$∫ (x^3+x^2+x+3)/((x^2+1)(x^2+3))$

The first step I did is distributed the denominator so that I can find out if I should use synthetic division. Which after doing this I discovered that the denominator's exponent is greater therefore I do not use synthetic division

= $∫ (x^3+x^2+x+3)/(x^4+4x^2+3)$ Therfore I solve.

$∫ (x^3+x^2+x+3)/((x^2+1)(x^2+3))$ =∫ $(Ax +B)/(x^2+1) + (Cx+D)/(x^2+3)$ then I distributed $(x^2+1)(x^2+1)$ on both sides and got.

$∫ (x^3+x^2+x+3) $= $(Ax +B)(x^2+3)+( Cx+D)(x^2+1)$

Then I solve for B.
x=0

$0+0+0+3 = (A(0)+B)(0+3)+(0+0)(0+1) =$

3=B If this is right how does one find A,CD? Do I make x =0 and B =3 and e verything else 0?