Question Number 159315 by cortano last updated on 15/Nov/21
$$\:\int\:\frac{{dx}}{\:\sqrt[{\mathrm{4}}]{\left({x}−\mathrm{1}\right)^{\mathrm{3}} \left({x}+\mathrm{2}\right)^{\mathrm{5}} }\:}\:? \\ $$
Commented by tounghoungko last updated on 15/Nov/21
$${Y}=\int\:\frac{{dx}}{\left({x}+\mathrm{2}\right)^{\mathrm{2}} \left(\frac{{x}−\mathrm{1}}{{x}+\mathrm{2}}\right)^{\mathrm{3}/\mathrm{4}} }\: \\ $$$${let}\:\sqrt[{\mathrm{4}}]{\frac{{x}−\mathrm{1}}{{x}+\mathrm{2}}}\:=\:{v}\:\Rightarrow{x}=\frac{\mathrm{2}{v}^{\mathrm{4}} +\mathrm{1}}{\mathrm{1}−{v}^{\mathrm{4}} } \\ $$$${Y}=\:\int\frac{\mathrm{1}}{\left(\frac{\mathrm{2}{v}^{\mathrm{4}} +\mathrm{1}+\mathrm{2}−\mathrm{2}{v}^{\mathrm{4}} }{\mathrm{1}−{v}^{\mathrm{4}} }\right)^{\mathrm{2}} {v}^{\mathrm{3}} }\:\left(\frac{\mathrm{12}{v}^{\mathrm{3}} }{\left(\mathrm{1}−{v}^{\mathrm{4}} \right)^{\mathrm{2}} }\right){dv} \\ $$$${Y}=\:\int\:\frac{\mathrm{12}}{\mathrm{9}}\:{dv}\:=\:\frac{\mathrm{4}}{\mathrm{3}}{v}\:+{c}\:=\:\frac{\mathrm{4}}{\mathrm{3}}\sqrt[{\mathrm{4}}]{\frac{{x}−\mathrm{1}}{{x}+\mathrm{2}}}\:+{c}\: \\ $$