How do you find ∫3x2−x(1−x)2(1−3x)dx using partial fractions?

1 Answer
Tom
Nov 26, 2015

[ln(1x−1)+1x−1]+C

Explanation:

∫3x2−x(1−x)2(1−3x)dx

∃A,∃B,∃C such as

−3x2−x(x−1)2(3x−1)=−Ax−1−B(x−1)2−C3x−1

(I factorized the denominator to do partial fraction)

Multiply both side by : (x−1)2(3x−1)

3x2−x=A(3x−1)(x−1)+B(3x−1)+C(x−1)2

3x2−x=A(1−4x+3x2)+B(3x−1)+C(1+x2−2x)

3x2−x=A−4Ax+3Ax2−B+3Bx+C+Cx2−2Cx

3x2−x=x2(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

∫3x2−x(x−1)2(3x−1)dx=−∫1x−1+1(x−1)2dx

Which is [ln(1x−1)+1x−1]+C

(Because aln(b)=ln(ba))

and −∫1u2=1u

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