Question

The value of the integral $\int\limits_{1 / 2}^2 \frac{\tan ^{-1} x}{x} d x$ 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)

$I =\int\limits_{1 / 2}^2 \frac{\tan ^{-1} x }{ x } dx$...(i)
Put $ x =\frac{1}{ t } dx =-\frac{1}{ t ^2} dt$
$ I =-\int\limits_2^{1 / 2} \frac{\tan ^{-1} \frac{1}{ t }}{\frac{1}{ t }} \cdot \frac{1}{ t ^2} dt =-\int\limits_2^{1 / 2} \frac{\tan ^{-1} \frac{1}{ t }}{ t } dt $
$ I =\int\limits_{1 / 2}^2 \frac{\cot ^{-1} t }{ t } dt =\int\limits_{1 / 2}^2 \frac{\cot ^{-1} x }{ x } dx \ldots . .(ii)$
Add both equation
$2 I= \int\limits_{1 / 2}^2 \frac{\tan ^{-1} x+\cot ^{-1} x}{x} d x=\frac{\pi}{2} \int\limits_{1 / 2}^2 \frac{d x}{x}=\frac{\pi}{2}(\ell \operatorname{nn} 2)_{1 / 2}^2$
$ =\frac{\pi}{2}\left(\ell \ln 2-\ell \ln \frac{1}{2}\right)=\pi \ell \operatorname{n} 2$
$I =\frac{\pi}{2} \ell \ln 2$