How do you integrate #(x^3-4x-10)/(x^2-x-6)# using partial fractions?

1 Answer
Apr 16, 2016

#ln|x-3| + 2ln|x+2| + c #

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#