If Fourier transform of $h(t)$ is $H(f)=\dfrac{2}{(3+i2\pi f)^2}$ and Fourier transform of $x(t)$ is $X(f)=e^{-2\pi f}$, then determine the convolution of $h(t)*x(t)$.
From the theorem $$f(t)*g(t)=\mathcal{F}^{-1}[F(\omega)G(\omega)]$$ I got $$\begin{align}h(t)*x(t)&=\mathcal{F}^{-1}\left[\dfrac{2e^{-2\pi f}}{(3+i2\pi f)^2}\right]\\&=\int_{-\infty}^{\infty}\frac{2e^{-2\pi f}}{(3+i2\pi f)^2}\cdot e^{i2\pi ft}df\\&=2\int_{-\infty}^{\infty}\frac{e^{-(2\pi-i2\pi t) f}}{(3+i2\pi f)^2}df\end{align}$$ but I'm stuck with the latter integral. Help is needed. Thanks.