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The paper here https://www.imsc.res.in/~rao/ramanujan/collectedpapers/Integral/Integral1.htm says that, $\int_{0}^{x}\frac{\arctan(t)}{t}dt-\int_{0}^{\frac{1}{x}}\frac{\arctan(t)}{t}dt=\frac{1}{2}\pi\ln(x)$

I tried verifying this result graphically using Desmos. But the graph seems oddly off. It looks like the answer must be $\pi\ln(x)$ and not $\frac{1}{2}\pi\ln(x)$. Here is the graph

Any help ? Thanks :)

Mathoholic99
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    Make the substitution $u = \frac{1}{t}$ in the second integral, and then use the fact that $\arctan(x) + \arctan(\frac{1}{x}) = \frac{\pi}{2}$. – Chris Aug 26 '17 at 08:49

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Hint. You may differentiate the function given by $$ f(x)=\int_{0}^{x}\frac{\arctan(t)}{t}dt-\int_{0}^{\frac{1}{x}}\frac{\arctan(t)}{t}dt, \qquad x>0, $$ obtaining $$ f'(x)=\frac{\arctan(x)}{x}-\left(-\frac1{x^2} \right)\frac{\arctan(\frac1x)}{\frac1x}=\frac{\arctan(x)+\arctan(\frac1x)}{x}, \quad x>0, $$ then one may recall that $$ \arctan x+\arctan\frac1x=\frac \pi2,\qquad x>0, $$that is $$ f'(x)=\frac \pi2 \: \frac1x, \quad x>0. $$Observe that $f(1)=0$.

I hope you can take it from here.

The case $x<0$ is similar.

Olivier Oloa
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