+\frac{7}{4} x+\frac{1}{2} \ln |x|-\frac{25}{8} \ln |2 x+3|+C \end{aligned} $$

Answer 2

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}