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

Question

Christopher Agresta · Accepted Answer

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

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#

Emily Ammerman · Answer

undefined To integrate $(x^3 - 4x - 10)/(x^2 - x - 6)$ using partial fractions, follow these steps:

1. Factor the denominator $x^2 - x - 6 = (x - 3)(x + 2)$.
2. Write the given expression as a sum of partial fractions: 
   $(x^3 - 4x - 10)/(x^2 - x - 6) = \frac{A}{x - 3} + \frac{B}{x + 2}$.
3. Multiply both sides by the denominator $(x^2 - x - 6)$ to clear the fractions.
4. After multiplying and simplifying, equate the numerators to find the values of $A$ and $B$.
5. Once you find $A$ and $B$, integrate each term separately.
6. The final result will be the sum of the integrals of $\frac{A}{x - 3}$ and $\frac{B}{x + 2}$.

Let me know if you need further clarification on any step!