Question

Let $f\left(x\right)={\int }_1^x\frac{{\tan }^{-1}t}{t}dt;(x>0)$. The value of $f\left(e^2\right)-f\left(\frac{1}{e^2}\right)$ is $\frac{m\pi }{8}$, then the value of $mis$

Answer 1

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