How do you find #int (3x^2-x)/((1 - x)^2(1 - 3x))dx# using partial fractions?

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Answer 1 · 2015-11-26T22:23:43

#[ln(1/(x-1))+1/(x-1)]+C#

#int (3x^2-x)/((1 - x)^2(1 - 3x))dx#

#EEA,EEB,EEC# such as

#-(3x^2-x)/((x-1)^2(3x-1)) = -A/(x-1)-B/(x-1)^2-C/(3x-1)#

(I factorized the denominator to do partial fraction)

Multiply both side by : #(x-1)^2(3x-1)#

#3x^2-x = A(3x-1)(x-1)+ B(3x-1) + C(x-1)^2#

#3x^2-x = A(1-4x+3x^2) + B(3x-1) + C(1+x^2-2x)#

#3x^2-x = A-4Ax+3Ax^2-B+3Bx+C+Cx^2-2Cx#

#3x^2-x = x^2(C+3A)+x(-4A+3B-2C) + A - B + C#

by identification

#C+3A = 3#
#-4A+3B-2C = -1#
#A-B+C = 0#

you find ( ;) )

#A = 1 #
#B = 1 #
#C = 0#

and then

#int(3x^2-x)/((x-1)^2(3x-1))dx = -int1/(x-1) + 1/(x-1)^2dx #

Which is #[ln(1/(x-1))+1/(x-1)]+C#

(Because #aln(b) = ln(b^a)#)

and #-int1/u^2 = 1/u#