The integral of $x^3/(x^2+4x+3)$

Question

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.

Answer 1

$$\frac{2x^3}{(x+1)(x+3)}=\frac{[(x+3)-(x+1)]x^3}{(x+1)(x+3)}=\frac{x^3}{x+1}-\frac{x^3}{x+3}$$

For $\displaystyle\int\dfrac{x^3}{x+a}dx,$ set $x+a=h$

Answer 2

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.