How do you integrate #(x/(x+2))#?
1 Answer
May 29, 2016
Explanation:
We can simplify this way is written:
#intx/(x+2)dx=int(x+2-2)/(x+2)dx=int((x+2)/(x+2)-2/(x+2)dx)#
#int(1-2/(x+2))dx=int1dx-2int1/(x+2)dx#
From here the integration is fairly simple:
#=x-2lnabs(x+2)+C#
