Is this the correct procedure for Integral Partial Fraction. $∫ (x^3+x^2+x+3)/((x^2+1)(x^2+3))$
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?

We do indeed have $$(Ax +B)(x^2+3)+( Cx+D)(x^2+1) = x^3 + x^2 + x + 3$$
If you put $x=0$, you get $$(A(0) +B)(0 + 3) + (C(0) + D)(0 + 1) = 3 \iff 3B + D = 3$$
Try using $x = (-1)$ and $x = 1$, and apply the same process. 
Then use any other constant that hasn't been used yet, say $x = 2$, using the same process. That will guarantee that you have four equations to solve for four unknowns: A, B, C, D.
