How do you integrate #int (1-x^2)/((x-9)(x+6)(x-2)) # using partial fractions?

1 Answer
Mar 31, 2018

#int(1-x^2)/((x-9)(x+6)(x-2))dx#

#=-16/21lnabs(x-9)-7/24lnabs(x+6)+3/56lnabs(x-2)+C#

Explanation:

Conveniently, the denominator in this problem is already factored.

To apply partial fraction decomposition, we want to express

#int(1-x^2)/((x-9)(x+6)(x-2))dx#

in the form

#int(A/(x-9)+B/(x+6)+C/(x-2))dx#

So we can equate

#(1-x^2)/((x-9)(x+6)(x-2))=A/(x-9)+B/(x+6)+C/(x-2)#

Multiplying both sides by the denominator of the left, we get:

#1-x^2=A(x+6)(x-2)+B(x-9)(x-2)+C(x-9)(x+6)#

Now let's expand the parentheses and group similar terms:

#1-x^2=A(x^2+4x-12)+B(x^2-11x+18)+C(x^2-3x-54)#

#1-x^2=(A+B+C)x^2+(4A-11B-3C)x+(-12A+18B-54C)#

Now we can compare the coefficients on both sides, and write a system of three equations:

#A+B+C=-1#
#4A-11B-3C=0#
#-12A+18B-54C=1#

Solving this system (whichever way you choose) gives:

#A=-16/21#

#B=-7/24#

#C=3/56#

So now we can rewrite the integral as:

#int(-16/21(1)/(x-9)-7/24(1)/(x+6)+3/56(1)/(x-2))dx#

#rArr-16/21int1/(x-9)dx-7/24int1/(x+6)dx+3/56int1/(x-2)dx#

Integrating, we get:

#-16/21lnabs(x-9)-7/24lnabs(x+6)+3/56lnabs(x-2)+C#