Answer
$\displaystyle\frac{\pi}{2\sqrt{3}} - \frac{π}{4}+\frac{1}{2}\ln(2) ≈ 0.468075$
Work Step by Step
$\displaystyle\int_1^\sqrt{3}\tan^{-1}(\frac{1}{x})dx = \int_1^\sqrt{3}\tan^{-1}(1)(\frac{1}{x})dx$
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Let $g' = 1$ and $f = \displaystyle \tan^{-1}(\frac{1}{x})$
thus $g = x$ and $f' = \displaystyle \frac {x}{1+x^2}$
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$(\displaystyle x\tan^{-1}(\frac{1}{x}))|_1^\sqrt{3} - \int_1^\sqrt{3}\frac {x}{1+x^2}dx, $ let $u = x^2+1 $ and $du=2x$ $dx$
$\displaystyle [\sqrt{3}\tan^{-1}(\frac{1}{\sqrt{3}})-(1)\tan^{-1}(1)] - \frac {1}{2}\int_2^4\frac{du}{u}$
$\displaystyle \sqrt{3}(\frac{π}{6})-\frac{π}{4}-\frac{1}{2}(\ln|u|)|_2^4$
$\displaystyle \frac{\pi \cdot \sqrt{3}}{2\cdot3}-\frac {\pi}{4}-\frac{1}{2}\ln|1+x^2|_1^\sqrt{3}$
$\displaystyle \frac{\pi}{2\sqrt{3}}-\frac {\pi}{4}-\frac{1}{2}[\ln(4)-\ln(2)]$
$\displaystyle\frac{\pi}{2\sqrt{3}} - \frac{π}{4}+\frac{1}{2}\ln(2) ≈ 0.468075$
