How do you integrate #f(x)=1/((x-2)(x-5)(x+3))# using partial fractions?
Since the factors on the denominator are linear then the numerators of the partial fractions will be constants , say A , B and C
#rArr 1/((x-2)(x-5)(x+3)) = A/(x-2) + B/(x-5) + C/(x+3) #
now, multiply through by (x-2)(x-5)(x+3)
so 1 = A(x-5)(x+3) + B(x-2)(x+3) + C(x-2)(x-5) ................................(1)
We now have to find the values of A , B and C. Note that if x = 2 , the terms with B and C will be zero. If x = 5 , the terms with A and C will be zero and if x = -3, the terms with A and B will be zero.
Making use of this fact , we obtain.
let x = 2 in (1) : 1 = -15A
# rArr A = -1/15 #
let x = 5 in (1) : 1 = 24B
#rArr B = 1/24 #
let x = -3 in (1) : 1 = 40C
# rArr C = 1/40 #
#rArr 1/((x-2)(x-5)(x+3)) = (-1/15)/(x-2) + (1/24)/(x-5) + (1/40)/(x+3)#
Integral now becomes.
#-1/15intdx/(x-2) + 1/24intdx/(x-5) + 1/40intdx/(x+3) #
# = 1/24ln|x-5| + 1/40ln|x+3| - 1/15ln|x-2| + c #