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# Choose the correct answer in $\Large \int \normalsize\frac{\large xdx}{\large (x-1)(x-2)}$ equals\begin{array}{1 1}(A)\;log\mid\frac{(x-1)^2}{x-2}\mid+\: C & (B)\;log\mid\frac{(x-2)^2}{x-1}\mid+\: C\\(C)\;log\mid\bigg(\frac{x-1}{x-2}\bigg)^2\mid+\: C & (D)\;log\mid(x-1)(x-2)\mid+ \: C\end{array}

Toolbox:
• $\;$(i)If the given function is a rational function of the form $\frac{px+q}{(x+a)(x+b)}$,then it can be resolved into its partial function as $\frac{A}{(x+a)}+\frac{B}{(x+b)}.$
• $(ii)\;\int\frac{dx}{(x+a)}=log|x+a|+c.$
Given $I=\int\frac{xdx}{(x-1)(x-2)}dx.$

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

x=A(x-2)+B(x-1)

Equating the coefficients of x we get

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

Equating the constant terms,

0=-2A-B-----(2)

Let us multiply equ(1) by 2 and add with equ(1)

2A+2B=2
-2A-B=0
______________
B=2
Substituting for B in equ(1)

A+2=1

Therefore A=-1.

Hence A=-1 and B=2.

$\frac{x}{(x-1)(x-2)}=\frac{-1}{(x-1)}+\frac{2}{(x-2)}.$

Hence integration of I is

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

On integrating we get

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

But $2log|x-2|=log|x-2|^2$.

$-Log |x-1|+log|x-2|^2+c.$

log a-log b=$log|\frac{a}{b}|$

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

So the correct answer is B.