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
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.