That's the correct idea, but you're making mistakes. Here's what I got:

$\displaystyle \frac{x^2+4}{x^4+16}=\frac{Ax+B}{x^2-2\sqrt{2}x+4}+\frac{Cx+D}{x^2+2\sqrt{2}x+4}$

$\displaystyle (Ax+B)(x^2+2\sqrt{2}x+4)+(Cx+D)(x^2-2\sqrt{2}x+4)=x^2+4$

$\displaystyle (A+C)x^3 + (2\sqrt{2}A - 2\sqrt{2}C + B + D)x^2 +$

$\displaystyle (2\sqrt{2}B - 2\sqrt{2}D + 4A + 4C)x + (4B+4D) = x^2+4$

We can equate all four coefficients:

$\displaystyle A+C=0$

$\displaystyle 2\sqrt{2}A - 2\sqrt{2}C + B + D=1$

$\displaystyle 2\sqrt{2}B - 2\sqrt{2}D + 4A + 4C=0$

$\displaystyle 4B+4D=4$

Next you can solve for A, B, C, and D - four equations in four unknowns. You should get $\displaystyle A=C=0$ and $\displaystyle B=D=\frac{1}{2}$. Now you can integrate the two functions - do you know how?

- Hollywood