Integrate the rational functions$\frac{3x-1}{(x-1)(x-2)(x-3)}$

Toolbox:
• $(i)\;$Form of the rational function$\frac{px+q}{(x-a)(x-b)(x-c)},a\neq b\neq c$
• $(ii)\;$Form of the partial function$\frac{A}{x-a}+\frac{B}{x-b}+\frac{C}{(x-c)}$
Given $I=\int\frac{3x-1}{(x-1)(x-2)(x-3)}.$

This can be written as:

$\frac{3x-1}{(x-1)(x-2)(x-3)}=\frac{A}{(x-1)}+\frac{B}{(x-2)}+\frac{C}{(x-3)}.$

$\Rightarrow 3x-1=A(x-2)(x-3)+B(x-1)(x-3)+c(x-1)(x-2)$.

Equating the coefficients of $x^2$,

0=A+B+C-----(1)

equating the coefficients of x,

3=-5A-5B-3C----(2)

Equating the constant terms,

-1=6A+3B+2C------(3)

Let us solve for A,B and C.

Multiply equ(1) by 5 and add with equ(2)
5A+5B+5C=0
-5A-4B-3C=3
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B+2C=3------(4)

Multiply equ(1) by 6 and subtract from equ(3)
6A+3B+2C=-1
6A+6B+6C=0
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-3B-4C=-1-------(5)

Multiply equ(4) by 3 and add equ(5)
3B+6C=9
-3B-4C=-1
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2C=8$\Rightarrow C=4.$
Substituting for C in equ(4) we get,

B+8=3$\Rightarrow B=-5.$

Substituting for B and C in equ(1)

A-5+4=0.

$\rightarrow A=1.$

Hence A=1,B=-5 and C=4.

Now substituting for A,B and C in I we get,

$I=\int\frac{1}{(x+1)}+\frac{(-5)}{(x-2)}+\frac{4}{(x-3)}dx.$

On separating the terms we get,

$I=\int\frac{1}{x-1}dx-\int\frac{5}{(x-2)}dx+\int\frac{4}{(x-3)}dx.$

On integrating we get,

$\;\;\;=log|x-1|-5log|x-2|+4log|x-3|+c.$

Hence $I=log|x-1|-5log|x-2|+4log|x-3|+c.$