MATH116 Lecture Notes - Lecture 3: Jea

• School: University of Waterloo
• Department: Mathematics
• Course: MATH116
• Professor: Fiona Dunbar

Document Summary

0 e y e2y dy x dx ln(x) 4. (cid:82) sin(ln(x))dx: (cid:82) ln(2x + 1)dx. First try substitution: let s = 2x + 1 ds = 2dx. 2 (cid:90) (cid:20) ln(2x + 1)dx = ln(s)ds. (1) (cid:90) Next try integration by parts: let u = ln(s) and dv = ds. Then du = 1 and s ds and v = s (cid:90) (cid:21) (cid:20) ln(s)ds = 2 (2) (2x + 1) + c. (cid:90) (cid:21) (cid:90) (cid:90) Try integration by parts rst: let u = ln(2x + 1) and dv = dx. Then du = 2 v = x and. 2x+1dx and ln(2x + 1)dx = x ln(2x + 1) . 2x+1 we use substitution: let u = 2x + 1 then du = 2dx, x = 1. So, (cid:90) (cid:90) (cid:18) u 1 (cid:90) u (4) (5) ln(2x + 1)dx = x ln(2x + 1) 1. 2 (2x + 1) ln(2x + 1) 1.