Question Number 51419 by Tawa1 last updated on 26/Dec/18
$$\int\:\frac{\sqrt{\mathrm{x}}}{\mathrm{1}\:+\:\:\sqrt[{\mathrm{3}}]{\mathrm{x}}}\:\mathrm{dx} \\ $$
Answered by Smail last updated on 27/Dec/18
$${let}\:{u}=\sqrt[{\mathrm{6}}]{{x}}\Rightarrow\mathrm{6}{u}^{\mathrm{5}} {du}={dx} \\ $$$${A}=\int\frac{\sqrt{{x}}}{\mathrm{1}+\sqrt[{\mathrm{3}}]{{x}}}{dx}=\int\frac{{u}^{\mathrm{3}} }{\mathrm{1}+{u}^{\mathrm{2}} }\left(\mathrm{6}{u}^{\mathrm{5}} {du}\right) \\ $$$$=\mathrm{6}\int\frac{{u}^{\mathrm{8}} }{\mathrm{1}+{u}^{\mathrm{2}} }{du} \\ $$$${u}^{\mathrm{8}} ={u}^{\mathrm{6}} \left({u}^{\mathrm{2}} +\mathrm{1}\right)−{u}^{\mathrm{4}} \left({u}^{\mathrm{2}} +\mathrm{1}\right)+{u}^{\mathrm{2}} \left({u}^{\mathrm{2}} +\mathrm{1}\right)−\left({u}^{\mathrm{2}} +\mathrm{1}\right)+\mathrm{1} \\ $$$$=\left({u}^{\mathrm{2}} +\mathrm{1}\right)\left({u}^{\mathrm{6}} −{u}^{\mathrm{4}} +{u}^{\mathrm{2}} −\mathrm{1}\right)+\mathrm{1} \\ $$$$\frac{{u}^{\mathrm{8}} }{{u}^{\mathrm{2}} +\mathrm{1}}={u}^{\mathrm{6}} −{u}^{\mathrm{4}} +{u}^{\mathrm{2}} −\mathrm{1}+\frac{\mathrm{1}}{{u}^{\mathrm{2}} +\mathrm{1}} \\ $$$${A}=\mathrm{6}\left(\frac{{u}^{\mathrm{7}} }{\mathrm{7}}−\frac{{u}^{\mathrm{5}} }{\mathrm{5}}+\frac{{u}^{\mathrm{3}} }{\mathrm{3}}−{u}+{tan}^{−\mathrm{1}} \left({u}\right)\right)+{C} \\ $$$$=\mathrm{6}\left(\frac{\left(\sqrt[{\mathrm{6}}]{{x}}\right)^{\mathrm{7}} }{\mathrm{7}}−\frac{\left(\sqrt[{\mathrm{6}}]{{x}}\right)^{\mathrm{5}} }{\mathrm{5}}+\frac{\left(\sqrt[{\mathrm{6}}]{{x}}\right)^{\mathrm{3}} }{\mathrm{3}}−\sqrt[{\mathrm{6}}]{{x}}+{tan}^{−\mathrm{1}} \left(\sqrt[{\mathrm{6}}]{{x}}\right)\right)+{C} \\ $$
Commented by Tawa1 last updated on 27/Dec/18
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$