How do you integrate # x/(x^3-x^2-6x)# using partial fractions?

1 Answer
Oct 2, 2016

#int x/(x^3-x^2-6x) dx = 1/5 ln abs(x-3)-1/5 ln abs(x+2) + C#

Explanation:

Quick version

Note that both the numerator and denominator are divisible by #x#, so we find:

#x/(x^3-x^2-6x) = 1/(x^2-x-6) = 1/((x-3)(x+2)) = 1/(5(x-3))-1/(5(x+2))#

So:

#int x/(x^3-x^2-6x) dx = int 1/(5(x-3))-1/(5(x+2)) dx#

#color(white)(int x/(x^3-x^2-6x) dx) = 1/5 ln abs(x-3)-1/5 ln abs(x+2) + C#

Notes

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

#color(white)(1/((x-3)(x+2))) = (A(x+2)+B(x-3))/((x-3)(x+2))#

#color(white)(1/((x-3)(x+2))) = ((A+B)x+(2A-3B))/((x-3)(x+2))#

Equating coefficients, we have:

#{ (A+B = 0), (2A-3B = 1) :}#

Adding #3# times the first equation to the second, we get:

#5A = 1#

Hence #A = 1/5# and #B=-1/5#

So:

#1/((x-3)(x+2)) = 1/(5(x-3)) - 1/(5(x+2))#

#color(white)()#
Alternatively, we could find #A# and #B# using Heaviside's cover up method:

#A = 1/(color(blue)(3)+2) = 1/5#

#B = 1/(color(blue)(-2)-3) = 1/(-5) = -1/5#

#color(white)()#
The method I actually used was:

If #1/((x-3)(x+2)) = A/(x-3)+B/(x+2)#

then when we express in terms of a common denominator we will need #A# and #B# to be the same size but opposite signs in order that the #x# term cancels out.

If try #1/(x-3)-1/(x+2)# then we get #2-(-3) = 2+3 = 5#, so that's #5# times too large.

Hence #A=1/5# and #B=-1/5#