I don't think he did an explicit integration; rather, he (probably) used some standard theorems about Fourier transforms and their inverses. For example, if you use a Fourier transform defined by
$$g(\theta) = ({\cal F}f)(\theta) \equiv \int_{-\infty}^{\infty} f(t) e^{-2 \pi i \theta t} \, dt, $$
then for piecewise smooth ## f ## which, together with all its derivatives, decay quickly as ##t \to \pm \infty##, we have
$$ ({\cal F}^{-1} g) (t) \equiv
\int_{-\infty}^{\infty} g(\theta) e^{+2 \pi i \theta t} \, d \theta = \frac{1}{2} \left[ f(t-) + f(t+)\right],$$
where ##f(t-)## and ##f(t+)## are the left-hand limits and right-hand limits of ##f(u)## as ##u \to t##.
In other words, at points ##t## where ##f(t)## is continuous we have ##({\cal F}^{-1} g) (t) = f(t)##, but at a point ##t## where ##f## has a "jump discontinuity" the inverse F.T. equals the average of the two ## f ## values on either side of the jump.
The ##f(t)## in your problem has jump discontinuities at ##t=-1## and ##t=1## but is continuous everywhere else (and it certainly is damped at large valus of ##|t|##), so that gives the result, up to some constants. The author uses a different definition/normalization for the Fourier transform (using ##i \omega t## instead of ##2 \pi i \theta t## in the exponent), so the inverse will have an extra multiplicative constant in it; you can work out the details for yourself.
Note: the theorem I cited above is not trivial, and may not be proved at all in an introductory course or an "engineering math" or "math for physics course". A brief proof outline is given in
https://en.wikipedia.org/wiki/Fourier_inversion_theorem
but it is very short and leaves out a lot of details for the reader to fill in if he/she can.
A longer and more detailed proof is in
https://www.ima.umn.edu/~miller/fouriertransform.pdf
but it is still not easy.