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

2 Answers
Sep 26, 2017

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

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

Explanation:

Note that:

#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#

Sep 26, 2017

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

Explanation:

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 #

First let us decompose the fraction into partial fractions, which will be of the 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))#

Leading to the identity:

# 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 #

Thus we have:

# I = int \ -3/(x+1) + 2/(x+2) + 2/(x+3) \ dx #

Which, we can now integrate to get:

# I = -3ln|x+1| + 2ln|x+2| + 2ln|x+3| + c #