Question

The value of the integral $\underset{1/2}{\overset{2}{\int }}\frac{{\tan }^{-1}x}{x}dx$ is equal to

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

Answer 1

Answer: (D)

$I=\underset{1/2}{\overset{2}{\int }}\frac{{\tan }^{-1}x}{x}dx$...(i)
Put $x=\frac{1}{t}dx=-\frac{1}{t^2}dt$
$I=-\underset{2}{\overset{1/2}{\int }}\frac{{\tan }^{-1}\frac{1}{t}}{\frac{1}{t}}\cdot \frac{1}{t^2}dt=-\underset{2}{\overset{1/2}{\int }}\frac{{\tan }^{-1}\frac{1}{t}}{t}dt$
$I=\underset{1/2}{\overset{2}{\int }}\frac{{\cot }^{-1}t}{t}dt=\underset{1/2}{\overset{2}{\int }}\frac{{\cot }^{-1}x}{x}dx\dots ..(ii)$
Add both equation
$2I=\underset{1/2}{\overset{2}{\int }}\frac{{\tan }^{-1}x+{\cot }^{-1}x}{x}dx=\frac{\pi }{2}\underset{1/2}{\overset{2}{\int }}\frac{dx}{x}=\frac{\pi }{2}(\ell \mathrm{nn}2{)}_{1/2}^2$
$=\frac{\pi }{2}\left(\ell \ln 2-\ell \ln \frac{1}{2}\right)=\pi \ell n2$
$I=\frac{\pi }{2}\ell \ln 2$