Let $f(t) = e^{2t}$ if $t < 0$ and $0$ otherwise.
So the Fourier transform of $f(t)$ will be:
$$F(\omega) = \int_{-\infty}^{0}e^{2t}e^{-i\omega t}dt = \int_{-\infty}^{0}e^{(2-i\omega)t} = \frac 1{2-i\omega}$$
What am I doing wrong? In my textbook it says that the answer is: $\frac {i}{2i+\omega}$
These are the same, since $$\frac{1}{2-i\omega}=\frac{i}{i(2-i\omega)}=\frac{i}{2i-i^2\omega}=\frac{i}{2i+\omega}.$$ You have it right.