Split the integral into two pieces:
$$\int_0^{+\infty} + \int_{-\infty}^0 \frac{e^{x(1 - i\omega)}}{1 + e^x}\ dx$$
Take the first integral, and collect $e^x$ at the denominator
$$\int_0^{+\infty} \frac{e^{x(1 - i\omega)}}{e^x(1 + e^{-x})} dx = \int_0^{+\infty} \frac{e^{-i\omega x}}{1 + e^{-x}}$$
Now you can use Geometric Series for the term
$$\frac{1}{1 + e^{-x}} = \sum_{k = 0}^{+\infty} (-1)^ke^{-kx}$$
Hence
$$\sum_{k = 0}^{+\infty} (-1)^k \int_0^{+\infty}e^{-x(k + i\omega)} dx = \sum_{k = 0}^{+\infty} (-1)^k \frac{1}{k + i\omega}$$
If you have a basic knowledge of special functions, this sum is called the generalized Lerch Phi function:
$$\Phi (-1,1,i \omega)$$
You can find tabulates everywhere.
For the second integral, due to the range, you don't collect anything, just use Geometric Series for
$$\frac{1}{1 + e^x} = \sum_{k = 0}^{+\infty} (-1)^k e^{kx}$$
hence
$$\sum_{k = 0}^{+\infty}(-1)^k \int_{-\infty}^0 e^{x(1 - i\omega + k)} = \sum_{k = 0}^{+\infty} \frac{(-1)^k}{1 - i\omega + k}$$
Again a Lerch Phi function:
$$\Phi (-1,1,1-i w)$$
Eventually:
$$\Phi (-1,1,i \omega) + \Phi (-1,1,1-i \omega)$$
By mathematical manipulations, using the definition and special cases, that sum is nothing but
$$\frac{1}{2} i \pi \tanh \left(\frac{\pi \omega}{2}\right)-\frac{1}{2} i \pi \coth \left(\frac{\pi \omega}{2}\right)$$
Or more easily
$$-i \pi\ \text{csch}(\pi \omega)$$
Best Answer
A good strategy for determining (Inverse) Fourier Transforms is to use Fourier Transform theorems and Fourier Transform table lookups to the fullest extent possible, before attempting integration.
Observe that
$$\dfrac{1}{(1+i\omega)^2} = i\dfrac{d}{d\omega} \dfrac{1}{1+i\omega}$$
and note that the Fourier Transform has a derivative theorem
$$\mathscr{F}\left\{x^n f(x)\right\}=\mathscr{F}\left\{x^n \mathscr{F}^{-1}\left\{ F(\omega)\right\}\right\}= i^n\dfrac{d^n}{d\omega^n} F(\omega)$$
A table lookup of Fourier Transforms yields
$$\mathscr{F}\left\{e^{-ax}H(x)\right\} = \dfrac{1}{\sqrt{2\pi}}\cdot\dfrac{1}{a+i\omega}$$
where $H(x)$ is the Heaviside unit step function. (That forward transform is easy to compute and verify for yourself.)
You can now say that
$$\begin{align*}\mathscr{F}^{-1}\left\{\dfrac{1}{(1+i\omega)^2}\right\} &= \mathscr{F}^{-1}\left\{ i\dfrac{d}{d\omega} \dfrac{1}{1+i\omega}\right\}\\ \\ &= x\mathscr{F}^{-1}\left\{ \dfrac{1}{1+i\omega}\right\}\\ \\ &= \sqrt{2\pi}x\mathscr{F}^{-1}\left\{\dfrac{1}{\sqrt{2\pi}}\cdot \dfrac{1}{1+i\omega}\right\}\\ \\ &= \sqrt{2\pi}xe^{-x}H(x)\\ \end{align*}$$