Question

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

Answer: (B)

We have,
$I=\underset{1/2014}{\overset{2014}{\int }}\frac{{\tan }^{-1}x}{x}dx\dots$(i)
Let $x=\frac{1}{t}$
$\Rightarrow dx=\frac{-1}{t^2}dt$
Now, $I=\underset{2014}{\overset{1/2014}{\int }}\frac{{\tan }^{-1}(1/t)}{1/t}\left(\frac{-1}{t^2}dt\right)$
$=\underset{1/2014}{\overset{2014}{\int }}\frac{{\cot }^{-1}t}{t}dt$
$=\underset{1/2014}{\overset{2014}{\int }}\frac{{\cot }^{-1}x}{x}dx\dots$(ii)
On adding Eqs. (i) and (ii), we get
$2I=\underset{1/2014}{\overset{2014}{\int }}\frac{\pi /2}{x}dx=\frac{\pi }{2}(\log x{)}_{1/2014}^{2014}$
$=\frac{\pi }{2}(\log 2014-\log 1/2014)$
$\therefore I=\frac{\pi }{4}\left(\log 2014-\log \frac{1}{2014}\right)$
$=\frac{\pi }{4}(\log 2014+\log 2014)$
$=\frac{\pi }{4}(2\log 2014)=\frac{\pi }{2}\log 2014$