1. 1 Example 3 Determine whether the improper integral converges or diverges. Solution This improper integral represents an area which is unbounded in six places: at the left where x approaches - , to the left of the vertical asymptote x=-1, to the right of the vertical asymptote x=-1, to the left of the vertical asymptote x=1, to the right of the vertical asymptote x=1 and at the right where x approaches + . Therefore we write The area represented by each of these integrals is unbounded at exactly one place. (Note that there is nothing special about the choices of –2, 0 and +2. Any number less than -1 could be used instead of –2, any number between –1 and +1 could be used instead of 0 and any number greater than +1 could be used instead of +2.)

2. 2 Before proceeding, we integrate the rational function Note that x 2 -1 factors as x 2 -1 = (x+1)(x-1). We use the method of partial fractions to write Add the fractions on the right and equate the numerator of this sum with the numerator on the left. Set x=-1 in the preceding equation: 1=A(-2)+0 = -2A and A = -1/2. Set x=1 in the preceding equation: 1=0+B(2) = 2B and B = 1/2. Hence

3. 3 Then Note that as B approaches 1 from the left, the values of approach 0 from the right. Hence Thus diverges, and therefore diverges.