I How did this integral come into being?

Substitute (3.4) into (3.2) to obtain
$$
F(\mathbf q) \propto \int e^{i\mathbf q \cdot \mathbf r} \int \frac{\rho(\mathbf r')}{|\mathbf r - \mathbf r'|} dv' dv = \int \rho(\mathbf r') \left( \int \frac{e^{i\mathbf q \cdot \mathbf r}}{|\mathbf r - \mathbf r'|} dv \right)dv'
$$
Compute the integral in the parentheses first by a change of variable ##\mathbf u = \mathbf r - \mathbf r'##,
$$
\int \frac{e^{i\mathbf q \cdot \mathbf r}}{|\mathbf r - \mathbf r'|} dv = e^{i\mathbf q \cdot \mathbf r'} \int \frac{e^{i\mathbf q \cdot \mathbf u}}{u} dv
$$
where now ##dv = u^2 \sin \theta du\,d\theta\,d\phi##.Here ##\theta## is the angle between ##\mathbf q## and ##\mathbf u##, thus ##\mathbf q \cdot \mathbf u = qu \cos \theta## (I think it should be cosine instead of sine as your source states). The integral above becomes
$$
e^{i\mathbf q \cdot \mathbf r'} \int e^{iqu \cos \theta} u \sin \theta du\,d\theta\,d\phi = -e^{i\mathbf q \cdot \mathbf r'} \int_0^\infty \int_0^\pi \int_0^{2\pi} e^{iqu \cos \theta} u du\,d\cos\theta\,d\phi
$$
Now solve this last integral.
Hint: Solve the integral over ##\theta## first. The integral over ##\phi## is trivial. As for the remaining integral over ##u##, employ a limiting condition by multiplying the integrand with a function of the form ##e^{-bu}##, then apply the limit ##b\to 0## after calculating the integral.

 

Question I have-(I tried working this earlier and got something similar to @blue_leaf77 calculation above), but I got that there should be a ## q^2 ## in the denominator of equation (3.5) which will be a ## q^3 ## in the denominator of equation (3.6). Did I compute it incorrectly? I'm double-checking my algebra. I'm hoping the author has (3.5) and (3.6) correct, but I'm not so sure. (If you compute blue_leaf77's last integral with the substitution ## u'=qu ##, you get a ## q^2 ## in the denominator.)

 

Yes you are right I think there should be ##1/q^2## in equation (3.5).