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Split the integral
into a sum of integrals that are each improper for only one reason.
The denominator of the integrand factors as
x2 – 2x – 15 = (x – 5)(x + 3).
Looking at this, and at the graph, we can see that the integrand is badly behaved at x = -3 and at x = 5. We need to split the integral at x = -3 and at some point between x = -3 and x = 5 (again, we may as well take x = 0). Then