MCQ
$\int \frac{x}{(x-1)(x-2)} d x$ बराबर है:
  • A
    log |(x - 1) (x - 2)| + C
  • B
    $\log \left|\frac{(x-2)^{2}}{x-1}\right|+C$
  • C
    $\log \left|\frac{(x-1)^{2}}{x-2}\right|+C$
  • D
    $\log \left|\left(\frac{x-1}{x-2}\right)^{2}\right|+\mathrm{C}$

Answer

माना $\frac{x}{(x-1)(x-2)}$ $=\frac{A}{(x-1)}+\frac{B}{(x-2)}$
$\Rightarrow$ x = A(x - 2) + B(x - 1) ...(i)
समी (i) में x = 1 और 2 रखने पर,
A = -1 और B = 2
$\therefore$ $\frac{x}{(x-1)(x-2)}$ $-\frac{1}{(x-1)}+\frac{2}{(x-2)}$
$\therefore$ $\int \frac{x}{(x-1)(x-2)} d x$ $=\int \frac{(-1)}{x-1} d x$ $+\int \frac{2}{x-2} d x$
= - log |x - 1| + 2 log |x - 2| + C
= - log |x - 1| + log |x - 2|2 + C
$=\log \left|\frac{(x-2)^{2}}{x-1}\right|+C$  ($\because$ log b - log a = log $\frac{b}{a}$)

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