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The value of the integral $$\int\limits_{1/2}^2 {{{{{\tan }^{ - 1}}x} \over x}dx} $$ is equal to :

JEE · Math · previous-year question

  1. A.$${\pi \over 2}{\log _e}2$$correct
  2. B.$${\pi \over 4}{\log _e}2$$
  3. C.$${1 \over 2}{\log _e}2$$
  4. D.$$\pi {\log _e}2$$

Answer

A. $${\pi \over 2}{\log _e}2$$

Explanation

$$I = \int\limits_{{1 \over 2}}^2 {{{{{\tan }^{ - 1}}x} \over x}dx} $$ ..... (i) $$x \to {1 \over x}$$ $$I = \int\limits_{{1 \over 2}}^2 {{1 \over x}{{\tan }^{ - 1}}{1 \over x}dx} $$ ..... (ii) $$2I = \int\limits_{{1 \over 2}}^2 {{1 \over x}\,.\,{\pi \over 2}dx} $$ $$ = \left. {{\pi \over 2}\ln x} \right|_{{1 \over 2}}^2 = \pi \ln 2$$ $$ \Rightarrow I = {\pi \over 2}\ln 2$$

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