How do you integrate #x/(x+1)dx#?
1 Answer
Jul 29, 2016
Explanation:
#int x/(x+1) dx#
#=int (x+1-1)/(x+1) dx#
#=int (1-1/(x+1)) dx#
#= x-ln abs(x+1)+C#
#int x/(x+1) dx#
#=int (x+1-1)/(x+1) dx#
#=int (1-1/(x+1)) dx#
#= x-ln abs(x+1)+C#