Q:

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?

A:

You should have
$2x^2 + 3x + 3 = A(x^2+1) + (Bx+C)(x+1)$  
Now, try suitable values of $x$ to find $A, B, C$.  For e.g. $x = -1$ will easily give you $A$.  Hint, there is another value which easily gives you $C$ after you know $A$.
In case no suitable $x$ values occur to you, just use a few simple values to get as many equations as the coefficients, and then solve a set of linear equations.

