Rational Decomposition 
Find the partial fractions for $\frac{3x^5+7x^4-7x^3-3x^2+x}{3x^2+10x+3}$

First, I recognize it's not a proper fraction. I also know the denominator is $(x+3)(3x+1)$. Now every time I try long division to convert to a proper fraction by hand I get $x^3-x^2+1+(\frac{-10x-3}{(x+3)(3x+1)})$, but this is wrong because I'm supposed to get $x^3-x^2+\frac{x}{(x+3)(3x+1)}$. 
I can't find the partial fraction just yet if I can't convert this into a proper fraction. When I tried I got very close, but not exactly correct at all. I wish I could show my work, but it seems displaying long division or synthetic division isn't doable on this website. 
Can anyone help? Or provide an alternative method that may not require so much long division? 

The first step you need to follow is the long division method, which will for sure give you $$\frac{3x^5+7x^4-7x^3-3x^2+x}{3x^2+10x+3}=\frac{(x^3-x^2)(3x^2+10x+3)+x}{(x+3)(3x+1)}=x^3-x^2+\frac{x}{(x+3)(3x+1)}$$Half work done!! Now we just need to break  $\frac{x}{(x+3)(3x+1)}$ into partial fractions. That's easy,$$\frac{x}{(x+3)(3x+1)}=\frac{\frac{3}{8}(3x+1)-\frac{1}{8}(x+3)}{(x+3)(3x+1)}=\frac{3}{8(x+3)}-\frac{1}{8(3x+1)}$$Thus, we have$$\frac{3x^5+7x^4-7x^3-3x^2+x}{3x^2+10x+3}=x^3-x^2+\frac{3}{8(x+3)}-\frac{1}{8(3x+1)}$$
