Come si integra # x / (x + 1) dx #?
Risposta:
#int x/(x+1) dx = x-ln abs(x+1)+C#
Spiegazione:
#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 = 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#