How do you integrate #int(x+1)/((x-9)(x+3)(x-2))# using partial fractions?

1 Answer
Apr 26, 2016

Set up a partial fraction decomposition and choose #x#-values to find the numerators of the partial fractions to get #int(x+1)/((x-9)(x+3)(x-2))dx=5/42lnabs(x-9)-1/30lnabs(x+3)-3/35lnabs(x-2)+C#.

Explanation:

Since all the factors in the denominator of #(x+1)/((x-9)(x+3)(x-2))# are linear (yay!), our partial fraction decomposition will be:
#A/(x-9)+B/(x+3)+C/(x-2)#

For now, let's ignore the integral and focus on decomposing this fraction:
#(x+1)/((x-9)(x+3)(x-2))=A/(x-9)+B/(x+3)+C/(x-2)#

#x+1=A(x+3)(x-2)+B(x-9)(x-2)+C(x-9)(x+3)#

Let #x=9# to find the value of #A#:
#9+1=A(9+3)(9-2)+B(9-9)(9-2)+C(9-9)(9+3)#
#10=84A#
#A=10/84=5/42#

Let #x=-3# to find the value of #B#:
#-3+1=A(-3+3)(-3-2)+B(-3-9)(-3-2)+C(-3-9)(-3+3)#
#-2=60B#
#B=-2/60=-1/30#

And let #x=2# to find the value of #C#:
#2+1=A(2+3)(2-2)+B(2-9)(2-2)+C(2-9)(2+3)#
#3=-35C#
#C=-3/35#

Therefore, #(x+1)/((x-9)(x+3)(x-2))=(5/42)/(x-9)-(1/30)/(x+3)-(3/35)/(x-2)#. Our integral is now:
#int(5/42)/(x-9)-(1/30)/(x+3)-(3/35)/(x-2)dx#

Using the sum rule, this becomes:
#int(5/42)/(x-9)dx-int(1/30)/(x+3)dx-int(3/35)/(x-2)dx#

Integrating these gives:
#5/42int1/(x-9)dx-1/30int1/(x+3)dx-3/35int1/(x-2)dx#
#5/42lnabs(x-9)-1/30lnabs(x+3)-3/35lnabs(x-2)+C#