How do you integrate #int ( 1/((x+1)^2+4)) dx# using partial fractions?

1 Answer
Nov 9, 2016

#1/2arctan((x+1)/2)+C#

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 #x+1=2tantheta#. This implies that #dx=2sec^2thetad theta#. Substituting in tells us:

#I=int1/(4tan^2theta+4)(2sec^2thetad theta)=1/2intsec^2theta/(1+tan^2theta)d theta#

Since #1+tan^2theta=sec^2theta#:

#I=1/2intd theta=1/2theta+C#

Undoing the substitution #x+1=2tantheta#:

#I=1/2arctan((x+1)/2)+C#