 To apply Koshy's integral formula on some contour integral, we need f of c to be analytic everywhere. But what if we can't find such a function? Then we chase the problem. We can either rewrite c or rewrite the integrand. For example, let's consider this integral where our path is the circle of radius 2 centered at the origin. We note the integrand will fail to be analytic where the denominator is zero, so solving...so the bad places are at z equal to plus or minus i. But the Koshy integral formula only allows for one bad place. So let's rewrite our contour. The upper half circle h with d toward d back to the start only includes one bad place. If we go along negative d and take the lower half circle l, we complete the path c. So our contour can be written as h plus d minus d plus l, and h plus d negative d plus l are contours that only include one bad point. So let h plus d be the upper half circle, and negative d plus l be the lower half circle. The additivity of the integral allows us to rewrite our integral as, since the only bad point in h plus d is z equal to i, we want to find f of z, where our integrand is f of z over z minus i. Since c squared plus 1 is z minus i times z plus i, we can rewrite, and comparing the two sides gives us f of z. Consequently, this first integral will be, similarly, the only bad point in negative d plus l, the lower half circle, is z equals negative i. So we can rewrite our integrand as, and we find, and so the value of the integral is zero, and we did all this work for nothing. Now math is often more than one way to solve a particular problem, so as an alternative we could find the partial fraction decomposition of our integrand, which will be, and so our integral will be, and now both integrals only contain one bad point. So for the first integral, the bad point is that z equals negative i, so we'll use f of z, one half i, which gives us, and remember one half i is a constant, so when you evaluate it at any point, you get one half i. For the second integral, the bad point is that z equal to i, and so we'll use f of z is negative one half i, and again we find, and we find that this time our integral has a value of zero, which is good because that's the value we got the previous time.