Finding Inverse Fourier transform: Without integrating How would you be able to do this without using integration?
Can't think of any way.

I tried doing it using the normal integration method. However this comes down to  very difficult integral that I could not solve and nor where many online calculators.
I would appreciate any help.
Many thanks
 A: If $$F(\omega)=\frac 1 {2+3i(\omega-4)}$$
then
$$(2-12i)F(\omega)+3i\omega F(\omega)=1 \tag{1}$$
$f$ and $F$ are connected by the following equalities
$$F(\omega)=\int_{\mathbb R}f(t)e^{-i\omega t}dt \text{  and the inverse Fourier transform } f(t)=\frac 1 {2\pi}\int_{\mathbb R}F(\omega)e^{i\omega t}d\omega$$
So you can verify that $\omega\mapsto i\omega F(\omega)$ is the Fourier transform of $f$. Also, $\omega\mapsto1$ is the Fourier transform of the Dirac delta. Going back to $(1)$, this implies that
$$(2-12i)f(t)+3f^\prime(t)=\delta(t)\tag{2}$$
The solution of the homogeneous equation $$(2-12i)f(t)+3f^\prime(t)=0$$
is, for some constant $C$, 
$$f(t)=Ce^{\frac{12i - 2}{3}t}$$
This inspires us to look for solutions to the non-homogeneous equation $(2)$ in the form of $$f(t)=e^{\frac{12i - 2}{3}t}H(t)$$ where $H$ is the Heaviside function, that is $H(t)=0$ if $t\leq 0$, and $H(t)=1$ if $t>0$.
Then $$
\begin{split}f^\prime(t)&=\frac{12i-2}3 e^{\frac{12i-2}3 t}H(t)+e^{\frac{12i-2}3 t}\delta(t)\\
&=\frac{12i-2}3 e^{\frac{12i-2}3 t}H(t)+\delta(t)\\
&=\frac{12i-2}3f(t)+\delta(t)
\end{split}$$
This proves that $$f(t)=e^{\frac{12i - 2}{3}t}H(t)$$ is the solution.
