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

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

$I=\underset{1/2014}{\overset{2014}{\int }}\frac{{\tan }^{-1}x}{x}dx......\left(1\right)$
Let
$x=\frac{1}{t}\Rightarrow dx=\frac{-dt}{t^2}\Rightarrow I=\underset{2014}{\overset{1/2014}{\int }}\frac{{\tan }^{-1}\left(\frac{1}{t}\right)}{\left(\frac{1}{t}\right)}(-\frac{1}{t^2})dt\Rightarrow I=-\underset{2014}{\overset{1/2014}{\int }}\frac{{\cot }^{-1}t}{t}dt\Rightarrow I=\underset{1/2014}{\overset{2014}{\int }}\frac{{\cot }^{-1}t}{t}dt......\left(2\right)$

from $\left(1\right)+\left(2\right)$
$\Rightarrow 2I=\underset{1/2014}{\overset{2014}{\int }}\frac{{\tan }^{-1}t+{\cot }^{-1}t}{t}dt\Rightarrow 2I=\underset{1/2014}{\overset{2014}{\int }}\frac{\frac{\pi }{2}}{t}dt\Rightarrow I=\frac{\pi }{4}[\log t{]}_{1/2014}^{2014}\Rightarrow I=\frac{\pi }{4}[\log 2014-\log \frac{1}{2014}]\Rightarrow I=\frac{\pi }{4}\left(2\log 2014\right)=\frac{\pi }{2}\log 2014$