How do you integrate #int(x+1)/((x+5)(x-1)(x-2))# using partial fractions?
Since the factors on the denominator are linear , the the numerators of the partial fractions will be constants , say A , B and C.
#(x+1)/((x+5)(x-1)(x-2)) = A/(x+5) + B/(x-1) + C/(x-2) #
multiply through by (x+5)(x-1)(x-2)
#x+1 = A(x-1)(x-2)+B(x+5)(x-2)+C(x+5)(x-1) ...............(1)#
The aim now is to find the values of A,B and C. Note that if x=1 then the terms with A and C will be zero. If x =2 the terms with A and B will be zero and if x = -5 the terms with B and C will be zero. This is the starting point in finding A , B and C.
let x = 1 in (1): 2 = -6B
# rArr B = -1/3#
let x = 2 in (1): 3 = 7C
# rArr C = 3/7 #
let x = -5 in (1): -4 = 42A
# rArr A = -2/21 #
Integral can now be written as :
#int(-2/21)/(x+5) dx -int (1/3)/(x-1) dx +int (3/7)/(x-2) dx #
#= 3/7ln|x-2| - 2/21ln|x+5| - 1/3ln|x-1| + c #