If $$f(x)=\int_1^x\frac{\tan^{-1}t}{t}\,\mathrm dt,$$ find $f(e^2)-f\left(\dfrac{1}{e^2}\right)$.
I think that Newton-Leibniz formula should be applied but the problem is that that if I apply it, I would get $f'(x)$ but I have to find the answer in terms of $f(e^2)$ which involves no derivatives.
How to go about it?
$\def\d{\mathrm{d}}$Note that $\arctan \dfrac{1}{x} = \dfrac{π}{2} - \arctan x$ for any $x > 0$, then\begin{align*} f\left( \frac{1}{x} \right) &= \int_1^{\textstyle \frac{1}{x}} \frac{\arctan t}{t} \,\d t = \int_1^x \frac{\dfrac{π}{2} - \arctan u}{\dfrac{1}{u}} · \left( -\frac{1}{u^2} \right) \,\d u\\ &= -\frac{π}{2} \int_1^x \frac{\d u}{u} + \int_1^x \frac{\arctan u}{u} \,\d u = -\frac{π}{2} \ln x + f(x). \end{align*}
Let $g(x)=f(x)-f(1/x)$. Then $$g'(x)=f'(x)+\frac{f'(1/x)}{x^2} =\frac{\tan^{-1}x}x+\frac{\tan^{-1}(1/x)}x=\frac{\pi}{2x}.$$ As $g(1)=0$ then $g(x)=\frac\pi2\ln x$. In particular, $g(e^2)=\pi$.
