I got a similar question when finding the partial fraction decomposition.
Here $i$ is imaginary number.
I set up the equation like this and I am confused here: $$ \frac{e^{ikx}}{(x-2i)(x+2i)}=\frac{Ai+B}{(x+2i)}+\frac{Ci+D}{(x-2i)} $$
$$ e^{ikx}=(Ai+B)(x-2i)+(Ci+D)(x+2i) $$
When $x = 2i$ $$ e^{-2k}=(Ci+D)(4i) $$
When $x = -2i$ $$ e^{2k}=(Ai+B)(-4i) $$
In the image example at the top, the partial fraction decomposition still includes the $e^{ikx}$ term. When you're trying to set up your own partial fractions, you're forgetting to include the $e^{ikx}$ on RHS. That's what's getting you into trouble here.
To calculate the right partial fraction decomposition, you should instead start out like $$\frac{e^{ikx}}{(x-2i)(x+2i)}=\frac{(Ai+B) e^{ikx}}{(x+2i)}+\frac{(Ci+D)e^{ikx}}{(x-2i)}$$ Or better yet, just remove the $e^{ikx}$ from both sides like $$\frac{1}{(x-2i)(x+2i)}=\frac{Ai+B}{(x+2i)}+\frac{Ci+D}{(x-2i)}$$ (but if you go this route remember to multiply all your terms by $e^{ikx}$ at the end!)