How do you integrate #int (x^2+4)/(x+2)# using substitution?
1 Answer
Oct 2, 2016
Explanation:
#I=int(x^2+4)/(x+2)dx#
Before performing any integration, let's rewrite this function:
#I=intx^2/(x+2)dx+int4/(x+2)dx#
Perform long division on
#I=intxdx-2intdx+8intdx/(x+2)#
The first two don't require substitution:
#I=x^2/2-2x+8intdx/(x+2)#
For the final integral, let
#I=(x^2-4x)/2+8int(du)/u#
This is a common integral:
#I=(x^2-4x)/2+8lnabsu+C#
Back-substitute
#I=(x^2-4x+16lnabs(x+2))/2+C#