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MHT CET · Maths · Indefinite Integration

\(\int \sec ^{-1} x d x=\)

  1. A \(x \sec ^{-1} x+\log \left|x+\sqrt{x^2-1}\right|+c\)
  2. B \(x \sec ^{-1} x-\log \left|x+\sqrt{x^2-1}\right|+c\)
  3. C \(x \sec ^{-1} x-\log \left|x+\sqrt{x^2+1}\right|+c\)
  4. D \(x \sec ^{-1} x+\log \left|x+\sqrt{x^2+1}\right|+c\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(x \sec ^{-1} x-\log \left|x+\sqrt{x^2-1}\right|+c\)

Step-by-step Solution

Detailed explanation

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