How do you integrate #7/(x^2+1)# using partial fractions?

1 Answer
Aug 21, 2016

#int 7/(x^2+1) dx = 7/2 ln(x+i) - 7/2 ln(x-i) + C#

Explanation:

This integral would normally be done using trigonometric substitution, but if you really want to use partial fractions to integrate this then you will need to use Complex coefficients:

#7/(x^2+1) = A/(x+i)+B/(x-i) = (A(x-i)+B(x+i))/(x^2+1)#

#=((A+B)x+(B-A)i)/(x^2+1)#

Equating coefficients:

#{ (A+B=0), ((B-A)i = 7) :}#

Multiply both sides of the second equation by #i# to get:

#A-B=7i#

Hence #A=7/2i# and #B=-7/2i#
So:

#int 7/(x^2+1) dx = int 7/(2(x+i))-7/(2(x-i)) dx#

#= 7/2 ln(x+i) - 7/2 ln(x-i) + C#