How do you integrate #(x^3-4x-10)/(x^2-x-6)# using partial fractions?
1 Answer
Explanation:
The first step is to factor the denominator of the function
#rArr (x^3-4x-10)/(x^2-x-6) = (x^3-4x-10)/((x-3)(x+2))# since the factors are linear then the coefficients of the partial fractions will be constants , say A and B. Writing the function in terms of it's partial fractions.
#(x^3-4x-10)/((x-3)(x+2)) = A/(x-3) + B/(x+2) # multiplying through by (x-3)(x+2)
#x^3-4x-10 = A(x+2) + B(x-3) ...................... (1)# We now have to find the values of A and B .Note that if x = -2 the term with A will be zero and if x = 3 the term with B will be zero. We can make use of this fact in finding A and B.
let x = -2 in (1) :- 10 = -5B
#rArrcolor(blue)" B = 2"# let x = 3 in (1) : 5 = 5A
#rArr color(blue)" A = 1 "#
#rArr (x^3-4x-10)/((x-3)(x+2)) = 1/(x-3) + 2/(x+2)# Integral becomes :
#int(dx)/(x-3) + int(2dx)/(x+2) #
#= ln|x-3| + 2ln|x+2| + c#
