The strategy for all these problems is the same: you use Cauchy's theorem which says that the integral of an analytic function round a closed contour is equal to times the sum of the residues of the function at the poles inside the contour.

For these three problems, the contour to be used is the same in each case. It consists of the interval [–R,R] of the real axis (where R is a large positive number), together with the semicircle of radius R in the upper half-plane. You should have been given a theorem saying that for certain analytic functions, the integral round the semicircular part of the contour goes to 0 as , so that the integral along the real line is equal to times the sum of the residues.

As an example, here's how to do problem b). The function has double poles at the points where the denominator vanishes. These are the points . Only one of those points, , is inside the contour. Using the limit formula for higher order poles, you can check that the residue at this point is . Now multiply by to get the answer for the integral round the contour. Finally, check that the function satisfies the condition for the integral round the semicircle to go to 0, and you have the answer for .

Do the other problems in the same way. For a), you will need to say that the function is even, so its integral from 0 to is half the integral over the whole line. For c), you can't use the function as it stands, because it does not satisfy the condition for the integral round the semicircle to go to 0. So replace it by the function and take the real part of the answer. (In fact, the answer should turn out to be real anyway, because its imaginary part would give the value of the integral , and that is 0 because the function is odd.)