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

Heres my attempt:

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

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

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

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

$\displaystyle A=\frac{1}{50}$

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

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

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

Where did I go wrong?