Q105 / 160 · WBJEE 2009 · Mathematics

$\int \frac{d x}{x(x+1)}$ equals where $c$ is an arbitrary constant.

  • A
    $\ln \left|\frac{x+1}{x}\right|+c$
  • Correct answer
    B
    $\ln \left|\frac{x}{x+1}\right|+c$
  • C
    $\ln \left|\frac{x-1}{x}\right|+c$
  • D
    $\ln \left|\frac{x-1}{x+1}\right|+c$

Solution

$$ \text { Hints : } \int \frac{d x}{x(x+1)}=\int\left(\frac{1}{x}-\frac{1}{x+1}\right) d x=\int \frac{d x}{x}-\int \frac{d x}{x+1}=\ln |x|-\ln |x+1|+C=\ln \left|\frac{x}{x+1}\right|+C $$

Hence, the correct answer is option: 'B'.

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