Q96 / 150 · MHT CET 2021 (24 Sep Shift 1) · Mathematics

$\int \sec ^{-1} x d x=$

  • A
    $x \sec ^{-1} x+\log \left|x+\sqrt{x^2-1}\right|+c$
  • Correct answer
    B
    $x \sec ^{-1} x-\log \left|x+\sqrt{x^2-1}\right|+c$
  • C
    $x \sec ^{-1} x-\log \left|x+\sqrt{x^2+1}\right|+c$
  • D
    $x \sec ^{-1} x+\log \left|x+\sqrt{x^2+1}\right|+c$

Solution

$\begin{aligned} & \text { Let } I=\int \sec ^{-1} x d x=\int \sec ^{-1} x \cdot d x \\ & =\left(x \sec ^{-1} x\right)-\int \frac{x}{x \sqrt{x^2-1}} d x \\ & =\left(x \sec ^{-1} x\right)-\log \left|x+\sqrt{x^2-1}\right|+C\end{aligned}$

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

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