How do you integrate #int (x+1)^2ln3x# by integration by parts method?

1 Answer
Alberto P.
Nov 3, 2016

#(x+1)^3/3ln3x-1/3(x^3/3+3/2x^2+3x+lnabs(x))+c#

Explanation:

#f^'(x)=(x+1)^2 =>f(x)=(x+1)^3/3#
#g(x)=ln3x => g'(x)=1/x#

#int(x+1)^2ln3x\ dx=(x+1)^3/3ln3x-1/3int(x+1)^3\ 1/x\ dx=#
#=(x+1)^3/3ln3x-1/3int(x^3+3x^2+3x+1)/x\ dx=#
#=(x+1)^3/3ln3x-1/3intx^2+3x+3+1/x\ dx=#
#=(x+1)^3/3ln3x-1/3(x^3/3+3/2x^2+3x+lnabs(x))+c#

Related questions
See all questions in Integration by Parts
Impact of this question
2277 views around the world
You can reuse this answer
Creative Commons License