Stuck on decomposing partial fraction I want to decompose the following, and I think got stuck in the thick of it $$ \dfrac{2x^3+3x+1}{(x+1)^2}$$
I tried like this:
OK, after advice from @Daniel Fischer  and @lab bhattacharjee I decided to use division:
first separated the equation as  $[\dfrac{2x^3+3x+1}{(x+1)}]\times \dfrac{1}{(x+1)} $
The I used polynomial division on  $[\dfrac{2x^3+3x+1}{(x+1)}]$ to get $ [2x^2-2x+3- \dfrac{2}{x+1}] \times \dfrac{1}{(x+1)} $
with me, $ [2x^2-2x+3- \dfrac{2}{x+1}] \times \dfrac{1}{(x+1)}$ reduces to $\dfrac{2x^2-2x+3}{x-1}-2  $
Is this still on course, so far?

Your polynomial long division is wrong: instead of factoring out a $1/(x+1)$, just divide
$$(2x^3 + 3x + 1)\div (x^2 + 2x + 1) = 2x - 4 + \frac{r(x)}{(x+1)^2}.$$
Now use partial fractions on the remainder term: 
$$r(x) = A(x+1) + B = (A)x + (A+B)$$
and equate the coefficients on both sides to get a system in $A$ and $B$. Solve the system.

 Answer: $r(x) = 9x + 5$, $A = 9$, $B = -4$

