What is the antiderivative of #(x+2)/(x+1)#?
1 Answer
Explanation:
Using substitution:
Set
#int(x+2)/(x+1)dx=int((u-1)+2)/udu=int(u+1)/udu#
#=int1+1/udu=u+lnu+C=x+1+lnabs(x+1)+C#
The constant
#x+lnabs(x+1)+C#
Using another more intuitive approach:
#int(x+2)/(x+1)dx=int(x+1+1)/(x+1)dx=int(x+1)/(x+1)+1/(x+1)dx#
#=int1+1/(x+1)dx#
Again, substitution can be applied or you could realize that the derivative of
Instead of using the method of splitting