Question Number 141531 by sarkor last updated on 20/May/21
Commented by mohammad17 last updated on 20/May/21
$${let}\:{z}^{\mathrm{6}} ={x}+\mathrm{1}\Rightarrow\mathrm{6}{z}^{\mathrm{5}} {dz}={dx} \\ $$$$ \\ $$$$\int\:\:\frac{\mathrm{6}{z}^{\mathrm{3}} \left({z}^{\mathrm{4}} +{z}\right)}{\left({z}+\mathrm{1}\right)}{dz}=\mathrm{6}\int\:\:\frac{{z}^{\mathrm{7}} +{z}^{\mathrm{4}} }{\left.{z}+\mathrm{1}\right)}{dz} \\ $$$$ \\ $$$$=\mathrm{6}\int{z}^{\mathrm{4}} \left({z}^{\mathrm{2}} −{z}+\mathrm{1}\right){dz}=\mathrm{6}\left(\frac{{z}^{\mathrm{7}} }{\mathrm{7}}−\frac{{z}^{\mathrm{6}} }{\mathrm{6}}+\frac{{z}^{\mathrm{5}} }{\mathrm{5}}\right)+{C} \\ $$$$ \\ $$$${set}:{z}=\sqrt[{\mathrm{6}}]{{x}+\mathrm{1}} \\ $$$$ \\ $$$$\therefore\int\:\:\frac{\sqrt[{\mathrm{3}}]{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }+\sqrt[{\mathrm{6}}]{{x}+\mathrm{1}}}{\:\sqrt{{x}+\mathrm{1}}+\sqrt[{\mathrm{3}}]{{x}+\mathrm{1}}}{dx}=\frac{\mathrm{6}}{\mathrm{7}}\sqrt[{\mathrm{6}}]{\left({x}+\mathrm{1}\right)^{\mathrm{7}} }−\left({x}+\mathrm{1}\right)+\frac{\mathrm{6}}{\mathrm{5}}\sqrt[{\mathrm{6}}]{\left({x}+\mathrm{1}\right)^{\mathrm{5}} }+{C} \\ $$$$ \\ $$$${by}:\langle{m}.{o}\rangle \\ $$$$ \\ $$