Question #6f533
1 Answer
Dec 15, 2016
Explanation:
Using the integration by parts formula
Let
#=uv - intvdu#
#=1/3x(2x+1)^(3/2)-1/3int(2x+1)^(3/2)dx#
#=1/3x(2x+1)^(3/2)-1/3[1/5(2x+1)^(5/2)]+C#
#=1/3x(2x+1)^(3/2)-1/15(2x+1)^(5/2)+C#
#=1/3(2x+1)^(3/2)[x-1/5(2x+1)]+C#
#=(2x+1)^(3/2)(1/5x-1/15)+C#