= Solution
Choose the <Fourier transform> convention $\widehat f(k)=\int_{\mathbb R}e^{-ikx}f(x)dx$, with inverse $(2\pi)^{-1}\int e^{ikx}\widehat f(k)dk$. For absolutely integrable $f_1,f_2$, <Fubini's theorem> permits the substitution $s=x-y$ in the convolution transform:
$$
\begin{aligned}
\widehat{f_1*f_2}(k)
&=\int_{\mathbb R}\int_{\mathbb R}e^{-ikx}f_1(y)f_2(x-y)dy\,dx\\
&=\left[\int_{\mathbb R}e^{-iky}f_1(y)dy\right]\left[\int_{\mathbb R}e^{-iks}f_2(s)ds\right].
\end{aligned}
$$
This proves the <convolution theorem> in the chosen convention.
Take $q(x)=\tfrac12e^{-|x|}$. Splitting its transform at zero gives $\widehat q(k)=\tfrac12[(1+ik)^{-1}+(1-ik)^{-1}]=(1+k^2)^{-1}$. Thus the desired inverse is $q*q$. For $x\geq0$, split the convolution integral at $y=0$ and $y=x$:
$$
(q*q)(x)=\frac14\left[\frac{e^{-x}}2+xe^{-x}+\frac{e^{-x}}2\right]=\frac14(1+x)e^{-x}.
$$
The result is even, so
$$
\boxed{\mathcal F^{-1}\!\left[\frac1{(1+k^2)^2}\right](x)=\frac14(1+|x|)e^{-|x|}.}
$$
Different normalization conventions alter the transform prefactors, which is why the convention was stated explicitly.
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