## Calculus 8th Edition

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