Question

The value of the integral $I={\int }_{1/2014}^{2014}\frac{{\tan }^{-1}x}{x}dx$ is

• A. $\frac{\pi }{4}\log 2014$
• B. $\frac{\pi }{2}\log 2014$
• C. $\pi \log 2014$
• D. $\frac{1}{2}\log 2014$

Answer 1

The correct option is B $\frac{\pi }{2}\log 2014$

$I={\int }_{1/2014}^{2014}\frac{{\tan }^{-1}x}{x}dx...\left(1\right)$

Let $x=\frac{1}{t}$

$dx=-\frac{1}{t^2}dt$

$I={\int }_{2014}^{1/2014}\frac{{\tan }^{-1}\left(\frac{1}{t}\right)}{\left(\frac{1}{t}\right)}(-\frac{1}{t^2})dt$

$I={\int }_{1/2014}^{2014}\frac{{\cot }^{-1}t}{t}dt....\left(2\right)$

from (1) and (2)

$2I={\int }_{1/2014}^{2014}\frac{\pi /2}{t}dt$

$I=\frac{\pi }{4}{\left(\log t\right)}_{1/2014}^{2014}=\frac{\pi }{4}(\log 2014-\log \frac{1}{2014})$

$=\frac{\pi }{4}\left(2\log 2014\right)=\frac{\pi }{4}\log 2014$