Question

$\underset{0}{\overset{1/\sqrt{3}}{\int }}{\tan }^{-1}\left(\frac{3x-x^3}{1-3x^2}\right)dx$

Answer 1

$$\begin{gathered} {\int }_0^{\frac{1}{\sqrt{3}}}{\tan }^{-1}\left(\frac{3x-x^3}{1-3x^2}\right)dx \\ \mathrm{Let}x=\tan \theta ,\mathrm{then}dx=se{c}^2\theta d\theta \\ \mathrm{When},x\to 0;\theta \to 0 \\ \mathrm{And}x\to \frac{1}{\sqrt{3}};\theta \to \frac{\mathrm{\pi }}{6} \\ \mathrm{Therefore}\mathrm{the}\mathrm{integral}\mathrm{becomes} \\ {\int }_0^{\frac{\mathrm{\pi }}{6}}{\tan }^{-1}\left(\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }\right)se{c}^2\theta d\theta \\ ={\int }_0^{\frac{\mathrm{\pi }}{6}}{\tan }^{-1}\left(\tan 3\theta \right)se{c}^2\theta d\theta \\ =3{\int }_0^{\frac{\mathrm{\pi }}{6}}\theta se{c}^2\theta d\theta \\ =3{\left[\theta \tan \theta \right]}_0^{\frac{\mathrm{\pi }}{6}}-3{\int }_0^{\frac{\mathrm{\pi }}{6}}\tan \theta d\theta \\ =3{\left[\theta \tan \theta \right]}_0^{\frac{\mathrm{\pi }}{6}}-3{\left[-\log \left(\cos \theta \right)\right]}_0^{\frac{\mathrm{\pi }}{6}} \\ =3\left(\frac{\mathrm{\pi }}{6}\times \frac{1}{\sqrt{3}}-0\right)+3\left[\log \frac{\sqrt{3}}{2}\right] \\ =\frac{\mathrm{\pi }}{2\sqrt{3}}+3\log \frac{\sqrt{3}}{2} \\ =\frac{\mathrm{\pi }}{2\sqrt{3}}-\frac{3}{2}\log \frac{4}{3} \end{gathered}$$