How do you integrate #(x^2-x-8)/((x+1)(x^2+5x+6))# using partial fractions?

Question

Isaiah Allen · Accepted Answer

# int \ (x^2-x-8)/((x+1)(x^2+5x+6)) \ dx = -3ln|x+1| + 2ln|x+2| + 2ln|x+3| + c #

We seek: # I = int \ (x^2-x-8)/((x+1)(x^2+5x+6)) \ dx # # \ \ = int \ (x^2-x-8)/((x+1)(x+2)(x+3)) \ dx # Let's start by breaking down the fraction into its component parts, which will have the following form: # (x^2-x-8)/((x+1)(x+2)(x+3)) -= A/(x+1) + B/(x+2) + C/(x+3) # # " " = (A(x+2)(x+3) + B(x+1)(x+3) + C(x+2)(x+1)) / ((x+1)(x+2)(x+3))# Getting to the answer: # x^2-x-8 -= A(x+2)(x+3) + B(x+1)(x+3) + C(x+2)(x+1) # Where #A,B# are constants that are to be determined. We can find them by substitutions (In practice we do this via the "cover up" method: Put # x = -1 => 1+1-8 = A(1)(2) => A = -3# Put # x = -2 => 4+2-8 = B(-1)(1) => B = 2 # Put # x = -3 => 9+3-8 = C(-1)(-2) => C = 2 # Consequently, we have: # I = int \ -3/(x+1) + 2/(x+2) + 2/(x+3) \ dx # which we can now combine to obtain: # I = -3ln|x+1| + 2ln|x+2| + 2ln|x+3| + c #

Evelyn Agard · Answer

#int (x^2-x-8)/((x+1)(x+2)(x+3))#

#= -3ln abs(x+1)+2ln abs(x+2)+2ln abs(x+3) + C#

Take note of this: #x^2+5x+6 = (x+2)(x+3)# So: #(x^2-x-8)/((x+1)(x+2)(x+3)) = A/(x+1)+B/(x+2)+C/(x+3)# We can find #A#, #B# and #C# using Oliver Heaviside's cover up method: #A = ((color(blue)(-1))^2-(color(blue)(-1))-8)/(((color(blue)(-1))+2)((color(blue)(-1))+3)) = (-6)/2 = -3# #B = ((color(blue)(-2))^2-(color(blue)(-2))-8)/(((color(blue)(-2))+1)((color(blue)(-2))+3)) = (-2)/(-1) = 2# #C = ((color(blue)(-3))^2-(color(blue)(-3))-8)/(((color(blue)(-3))+1)((color(blue)(-3))+2)) = 4/2 = 2# So: #int (x^2-x-8)/((x+1)(x+2)(x+3))# #= int (-3)/(x+1)+2/(x+2)+2/(x+3) color(white)(.)dx# #= -3ln abs(x+1)+2ln abs(x+2)+2ln abs(x+3) + C#

Samantha Adkison · Answer

undefined To integrate the function $(x^2 - x - 8)/((x + 1)(x^2 + 5x + 6))$ using partial fractions, follow these steps:

1. Factor the denominator $(x + 1)(x^2 + 5x + 6)$ into linear factors:
$(x + 1)(x + 2)(x + 3)$.

2. Write the given fraction as a sum of partial fractions:
$\frac{A}{x + 1} + \frac{B}{x + 2} + \frac{C}{x + 3}$.

3. Multiply both sides of the equation by the common denominator $(x + 1)(x + 2)(x + 3)$ to clear the fractions.

4. Expand and simplify the equation to solve for $A$, $B$, and $C$.

5. Once you have found the values of $A$, $B$, and $C$, rewrite the original fraction with the partial fractions.

6. Integrate each partial fraction separately.

7. Finally, add the integrals of the partial fractions to find the overall integral of the given function.