Question Number 70686 by mathmax by abdo last updated on 06/Oct/19
$${find}\:\int\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{6}} } \\ $$
Answered by kaivan.ahmadi last updated on 06/Oct/19
Answered by MJS last updated on 07/Oct/19
$$\int\frac{{dx}}{{x}^{\mathrm{6}} +\mathrm{1}}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\int\frac{{dx}}{{x}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{6}}\int\frac{\sqrt{\mathrm{3}}{x}+\mathrm{2}}{{x}^{\mathrm{2}} +\sqrt{\mathrm{3}}{x}+\mathrm{1}}{dx}−\frac{\mathrm{1}}{\mathrm{6}}\int\frac{\sqrt{\mathrm{3}}{x}−\mathrm{2}}{{x}^{\mathrm{2}} −\sqrt{\mathrm{3}}{x}+\mathrm{1}}{dx}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\mathrm{arctan}\:{x}\:+ \\ $$$$\:\:\:\:\:+\frac{\sqrt{\mathrm{3}}}{\mathrm{12}}\mathrm{ln}\:\left({x}^{\mathrm{2}} +\sqrt{\mathrm{3}}{x}+\mathrm{1}\right)\:+\frac{\mathrm{1}}{\mathrm{6}}\mathrm{arctan}\:\left(\mathrm{2}{x}+\sqrt{\mathrm{3}}\right)\:− \\ $$$$\:\:\:\:\:−\frac{\sqrt{\mathrm{3}}}{\mathrm{12}}\mathrm{ln}\:\left({x}^{\mathrm{2}} −\sqrt{\mathrm{3}}{x}+\mathrm{1}\right)\:+\frac{\mathrm{1}}{\mathrm{6}}\mathrm{arctan}\:\left(\mathrm{2}{x}−\sqrt{\mathrm{3}}\right)\:+{C} \\ $$