<s>[INST] Partial Fraction Decomposition $\frac{2x^2+3x+3}{(x+1)(x^2+1)}$ How would one go about decomposing this fraction?
$$ 
2x^2+3x+3\over
(x+1)(x^2+1)
$$
Here is what I have so far:
$$
{2x^2+3x+3\over
(x+1)(x^2+1)
} =
 { A\over x+1 } + { Bx+C\over x^2+1} $$
$$
{2x^2+3x+3\over
(x+1)(x^2+1)
} =
 { A(x^2+1) } + (Bx+C)(x+1) $$
I know how to solve for A - it's 1. But how do I get B and C?
 [/INST] $$
\begin{align}
2x^2 + 3x +3 &= A(x^2 + 1) + (Bx+C)(x+1)\\
&= x^2(A+B) + x(B+C) + A+C
\end{align}
$$
By comparing the corresponding co-efficients  :
$$
A+B = 2;\\
B+C = 3;\\
C+A = 3;
$$
By solving, $A = 1$, $B = 1$, $C = 2$.
</s>