Fourier Transform help
There is an important trick to solve such problems. If you have a standard transform pair and you just interchange the variables \$t\$ and \$\omega\$, then you can just use your table if you know the following:

$$\mathcal{F}\{f(t)\}=F(\omega)\Longrightarrow\mathcal{F}\{F(t)\}=2\pi f(-\omega)\tag{1}$$

This is a consequence of the fact that the Fourier transform and the inverse transform are essentially identical, apart from the factor \$2\pi\$ and a minus sign in the exponent. This is exactly what you see in (1): you get a factor of \$2\pi\$ and you have to invert the frequency variable \$\omega\$.

So in your case the standard transform pair is

$$\mathcal{F}\{e^{-t}u(t)\}=\frac{1}{1+j\omega}$$

from which you get using (1)

$$\mathcal{F}\left\{\frac{1}{1+jt}\right\}=2\pi e^{\omega}u(-\omega)\tag{2}$$

PS: The mistake in your calculations is in the last derivative. You compute the derivative as if you considered the function \$e^{-\omega}\$ instead of \$e^{-|\omega|}\$. The correct derivative is

$$\pi\text{sign}(\omega)e^{-|\omega|}$$

which gives the answer

$$X(\omega)=\pi e^{-|\omega|}(1-\text{sign}(\omega))=2\pi e^{-|\omega|}u(-\omega)=2\pi e^{\omega}u(-\omega)$$

which is of course identical to (2).