Jump to content

Science:Math Exam Resources/Courses/MATH101/April 2011/Question 04 (b)/Solution 3

From UBC Wiki

We apply integration by parts with u=x22 and dv=2x(1+x2)12dx. Then du=xdx and v=23(1+x2)32. Thus

∫x31+x2dx=∫x22⋅2x(1+x2)12dx=x2223(1+x2)32−∫x23(1+x2)32dx=x2223(1+x2)32−1325(1+x2)52+C=x23(1+x2)32−215(1+x2)52+C

Note: The fastest way to confirm that this is the same answer as we get in the solutions above, check that they both simplify to 1+x2(−215+115x2+15x4)+C.