Question

Let $f\left(x\right)={\int }_1^x\frac{{\tan }^{-1}t}{t}dt$ $(x>0)$ then $f\left(e^2\right)-f\left(\frac{1}{e^2}\right)$ is

• A. $\frac{\pi }{2}$
• B. $\pi$
• C. $2\pi$
• D. $\frac{\pi }{4}$

Answer 1

The correct option is B $\pi$

$f\left(e^2\right)-f\left(\frac{1}{e^2}\right)$
$={\int }_1^{e^2}\frac{{\tan }^{-1}t}{t}dt-{\int }_1^{\frac{1}{e^2}}\frac{{\tan }^{-1}t}{t}dt\text{For second integral substitute}t=\frac{1}{z}\Rightarrow dt=-\frac{1}{z^2}dz{\int }_1^{e^2}\frac{{\tan }^{-1}t}{t}dt+{\int }_1^{e^2}\frac{{\tan }^{-1}\left(\frac{1}{z}\right)}{z}dz\text{For second integral put}z=t\text{integral will become}{\int }_1^{e^2}\frac{{\tan }^{-1}t}{t}dt+{\int }_1^{e^2}\frac{{\tan }^{-1}\left(\frac{1}{t}\right)}{t}dt\text{We know that,}{\tan }^{-1}\left(\frac{1}{t}\right)={\cot }^{-}\left(t\right)\text{After adding integrals we get,}={\int }_1^{e^2}\frac{\frac{\pi }{2}}{t}dt=\frac{\pi }{2}.\left(2\right)=\pi$