Z x2 4 x dx becomes d sec tan (a) 4z tan3 (b) 2z tan2 d (c) 8z sec3 (d) 4z cos (e) 2z cos2 sin d . Solution: b let x = 2 sec so that x2 4 = 4 tan2 and dx = 2 sec tan d . 2 sec tan d = 2z tan2 d . Consider the following improper integrals: z 1. Which of these integrals converges? (a) i, ii, and iii (b) i and iii only (c) i only (d) ii only (e) iii only. Since the interval of integration for ii contains [0, 1], ii also diverges . By process of elimination we rule out a d, so the answer must be e. however, we also compute. + 1(cid:21) = 1, and iii converges . The partial fraction decomposition of has the form. 1 (x2 1)2 (a) (b) (c) (d) (e) B (x 1)2 + (x 1)2 +
