How do you integrate #int ( 1/((x+1)^2+4)) dx# using partial fractions?
1 Answer
Nov 9, 2016
Explanation:
#I=int1/((x+1)^2+4)dx#
This isn't a job for partial fractions, it's a job for a trig substitution. Let
#I=int1/(4tan^2theta+4)(2sec^2thetad theta)=1/2intsec^2theta/(1+tan^2theta)d theta#
Since
#I=1/2intd theta=1/2theta+C#
Undoing the substitution
#I=1/2arctan((x+1)/2)+C#