<s>[INST] 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?
 [/INST] As the highest power of numerator is greater that that of denominator
using Partial Fraction Decomposition,
$$ \dfrac{2x^3+3x+1}{(x+1)^2}=2x+A+\frac B{x+1}+\frac C{(x+1)^2}$$
On multiplication by $(x+1)^2$ we get $$2x^3+3x+1=(x+1)^2(2x+A)+B(x+1)+C$$
$$\implies 2x^3+3x+1=2x^3+x^2(4+A)+x(2+2A+B)+A+B+C$$ 
Now compare the coefficients of the different powers of $x$ to find $A,B,C$
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