MAT137Y1 Midterm: 2008 Test 3 solution

• School: UTSG
• Department: Mathematics
• Course: MAT137Y1
• Professor: N/ A

Document Summary

Mat137y, 2008 2009 winter session, solutions to term test 3: evaluate the following integrals. (cid:90) ln8 ex . Let u = 8 + ex, then du = ex dx, so (cid:90) ln8 ex . 0 dx x2(x + 1)2 . (10%) (ii) A x x3(a +c) + x2(2a + b +c + d) + x(a + 2b) + b (x + 1)2 = x + 1 x2(x + 1)2. Ax(x + 1)2 + b(x + 1)2 +cx2(x + 1) + dx2 x2(x + 1)2. Matching coef cients, we have b = 1, a = 2, c = 2, d = 1. Therefore, (cid:90) (cid:90) dx x2(x + 1)2 = = 2ln|x| 1 x (x + 1)2 dx x + 1. +c = 2ln (cid:12)(cid:12)(cid:12)(cid:12)x + 1 x (cid:12)(cid:12)(cid:12)(cid:12) 1 x. Taking out sec2 , (cid:90) sec6 d = (cid:90) (cid:90) sec4 sec2 d = (tan2 + 1)2 sec2 d .