<s>[INST] 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!
 [/INST] $$5x^3+19x^2+27x-3=A(x+3)(x^2+3)+B(x^2+3)+(Cx+D)(x+3)^2=$$
$$=(A+C)x^3+(3A+B+6C)x^2+(3A+6D+9C)x+(9A+3B+9D)$$
equating the coefficients next to the same power of $x$ we get following system
$$A+C=5$$
$$3A+B+6C=19$$
$$3A+6D+9C=27$$
$$9A+3B+9D=-3$$
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