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