Question

What is the value of the following integral?
${\int }_{1/2014}^{2014}\frac{{\tan }^{-1}x}{x}$dx

• 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)

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

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

Let $x=\frac{1}{y},$ so $dx=\frac{-dy}{y^2}$

$\therefore I=\int y{\tan }^{-1}\left(\frac{1}{y}\right)\left(\frac{-dy}{y^2}\right)$

$=\int {\tan }^{-1}\frac{\left(\frac{1}{y}\right)}{y}dy$

Changing integration variable to $x$ and adding to $\left(1\right)$ gives

$\Rightarrow 2I=\int \frac{\left({\tan }^{-1}\left(\frac{1}{x}\right)+{\tan }^{-1}x\right)}{x}dx$

But ${\tan }^{-1}\left(\frac{1}{x}\right)+{\tan }^{-1}x=\frac{\pi }{2}$ for $x>0$

$\therefore 2I={\int }_{1/a}^a\left(\frac{1}{x}\right)\frac{dx}{x}=\frac{\pi }{2}(\log a-\log \frac{1}{a})=\pi \log a$

Setting $a=2014$ gives $I=\frac{\pi }{2}\log 2014.$

Hence, the answer is $\frac{\pi }{2}\log 2014.$