How do you use partial fraction decomposition to decompose the fraction to integrate #(x^2-x-8)/((x+1)(x^2+5x+6))#?

Question

Luke Alvarez · Accepted Answer

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

#(x^2-x-8)/((x+1)(x^2+5x+6))#

First we need to finish factoring the denominator:

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

Now we want #A, B, "and " C# to get:

# A/(x+1) + B/(x+2) + C/(x+3) = (x^2-x-8)/((x+1)(x+2)(x+3))#

So we need:

#A(x+2)(x+3)+B(x+1)(x+3)+C(x+1)(x+2) = x^2-x-8#

The left side can be rewritten:

#Ax^2+5Ax+6A+Bx^2+4Bx+3B+Cx^2+3Cx+2C#

and then:

#(A+B+C)x^2 +(5A+4B+3C)x+(6A+3B+2C)=x^2-x-8#

So we need to solve the system:

#A+B+C=1#

#5A+4B+3C=-1#

#6A+3B+2C=-8#

Multiplying the first equation by #-3# and adding to the second. Then the first times #-2# and add to the third, gets us:

#2A+B=-4#

#4A+B=-10#

So #2A = -6# and #A = -3#, which gets us #B = 2# and together these give us #C=2#.

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

Axel Alvarez · Answer

undefined To decompose the fraction $\frac{x^2 - x - 8}{(x+1)(x^2 + 5x + 6)}$ using partial fraction decomposition, first factor the denominator into linear terms:

$x^2 + 5x + 6 = (x + 2)(x + 3)$

Then, write the expression as the sum of two fractions with undetermined constants:

$\frac{x^2 - x - 8}{(x+1)(x^2 + 5x + 6)} = \frac{A}{x+1} + \frac{B}{x+2} + \frac{C}{x+3}$

Next, clear the denominators by multiplying both sides by $(x+1)(x+2)(x+3)$ to solve for $A$, $B$, and $C$.

Finally, integrate each term separately once the decomposition is complete.