How do you use partial fraction decomposition to decompose the fraction to integrate #3/(x^2-6x+8)#?

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Answer 1 · 2015-06-16T18:01:45

#3int1/(x^2-6x+8)dx = 3int -1/(2(x-2))+1/(2(x-4)#

#3int1/(x^2-6x+8)dx#

You need to factorize the denominator before using partial fraction decomposition

note : #x^2-6x+8 = (x - 2)(x-4)#

#3int1/((x-2)(x-4))dx = A/(x-2) + B/(x-4)#

#= (A(x-4))/((x-2)(x-4)) + (B(x-2))/((x-4)(x-2))#

#= Ax - 4A + Bx-2B = x(A+B) - 4A - 2B#

#A + B = 0#
#-4A - 2B = 1#

So

#A = -1/2#
#B = 1/2#

Then

#3int1/(x^2-6x+8)dx = 3int -1/(2(x-2))+1/(2(x-4)#