How do you use partial fraction decomposition to decompose the fraction to integrate #(x^2-5x+6)/(x^3-x^2+2x)#?

1 Answer
Jun 19, 2015

See below

Explanation:

The denominator factors as: #x(x^2-x+2)# and cannot be factored further using real numbers (discriminant is negative). The integrand cannot be reduced, so we proceed:

Find, #A, B, and C# to make:

#A/x +(Bx+C)/(x^2-x+2) = (x^2-5x+6)/(x(x^2-x+2))#

We need:
#Ax^2-Ax+2A+Bx^2+Cx = x^2-5x+6#

Solve the system:
#A+B=1#
#-A+C=-5#
#2A=6#

Obviously, #A=3# and this makes #B=-2# and #C=-2#

The partial fraction decomposition is:

#3/x -(2x+2)/(x^2-x+2) = (x^2-5x+6)/(x(x^2-x+2))#.

If you have time check by combining the fractions on the left to make sure you've made no mistakes.

Now integrate:

#int (x^2-5x+6)/(x^3-x^2+2x)) dx = int (3/x)dx - int (2x+2)/(x^2-x+2) dx#

The first integral is easy. I'd split the second using:

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

#int (2x-1)/(x^2-x+2) dx# is a simple substitution and then

#int 3/(x^2-x+2) dx# is an inverse tangent with some constants to figure out by completing the square and substituting.