Question Number 84733 by manr last updated on 15/Mar/20
$$\int\frac{{x}^{\mathrm{4}} +\mathrm{1}}{{x}^{\mathrm{6}} +\mathrm{1}}{dx} \\ $$
Answered by TANMAY PANACEA last updated on 15/Mar/20
$$ \\ $$$$\int\frac{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}{x}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{4}} −{x}^{\mathrm{2}} +\mathrm{1}\right)}{dx} \\ $$$$\int\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}^{\mathrm{4}} −{x}^{\mathrm{2}} +\mathrm{1}}−\mathrm{2}\int\frac{{x}^{\mathrm{2}} {dx}}{{x}^{\mathrm{6}} +\mathrm{1}} \\ $$$$\int\frac{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}{\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\mathrm{1}\right)}−\frac{\mathrm{2}}{\mathrm{3}}\int\frac{{d}\left({x}^{\mathrm{3}} \right)}{\left({x}^{\mathrm{3}} \right)^{\mathrm{2}} +\mathrm{1}} \\ $$$${I}={I}_{\mathrm{1}} −{I}_{\mathrm{2}} \\ $$$${I}_{\mathrm{1}} =\int\frac{{d}\left({x}−\frac{\mathrm{1}}{{x}}\right)}{\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} +\mathrm{1}}={tan}^{−\mathrm{1}} \left({x}−\frac{\mathrm{1}}{{x}}\right)+{C}_{\mathrm{1}} \\ $$$${I}_{\mathrm{2}} =\frac{\mathrm{2}}{\mathrm{3}}\int\frac{{d}\left({x}^{\mathrm{3}} \right)}{\left({x}^{\mathrm{3}} \right)^{\mathrm{2}} +\mathrm{1}}=\frac{\mathrm{2}}{\mathrm{3}}{tan}^{−\mathrm{1}} \left({x}^{\mathrm{3}} \right) \\ $$$${I}={tan}^{−} \left({x}−\frac{\mathrm{1}}{{x}}\right)+\frac{\mathrm{2}}{\mathrm{3}}{tan}^{−\mathrm{1}} \left({x}^{\mathrm{3}} \right)+{C} \\ $$$$ \\ $$