Question Number 53357 by gunawan last updated on 20/Jan/19
$$\int\:\:\frac{\left({x}^{\mathrm{4}} −{x}\right)^{\mathrm{1}/\mathrm{4}} }{{x}^{\mathrm{5}} }\:{dx}\:= \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 21/Jan/19
$$\int\frac{\left\{{x}^{\mathrm{4}} \left(\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right)\right\}^{\frac{\mathrm{1}}{\mathrm{4}}} }{{x}^{\mathrm{5}} }{dx} \\ $$$$\int\frac{\left(\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right)^{\frac{\mathrm{1}}{\mathrm{4}}} }{{x}^{\mathrm{4}} }{dx} \\ $$$${t}^{\mathrm{4}} =\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{3}} } \\ $$$$\mathrm{4}{t}^{\mathrm{3}} {dt}=\left(\mathrm{0}+\frac{\mathrm{3}}{{x}^{\mathrm{4}} }\right){dx} \\ $$$$\frac{\mathrm{4}{t}^{\mathrm{3}} }{\mathrm{3}}{dt}=\frac{{dx}}{{x}^{\mathrm{4}} } \\ $$$$\int\frac{\left({t}^{\mathrm{4}} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} ×\frac{\mathrm{4}{t}^{\mathrm{3}} }{\mathrm{3}}{dt}}{\mathrm{1}}× \\ $$$$\frac{\mathrm{4}}{\mathrm{3}}\int{t}^{\mathrm{4}} {dt} \\ $$$$\frac{\mathrm{4}}{\mathrm{15}}×{t}^{\mathrm{5}} +{c} \\ $$$$\frac{\mathrm{4}}{\mathrm{15}}×\left(\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right)^{\frac{\mathrm{5}}{\mathrm{4}}} +{c} \\ $$