$\frac{\mathrm{\θ}}{1\+r^2}$

$\frac{\mathrm{\θ}}{1\+r^2}$

$0$

$0$

$\mathrm{\theta }$

$\mathrm{\θ}$

$0$

$0$

$\frac{\mathrm{\pi }}{3}$

$\frac{1}{3}\mathrm{\π}$

$\mathrm{Student}\left[\mathrm{MultivariateCalculus}\right]\left[\mathrm{MultiInt}\right]\left(\,r\=..\,\mathrm{\θ}\=..\,\mathrm{coordinates}\=\mathrm{polar}\left[r\,\mathrm{\θ}\right]\,\mathrm{output}\=\mathrm{integral}\right)$

${\∫}_0^{\frac{1}{3}\mathrm{\π}}{\∫}_0^{\mathrm{\θ}}\frac{\mathrm{\θ}r}{1\+r^2}\ⅆr\ⅆ\mathrm{\θ}$

$\mathrm{Student}\left[\mathrm{MultivariateCalculus}\right]\left[\mathrm{MultiInt}\right]\left(\,r\=..\,\mathrm{\θ}\=..\,\mathrm{coordinates}\=\mathrm{polar}\left[r\,\mathrm{\θ}\right]\right)$

$-\frac{1}{2}\ln \left(3\right)\+\frac{1}{4}\ln \left(9\+{\mathrm{\π}}^2\right)-\frac{1}{18}{\mathrm{\π}}^2\ln \left(3\right)\+\frac{1}{36}{\mathrm{\π}}^2\ln \left(9\+{\mathrm{\π}}^2\right)-\frac{1}{36}{\mathrm{\π}}^2$

$\mathrm{Student}\left[\mathrm{MultivariateCalculus}\right]\left[\mathrm{MultiInt}\right]\left(\,r\=..\,\mathrm{\θ}\=..\,\mathrm{coordinates}\=\mathrm{polar}\left[r\,\mathrm{\θ}\right]\,\mathrm{output}\=\mathrm{steps}\right)$

$\begin{array}{ccc}\multicolumn{3}{c}{{\int }_0^{\frac{\mathrm{\pi }}{3}}{\int }_0^{\mathrm{\theta }}\frac{\mathrm{\theta }r}{1+r^2}\ⅆr\ⅆ\mathrm{\theta }}\\ & \text{=}& {\int }_0^{\frac{\mathrm{\pi }}{3}}\left(\genfrac{}{}{0ex}{}{\frac{\mathrm{\theta }\ln \left(1+r^2\right)}{2}}{\phantom{r=0..\mathrm{\theta }}}|\genfrac{}{}{0ex}{}{\phantom{\frac{\mathrm{\theta }\ln \left(1+r^2\right)}{2}}}{r=0..\mathrm{\theta }}\right)\ⅆ\mathrm{\theta }\hfill \\ & \text{=}& {\int }_0^{\frac{\mathrm{\pi }}{3}}\frac{\ln \left(1+{\mathrm{\theta }}^2\right)\mathrm{\theta }}{2}\ⅆ\mathrm{\theta }\hfill \\ & \text{=}& \genfrac{}{}{0ex}{}{\left(\frac{\left(1+{\mathrm{\theta }}^2\right)\ln \left(1+{\mathrm{\theta }}^2\right)}{4}-\frac{1}{4}-\frac{{\mathrm{\theta }}^2}{4}\right)}{\phantom{\mathrm{\theta }=0..\frac{\mathrm{\pi }}{3}}}|\genfrac{}{}{0ex}{}{\phantom{\left(\frac{\left(1+{\mathrm{\theta }}^2\right)\ln \left(1+{\mathrm{\theta }}^2\right)}{4}-\frac{1}{4}-\frac{{\mathrm{\theta }}^2}{4}\right)}}{\mathrm{\theta }=0..\frac{\mathrm{\pi }}{3}}\hfill \end{array}$

$-\frac{1}{2}\ln \left(3\right)\+\frac{1}{4}\ln \left(9\+{\mathrm{\π}}^2\right)-\frac{1}{18}{\mathrm{\π}}^2\ln \left(3\right)\+\frac{1}{36}{\mathrm{\π}}^2\ln \left(9\+{\mathrm{\π}}^2\right)-\frac{1}{36}{\mathrm{\π}}^2$