Question Number 158742 by cortano last updated on 08/Nov/21
Commented by HongKing last updated on 08/Nov/21
$$=\:\frac{\mathrm{2}}{\mathrm{5}}\:\left(\mathrm{x}^{\mathrm{6}} \:+\:\mathrm{x}^{\mathrm{4}} \:+\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{4}} }\right)^{\frac{\mathrm{5}}{\mathrm{4}}} +\:\mathbb{C} \\ $$
Answered by tounghoungko last updated on 08/Nov/21
$$\int\:\frac{{x}^{\mathrm{5}} \left(\mathrm{3}{x}^{\mathrm{5}} +\mathrm{2}{x}^{\mathrm{3}} −\mathrm{2}{x}^{−\mathrm{5}} \right){x}\:\sqrt[{\mathrm{4}}]{{x}^{\mathrm{6}} +{x}^{\mathrm{4}} +{x}^{−\mathrm{4}} }}{{x}^{\mathrm{6}} }\:{dx} \\ $$$$=\int\:\left(\mathrm{3}{x}^{\mathrm{5}} +\mathrm{2}{x}^{\mathrm{3}} −\mathrm{2}{x}^{−\mathrm{5}} \right)\:\sqrt[{\mathrm{4}}]{{x}^{\mathrm{6}} +{x}^{\mathrm{4}} +{x}^{−\mathrm{4}} }\:{dx} \\ $$$${let}\:\lambda={x}^{\mathrm{6}} +{x}^{\mathrm{4}} +{x}^{−\mathrm{4}} \:\Rightarrow{d}\lambda=\mathrm{6}{x}^{\mathrm{5}} +\mathrm{4}{x}^{\mathrm{3}} −\mathrm{4}{x}^{−\mathrm{5}} \:{dx} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}}\int\:\lambda^{\mathrm{1}/\mathrm{4}} \:{d}\lambda=\:\frac{\mathrm{2}}{\mathrm{5}}\lambda^{\mathrm{5}/\mathrm{4}} \:+{c} \\ $$$$=\frac{\mathrm{2}}{\mathrm{5}}\sqrt[{\mathrm{4}}]{\left({x}^{\mathrm{6}} +{x}^{\mathrm{4}} +{x}^{−\mathrm{4}} \right)^{\mathrm{5}} }\:+\:{c} \\ $$$$=\frac{\mathrm{2}}{\mathrm{5}}\:\sqrt[{\mathrm{4}}]{\frac{\left({x}^{\mathrm{10}} +{x}^{\mathrm{8}} +\mathrm{1}\right)^{\mathrm{5}} }{{x}^{\mathrm{20}} }}\:+{c} \\ $$$$=\frac{\mathrm{2}\:\sqrt[{\mathrm{4}}]{\left({x}^{\mathrm{10}} +{x}^{\mathrm{8}} +\mathrm{1}\right)^{\mathrm{5}} }}{\mathrm{5}{x}^{\:\mathrm{5}} }\:+\:{c}\: \\ $$