How do you integrate #int(x)/((x+3)(x+6)(x+2))# using partial fractions?

1 Answer
Steve M
Nov 29, 2016

# int x/((x+3)(x+6)(x+2)) dx= ln|x+3| - 1/2ln|x+6| - 1/2ln|x+2| + C#

Explanation:

The Partial Fraction decomposition of the integrand will be of the form:

# x/((x+3)(x+6)(x+2)) -= A/(x+3) + B/(x+6) + C/(x+2) #
# :. x/((x+3)(x+6)(x+2)) = (A(x+6)(x+2) + B(x+3)(x+2) + C(x+3)(x+6))/((x+3)(x+6)(x+2)) #
# :. x = A(x+6)(x+2) + B(x+3)(x+2) + C(x+3)(x+6) #

Put # { (x=-3, =>-3=A(3)(-1), => A=1), (x=-6,=>-6=B(-3)(-4),=>B=-1/2),(x=-2,=>-2=C(1)(4),=>C=-1/2) :} #

Hence,

# x/((x+3)(x+6)(x+2)) -= 1/(x+3) - 1/(2(x+6)) - 1/(2(x+2)) #

So,

# int x/((x+3)(x+6)(x+2)) dx= int 1/(x+3) - 1/(2(x+6)) - 1/(2(x+2)) dx#
# :. int x/((x+3)(x+6)(x+2)) dx= ln|x+3| - 1/2ln|x+6| - 1/2ln|x+2| + C#

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