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

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$\frac{A}{x-a}+\frac{B}{x-b}+\frac{C}{(x-c)}$
• $(iii)\;\int\frac{dx}{(x-a)}=log|x-a|+c.$
Given $I=\int\frac{x}{(x-1)(x-2)(x-3)}dx.$

This can be written as:

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

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

Equating the coefficients of $x^2$,

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

equating the coefficients of x,

1=-5A-4B-3C-------(2)

Equating the constant terms,

0=6A+3B+2C------(3)

Multiply equ(1) by 5 and subtract equ(1) after multiplying by 5 from (2)

we get B+2C=1-------(4)

similarly multiply equ(1) by 6 and subtract it from (3) we get,

3B+4C=0-----(5)

Multiply equ(4) by 3 and subtract from equ(5)

we get $C=\frac{3}{2}$ and B=-2.

Substitute for B and C in equation (1)

We get,

A-2+$\frac{3}{2}$=0.

$\Rightarrow A=2-\frac{3}{2}=\frac{1}{2}.$

Hence $A=\frac{1}{2};B=-2$ and $C=\frac{3}{2}.$

Hence $I=\frac{1}{2}\int\frac{dx}{x-1}-2\int\frac{dx}{x-2}+\frac{3}{2}\int\frac{dx}{x-3}.$

On integrating we get,

$\;\;\;=\frac{1}{2}log|x-1|-2log|x-2|+\frac{3}{2}log|x-3|+c.$

answered Feb 5, 2013