JEE MAIN - Mathematics (2023 - 29th January Evening Shift - No. 13)
The value of the integral $$\int\limits_{1/2}^2 {{{{{\tan }^{ - 1}}x} \over x}dx} $$ is equal to :
$${\pi \over 2}{\log _e}2$$
$${\pi \over 4}{\log _e}2$$
$${1 \over 2}{\log _e}2$$
$$\pi {\log _e}2$$
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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