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

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Answer 1 · 2016-09-12T18:55:52

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

This cannot be expressed using partial fractions. However, we can integrate this using trigonometric substitutions.

#intdx/((x+1)^2+4)#

Let #x+1=2tantheta#. Thus, #dx=2sec^2thetad theta#. Substituting, this gives us:

#=int(2sec^2thetad theta)/(4tan^2theta+4)#

Factoring:

#=1/2int(sec^2thetad theta)/(tan^2theta+1)#

Recall that #tan^2theta+1=sec^2theta#:

#=1/2int(sec^2thetad theta)/sec^2theta#

#=1/2intd theta#

#=1/2theta+C#

From #x+1=2tantheta# we see that #theta=arctan((x+1)/2)#:

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