The integral of $x^3/(x^2+4x+3)$ I'm stumped in solving this problem. Every time I integrate by first dividing the $x^3$ by $x^2+4x+3$ and then integrating $x- \frac{4x^2-3}{x+3)(x+1)}$ using partial fractions, I keep getting the wrong answer. 

Hint
If you start doing the long division, you should arrive to $$\frac{x^3}{x^2+4 x+3}=x-4+\frac{13 x+12}{x^2+4 x+3}$$ For the last term, decompose in simple fractions and get $$\frac{x^3}{x^2+4 x+3}=x-4-\frac{1}{2 (x+1)}+\frac{27}{2 (x+3)}$$
I am sure that you can take from here.
