MHT CET · Maths · Indefinite Integration
\(\int \sec ^{-1} x d x=\)
- A \(x \sec ^{-1} x+\log \left|x+\sqrt{x^2-1}\right|+c\)
- 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 \(x \sec ^{-1} x+\log \left|x+\sqrt{x^2+1}\right|+c\)
Answer & Solution
Correct Answer
(B) \(x \sec ^{-1} x-\log \left|x+\sqrt{x^2-1}\right|+c\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \text { Let } I=\int \sec ^{-1} x d x=\int \sec ^{-1} x \cdot d x \\ & =\left(x \sec ^{-1} x\right)-\int \frac{x}{x \sqrt{x^2-1}} d x \\ & =\left(x \sec ^{-1} x\right)-\log \left|x+\sqrt{x^2-1}\right|+C\end{aligned}\)
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