Q51 / 155 · WBJEE 2018 · Mathematics
The value of the integral $I=\int_{1 / 2014}^{2014} \frac{\tan ^{-1} x}{x} d x$ is
- A$\frac{\pi}{4} \log 2014$
- Correct answerB$\frac{\pi}{2} \log 2014$
- C$\pi \log 2014$
- D$\frac{1}{2} \log 2014$
Solution
We have,
$$
I=\int_{1 / 2014}^{2014} \frac{\tan ^{-1} x}{x} d x
$$
Let
$$
x=\frac{1}{t}
$$
$\Rightarrow$
$$
d x=\frac{-1}{t^{2}} d t
$$
Now,
$$
\begin{aligned}
I &=\int_{2014}^{1 / 2014} \frac{\tan ^{-1}(1 / 7)}{1 / t}\left(\frac{-1}{t^{2}} d t\right) \\
&=\int_{1 / 2014}^{2014} \frac{\cot ^{-1} t}{t} d t \\
&=\int_{1 / 2014}^{2014} \frac{\cot ^{-1} x}{x} d x
\end{aligned}
$$
On adding Eqs. (i) and (ii), we get
$$
\begin{aligned}
2 I &=\int_{1 / 2014}^{2014} \frac{\pi / 2}{x} d x=\frac{\pi}{2}(\log x)_{1 / 2014}^{2014} \\
&=\frac{\pi}{2}(\log 2014-\log 1 / 2014)
\end{aligned}
$$
$\Rightarrow$
$$
\begin{aligned}
I &=\frac{\pi}{4}\left(\log 2014-\log \frac{1}{2014}\right) \\
&=\frac{\pi}{4}(\log 2014+\log 2014) \\
&=\frac{\pi}{4}(2 \log 2014)=\frac{\pi}{2} \log 2014
\end{aligned}
$$
Hence, the correct answer is option: 'B'.