Question

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

Answer: (B)

$\frac{\pi}{2} \log 2014$

Answer 2

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