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Question Number 152273 by peter frank last updated on 27/Aug/21

$$\int \frac{\tan \theta +\tan {}^3\theta }{1+\tan {}^3\theta }\mathrm{d}\theta$$

Answered by qaz last updated on 27/Aug/21

$$\int \frac{\tan \theta +\tan {}^3\theta }{1+\tan {}^3\theta }\mathrm{d}\theta$$$$=\int \frac{\tan \theta }{1+\tan {}^3\theta }\mathrm{d}\left(\tan \theta \right)$$$$=\int \frac{\mathrm{xdx}}{1+{\mathrm{x}}^3}$$$$=\int \frac{\mathrm{x}}{(\mathrm{x}+1)({\mathrm{x}}^2-\mathrm{x}+1)}\mathrm{dx}$$$$=\frac{1}{3}\int (\frac{\mathrm{x}+1}{{\mathrm{x}}^2-\mathrm{x}+1}-\frac{1}{\mathrm{x}+1})\mathrm{dx}$$$$=\frac{1}{6}\int \frac{2\mathrm{x}-1+3}{{\mathrm{x}}^2-\mathrm{x}+1}\mathrm{dx}-\frac{1}{3}\int \frac{1}{\mathrm{x}+1}\mathrm{dx}$$$$=\frac{1}{6}\ln \mid {\mathrm{x}}^2-\mathrm{x}+1\mid +\frac{1}{6}\cdot \frac{1}{2\cdot \sqrt{\frac{3}{4}}}{\tan }^{-1}\frac{\mathrm{x}-\frac{1}{2}}{\sqrt{\frac{3}{4}}}-\frac{1}{3}\ln \mid \mathrm{x}+1\mid +\mathrm{C}$$$$=\frac{1}{6}\ln \mid \frac{{\mathrm{x}}^2-\mathrm{x}+1}{{(\mathrm{x}+1)}^2}\mid +\frac{1}{6\sqrt{3}}{\tan }^{-1}\frac{2\mathrm{x}-1}{\sqrt{3}}+\mathrm{C}$$$$=\frac{1}{6}\ln \mid \frac{\tan {}^2\theta -\tan \theta +1}{{(\tan \theta +1)}^2}\mid +\frac{1}{6\sqrt{3}}{\tan }^{-1}\frac{2\tan \theta -1}{\sqrt{3}}+\mathrm{C}$$

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