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$\int \frac{x d x}{(x-1)(x-2)}$ equals

A. $\log \left|\frac{(x-1)^{2}}{x-2}\right|+C$

B. $\log \left|\frac{(x-2)^{2}}{x-1}\right|+\mathrm{C}$

C. $\log \left|\left(\frac{x-1}{x-2}\right)^{2}\right|+\mathrm{C}$

D. $\log |(x-1)(x-2)|+\mathrm{C}$


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

$x=A(x-2)+B(x-1)$    ...(1)

Substituting x = 1 and 2 in (1), we obtain

A = −1 and B = 2

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

$\Rightarrow \int \frac{x}{(x-1)(x-2)} d x=\int\left\{\frac{-1}{(x-1)}+\frac{2}{(x-2)}\right\} d x$

$=-\log |x-1|+2 \log |x-2|+\mathrm{C}$

$=\log \left|\frac{(x-2)^{2}}{x-1}\right|+\mathrm{C}$

Hence, the correct answer is B.

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