How do you find the indefinite integral of ∫ ( x^2 - 9x + 14 )dx / (x^2-4x+3) (x-4) ?

1 Answer
Apr 16, 2018

# ln|(x-1)|+2ln|(x-3)|-2ln|(x-4)|+C, or, ln|{(x-1)(x-3)^2}/(x-4)^2|+C.#

Explanation:

Suppose that, #I=int(x^2-9x+14)/{(x^2-4x+3)(x-4)}dx#.

# :. I=int(x^2-9x+14)/{(x-3)(x-1)(x-4)}dx#.

We will decompose the Integrand #(x^2-9x+14)/{(x-3)(x-1)(x-4)}#

using the Method of Partial Fraction.

To this end, we let, for some #A,B,C in RR#,

# (x^2-9x+14)/{(x-3)(x-1)(x-4)}=A/(x-1)+B/(x-3)+C/(x-4)#.

We use Heaviside's Cover Up Method to determine #A,B,C in RR#.

#:. A=[(x^2-9x+14)/{(x-3)(x-4)}]_(x=1)=(1-9+14)/{(-2)(-3)}=1#,

# B=[(x^2-9x+14)/{(x-1)(x-4)}]_(x=3)=(9-27+14)/{(2)(-1)}=2#,

# C=[(x^2-9x+14)/{(x-3)(x-1)}]_(x=4)=(16-36+14)/{(1)(3)}=-2#.

#:. I=int{1/(x-1)+2/(x-3)-2/(x-4)}dx#,

#=ln|(x-1)|+2ln|(x-3)|-2ln|(x-4)|#.

# rArr I=ln|(x-1)|+2ln|(x-3)|-2ln|(x-4)|+C, or, #

# I=ln|{(x-1)(x-3)^2}/(x-4)^2|+C.#

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