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Question Number 51456 by Tawa1 last updated on 27/Dec/18

$$\int \frac{{\mathrm{x}}^8}{{\mathrm{x}}^6+64}\mathrm{dx}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 27/Dec/18

$$x^3=2^{3}tan\theta$$$$x=2{\left(tan\theta \right)}^{\frac{1}{3}}$$$$dx=\frac{2}{3}\times {\left(tan\theta \right)}^{\frac{-2}{3}}\times {sec}^2\theta$$$$\int \frac{2^8\times {\left(tan\theta \right)}^{\frac{8}{3}}\times \frac{2}{3}\times {\left(tan\theta \right)}^{\frac{-2}{3}}{sec}^2\theta }{2^6{tan}^2\theta +2^6}d\theta$$$$=\frac{2^3}{3}\int {\left(tan\theta \right)}^{\frac{8}{3}-\frac{2}{3}}d\theta$$$$=\frac{8}{3}\int {tan}^2\theta d\theta$$$$=\frac{8}{3}[\int ({sec}^2\theta -1)d\theta$$$$=\frac{8}{3}[tan\theta -\theta ]+c$$$$=\frac{8}{3}[\frac{x^3}{8}-{tan}^{-1}\left(\frac{x^3}{8}\right)]+c$$

Commented by Tawa1 last updated on 27/Dec/18

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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