I need to find the inverse Fourier transform of $$g(k)=\frac{1}{\sqrt{2\pi}(ik+1)(ik+2)}$$ where $$y(t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}g(k)e^{ikt}dk.$$
Is there a trick to doing this integral? I feel like I shouldn't evaluate this directly: $$y(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{e^{ikt}}{-k^2+3ik+2}dk$$
HINT : $$\frac{1}{(ik+1)(ik+2)}= \frac{1}{ik+1}-\frac{1}{ik+2} $$ The inverse Fourier transform of $\frac{1}{ik+1}$ is $\sqrt{2\pi}e^t \text{H} (-t)$
The inverse Fourier transform of $\frac{1}{ik+2}$ is $\sqrt{2\pi}e^{2t} \text{H}(-t)$
H is the Heaviside step function.