Partial Fraction Decomposition Problem... half answered... $$\int \frac{5x^3+19x^2+27x-3}{(x+3)^2(x^2+3)}dx$$
I know I will be using partial fraction decomposition on this problem, at least it seems that way. so far, what I have is this:
$$\frac{5x^3+19x^2+27x-3}{(x+3)^2(x^2+3)}=\frac{A}{x+3}{}+\frac{B}{(x+3)^2}+\frac{Cx+D}{x^2+3}$$
Multiplying by the LCD : $(x+3)^2(x^2+3)$
I am left with :
$$5x^3+19x^2+27x-3=A(x+3)(x^2+3)+B(x^2+3)+(Cx+D)(x+3)^2$$
By setting $x=-3:B=-4$ 
Now is where I am running into trouble. Now that I can substitute B into the original decomposition equation, There is no value of x that will leave only one variable to solve for. Please lend me a hand you guys(and girls). Thanks!

I'll denote $f(x)$ this rational function. Here is an alternative to the expanding/equating strategy.
Limit trick: multiply both sides by $x$, compare the degrees, and let $x$ tend to $+\infty$:
$$
\lim_{x\rightarrow+\infty}xf(x)=5=A+C.
$$
Substitution trick: pick a small value in the domain of $f$. Here $0$ is perfect:
$$
f(0)=\frac{-3}{27}=\frac{A}{3}+\frac{B}{3}+\frac{D}{3}
$$
We need one more equation. Let's try $-2$, it's not too bad.
$$
f(-2)=\frac{-21}{7}=A+B+\frac{D-2C}{7}.
$$
Now, no miracle, there is a $4\times 4$ system to solve using also $B=-4$.
