# Thread: partial fraction decomposition help

1. ## partial fraction decomposition help

$\frac{1}{(x^2+1)(x+3)^2}$

Heres my attempt:

$\frac{1}{(x^2+1)(x+3)^2}=\frac{Ax+B}{x^2+1}+\frac{ C}{x+3}+\frac{D}{(x+3)^2}$

$1=(Ax+B)(x+3)^2+C(x^2+1)(x+3)+D(x^2+1)$

$1=Ax^3+6Ax^2+9Ax+Bx^2+6Bx+9B+Cx^3+3Cx^2+Cx+3C+Dx^2 +D$

9B+3C+D=1
9A+6B+C=0
6A+B+3C+D=0
A+C=0

This is what I get for A, B C, and D
$A=\frac{1}{50}$

$B=\frac{7}{50}$

$C=- \frac{1}{50}$

$D=- \frac{1}{5}$

Where did I go wrong?

2. It seems you missed something while solving simultaneously...

Anyway, if you take from here:

$1=(Ax+B)(x+3)^2+C(x^2+1)(x+3)+D(x^2+1)$

Putting x = -3 cancels A, B and C all at once, giving:

$1 = D((-3)^2 +1)$

And

$D = \frac{1}{10}$

3. your right i made one tiny mistake which messed up the whole thing. >.<

4. I agree. One slight mistake can ruin the whole thing since there are so many 'variables' to find...

Anyway, I then used wolframalpha and got the partial fractions:

http://www.wolframalpha.com/input/?i=\frac{1}{(x^2%2B1)(x%2B3)^2}