How do you integrate #(x+2)/(x+3)#?
1 Answer
Mar 1, 2018
Well, presumably by simplifying it first.
#(x+2)/(x+3) = 1 - 1/(x+3)#
Therefore:
#color(blue)(int (x+2)/(x+3)dx)#
#= color(blue)(x - ln|x+3| + C)#
