Fourier transform $\frac{1}{t^2+6t+10}$ I want to solve the equation $$\int^{\infty}_{-\infty}\frac{u(s)}{1+4(t-s)^2} ds= \frac{1}{t^2+6t+10}$$ using $$\mathcal{F}[\frac{1}{1+t^2}] = \pi e^{-|w|} $$
The left hand side is a convolution which I can solve using the hint. However I don't know how I Fourier Transform the right side. 
When I fractionize the term, I get $$\frac{1}{t+(3+i)}*\frac{1}{t+(3-i)}$$
Now I don't know were to go with it. I can see that the term is an absolute square of either of the terms. My idea is to continue with the parseval identity, thought I still don't know exactly how to use it. Any help would be appreciated. 

Let $v(s)=\frac{1}{1+4s^2}$. Your equation can be written as:
$$ (u*v)(t) = \frac{1}{t^2+6t+10}\tag{1} $$
and by switching to Fourier transforms (defined through $\mathcal{F}(u)(\xi)=\widehat{u}(\xi)=\int_{-\infty}^{+\infty}u(x)e^{-2\pi i x \xi}\,dx$) we have:
$$ \widehat{u}(\xi)\cdot \frac{\pi}{2} e^{-\pi|\xi|}= \widehat{u}(\xi) \cdot \widehat{v}(\xi) = \mathcal{F}\left(\frac{1}{(t+3)^2+1}\right)(\xi)=\pi e^{-2\pi|\xi|}\cdot e^{6\pi i\xi}\tag{2}$$
that implies:
$$ \widehat{u}(\xi) = 2e^{-\pi|\xi|}\cdot e^{6\pi i\xi} \tag{3}$$
and by Fourier inversion:
$$ \color{red}{u(t)} = \frac{2}{\pi}\cdot \frac{2}{1+(2\cdot(t+3))^2} = \color{red}{\frac{4}{\pi}\cdot\frac{1}{1+(2t+6)^2}}.\tag{4}$$
