How do you integrate #int (2x - 1) / [(x - 1)^3 (x - 2)]# using partial fractions?

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Answer 1 · 2017-07-09T16:46:22

I decomposed integrand into basic fractions,

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

#(2x-1)=A*(x^3-4x^2+5x-2)+B*(x^2-3x+2)+C*(x-2)+D*(x^3-3x^2+3x-1)#

Let #x=1#, #-C=1# or #C=-1#.

Let #x=2#, #D=3#.

Differentiate both sides,

#2=A*(3x^2-8x+5)+B*(2x-3)+C+D*(3x^2-6x+3)#

Let #x=1#, #-B+C=2# or #-B-1=2#. Hence #B=-3#

Differentiate both sides,

#0,=A*(6x-8)+2B+D*(6x-6)#

Let #x=1#, #-2A+2B=0# or #A=B=-3#.

Thus,

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

#=int-3/(x-1)dx+int-3/(x-1)^2dx++int-1/(x-1)^3dx+int3/(x-2)dx#

#=-3Ln(x-1)+3/(x-1)+1/2*(x-1)^(-2)+3Ln(x-2)+C#

#=3/(x-1)+1/2*(x-1)^(-2)+3Ln((x-2)/(x-1))+C#

I decomposed integrand into basic fractions.