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.

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