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MHT CET · Maths · Inverse Trigonometric Functions

\(\int_1^3\left[\tan ^{-1}\left(\frac{x}{x^2-1}\right)+\tan ^{-1}\left(\frac{x^2-1}{x}\right)\right] d x=\)

  1. A \(\pi\)
  2. B \(\frac{\pi}{4}\)
  3. C \(\frac{\pi}{2}\)
  4. D \(2 \pi\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\pi\)

Step-by-step Solution

Detailed explanation

\(\text { Let } I=\int_1^3\left[\tan ^1\left(\frac{x}{x^2-1}\right)+\tan ^{-1}\left(\frac{x^2-1}{x}\right)\right]\) \(d x \)
\( \int_1^3\left[\tan ^{-1}\left(\frac{x}{x^2-1}\right)+\cot ^{-1}\left(\frac{x}{x^2-1}\right)\right] d x \)
\( =\int_1^3\left(\frac{\pi}{2}\right) d x=\frac{\pi}{2} \int_1^3 d x=\frac{\pi}{2}[x]_1^3=\pi\)