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

Factorize the denominator:

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

Before performing partial fraction decomposition of the rational function, the numerator must have lower degree than the denominator, so split the function as:

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

Now:

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

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

#5x+2 = Ax-3A+Bx-2B#

#5x+2 = (A+B)x- (3A+2B)#

#{(A+B=5),(3A+2B=-2):}#

#{(A=-12),(B=17):}#

So:

#(5x+2)/(x^2-5x+6) = 17/(x-3)-12/(x-2)#

#(x^2+8)/(x^2-5x+6) = 1 + 17/(x-3)-12/(x-2)#

#int (x^2+8)/(x^2-5x+6) dx = int dx + 17 int dx/(x-3) -12int dx/(x-2)#

#int (x^2+8)/(x^2-5x+6) dx = x + 17 lnabs(x-3) -12ln abs(x-2)+C#

Dividing by the denominator leads to

# {x^2+8}/{x^2−5x+6} = 1+{5x+2}/{x^2−5x+6} #

Since #x^2−5x+6 = (x-3)(x-2)# we can write

#{5x+2}/{x^2−5x+6} = A/{x-3}+B/{x-2}#

Multiplying both sides by #x^2−5x+6# leads to

#5x+2 = A(x-2) + B(x-3)#

Substituting #x=3 # and #x=2# in turn leads to the result #A=17, B=-12#, so that

# {x^2+8}/{x^2−5x+6} = 1+17/{x-3}-12/{x-2} #

Integrating this leads to the quoted result quickly!