Hey fpb.
Do you know how the residue theorem is used to find some real valued integrals?
Method of Residues. Residue theorem. Evaluation of real definite integrals. Cauchy principal value. Summation of series.
I'd take a look there (and at some other resources) to look at how to start.
If you have been given a problem like this, whoever gave you the problem expects you to know that! Did you look at the web site chiro gave? It specifically says that the residue of f(z), at pole, z= a, of order m, is given by $\displaystyle \lim_{z\to a}\frac{1}{(m-1)!}\frac{d^{m-1} (z- a)^mf(z)}{dz^{m-1}}$.
(More fundamentally, if you were to expand a function, f, in a "Laurent series" (a power series like a Taylor series but including negative powers) around a pole of order m at z= a, it would involve negative powers down to $\displaystyle (z- a)^{-m}$. The 'residue' is the coefficient of $\displaystyle (z- a)^{-1}$.)
And, of course, the whole point of "residue" is that integral of a function around a closed path with poles of the function in the interior is the $\displaystyle 2\pi i$ times the sum of the residues at those poles. A pretty standard method for using residues to find an integral like this would be to take you path from the origin, along the positive real axis with the exception of a semicircular section of radius "r" about the pole, then out to x= R, along the semi-circular path from R to iR, then down the imaginary axis to the origin. Taking the limit as R goes to infinity, the integral on the semi-circular path goes to 0. Taking the limit as r goes to 0 gives the integral on the real axis as the integral you are trying to find. The "hard" part is probably finding the integral down the imaginary axis.
One key thing to note that hasn't really been mentioned is that given the correct contour Halls mentioned, the closed quarter circle in the 1st quadrant, there are no poles of $f(z)$ within the contour.
So the the closed contour integral is equal to 0.
Since you only want the integral along the real axis you still have to at least compute the integral along the imaginary axis to use this fact. And that's just accepting that the integral along the circular path goes to 0 as $R \to \infty$ (it does)
So the point is that for this particular problem residues don't really help you any.
below is a Mathematica sheet that works out the various parts of the integral.
The second integral is the one the problem asks for and the resulting value is $\dfrac {1}{5005}$
I don't see any way that residues would have helped.