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?

You have to use what is called a coefficient matrix.
So as you said we have that:
$$∫ \frac{x^3+x^2+x+3}{(x^2+1)(x^2+3)}dx =∫ \frac{Ax +B}{x^2+1} + \frac{Cx+D}{x^2+3}dx$$
Then multiplying both sides we get that:
$$x^3+x^2+x+3 = (x^2+3)(Ax+B) + (x^2+1)(Cx+D) = Ax^3+Bx^2+3Ax+3B + Cx^3+Dx^2+Cx+D$$
Now we analyze the coefficients on the LHS and RHS for each power of $x$ and the constants.
So for example we have for $x$ coefficients:
$$3A + C = 1$$
Now, we construct our equations:
For constants:
$$3B+D = 3$$
For $x$ coefficients:
$$3A + C = 1$$
For $x^2$ coefficients:
$$B + D = 1$$
For $x^3$ coefficients:
$$A + C = 1$$
Now construct your matrix and solve for $A$, $B$, $C$, and $D$.
EDIT
You don't have to use a matrix, using simple algebra can yield your answer.
Subtracting the equations gives us $A=D=0$, $B=1$ and $C=1$
Therefore, our partial fraction decomposition is:
$$ \frac{x^3+x^2+x+3}{(x^2+1)(x^2+3)} = \frac{1}{x^2+1} + \frac{x}{x^2+3}$$
