How do you write the partial fraction decomposition of the rational expression # x/(x^3-x^2-6x)#?

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Answer 1 · 2016-07-29T20:03:43

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

with exclusion #x != 0#

Both the numerator and the denominator are divisible by #x#, so we can simplify (noting #x=0# as an excluded value):

#x/(x^3-x^2-6x)#

#=1/(x^2-x-6)#

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

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

with exclusion #x != 0#

Use Heaviside's cover-up method to find:

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

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

So:

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

with exclusion #x != 0#