$\int\frac{3x^2-x-6}{(x+3)(x^2+x+2)}dx$
I started with $\frac{A}{x+3}+\frac{Bx+C}{x^2+x+2}$
If my first step is right, then I get,
$A(x^2+x+2)+(Bx+C)(x+3)=3x^2-x-6$,
If my working is correct, how do I find the values of A,B,C?
Do I put x = -3 into the function to find A first?
You are off to a great start so far. The next step is to get three equations, since you have three unknowns. Expand the quantity $A(x^2+x+2)+(Bx+C)(x+3)$. You should find that $$A(x^2+x+2)+(Bx+C)(x+3) = Ax^2+Ax+2A+Bx^2+3Bx+Cx+3C$$ Next observe that all the $x^2$ terms must sum up to $3x^2$, so you can make the first equation $$Ax^2+Bx^2 = 3x^2 \\ \implies A+B = 3$$ Next, note that all the $x$ terms must add up to $-x$, so you get the second equation $$Ax+3Bx +Cx = -x \\ \implies A+3B+C = -1$$ Lastly, all the constant terms must sum to $-6$, so we get the third equation $$2A+3C = -6$$ Using these three equations, you can solve for your three unknowns, $A,B,C$.