How do you integrate #(3x)/((x+2)(x-1))# using partial fractions?

1 Answer
Nov 21, 2016

# int (3x)/((x+2)(x-1)) dx = ln(A(x+2)^2|x-1|) #

Explanation:

Let us find the partial fraction decomposition of the integrand:

Let # (3x)/((x+2)(x-1)) -= A/(x+2) + B/(x-1) #
# :. (3x)/((x+2)(x-1)) = ( A(x-1) + B(x+2) ) / ((x+2)(x-1)) #
# :. 3x = A(x-1) + B(x+2) #

This is an identity and valid #AA x in RR#
Put #x=-2 =>-6=-3A + 0 => A=2#
Put #x=1 => 3=0+B(3) => B=1#

So the partial fraction decomposition is:

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

And so the integral can be written as:

# int (3x)/((x+2)(x-1)) dx = int 2/(x+2) + 1/(x-1) dx #
# :. int (3x)/((x+2)(x-1)) dx = 2int 1/(x+2) dx+ int1/(x-1) dx #
# :. int (3x)/((x+2)(x-1)) dx = 2ln|x+2| + ln|x-1| + lnA # (#lnA#=constant)
# :. int (3x)/((x+2)(x-1)) dx = ln(A(x+2)^2|x-1|) #