How do you integrate #int (3 - 8x)/(x(1 - x))# using partial fractions?
3ln|x| + 5ln|1 - x| + c
First write the expression in terms of it's partial fractions.
Since the factors on the denominator are linear then the numerators of the partial fractions will be constants, say A and B.
#(3 -8x)/(x(1-x)) = A/x + B/(1-x) #
now multiply both sides by x(1-x)
so 3-8x = A(1-x) + Bx ................................(1)
The aim now is to find the values of A and B . Note that if x = 0 , the term with B will be zero and if x = 1 the term with A will be zero. This is the starting point in finding A and B.
let x = 0 in (1) : 3 = A
let x = 1 in (1) : -5 = B
integral can now be written as :
#int3/x dx - int5/(1-x) dx #
#= 3ln|x| + 5ln|1 - x| + c #( c is constant of integration )