Question Number 101747 by Dwaipayan Shikari last updated on 04/Jul/20 | ||
$$\int\frac{{x}^{\frac{−\mathrm{1}}{\mathrm{2}}} }{\mathrm{1}+{x}^{\frac{\mathrm{1}}{\mathrm{3}}} }{dx} \\ $$ | ||
Answered by bemath last updated on 04/Jul/20 | ||
$${set}\:{x}\:=\:{t}^{\mathrm{6}} \:\Rightarrow\mathrm{dx}\:=\:\mathrm{6t}^{\mathrm{5}} \:\mathrm{dt}\: \\ $$$$\int\:\frac{\mathrm{t}^{−\mathrm{3}} }{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\:.\mathrm{6t}^{\mathrm{5}} \:\mathrm{dt}\:=\:\int\:\frac{\mathrm{6t}^{\mathrm{2}} }{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\:\mathrm{dt} \\ $$$$\mathrm{set}\:\mathrm{t}\:=\:\mathrm{tan}\:\theta \\ $$$$\int\:\frac{\mathrm{6tan}\:^{\mathrm{2}} \theta\:\mathrm{sec}\:^{\mathrm{2}} \:\theta\:\mathrm{d}\theta}{\mathrm{sec}\:^{\mathrm{2}} \theta}\:=\:\mathrm{6tan}\:\theta−\mathrm{6tan}^{−\mathrm{1}} \left(\theta\right)+\mathrm{C} \\ $$$$=\:\mathrm{6t}−\mathrm{6tan}^{−\mathrm{1}} \left(\mathrm{t}\right)\:+\:\mathrm{c} \\ $$$$=\mathrm{6}\:\sqrt[{\mathrm{6}\:}]{{x}}\:−\:\mathrm{6tan}\:\left(\sqrt[{\mathrm{6}\:}]{{x}}\right)\:+\:{c}\:\bigstar \\ $$ | ||
Commented by Dwaipayan Shikari last updated on 04/Jul/20 | ||
$$\boldsymbol{{Great}}! \\ $$ | ||
Commented by bemath last updated on 04/Jul/20 | ||
$$\mathrm{thank}\:\mathrm{you} \\ $$ | ||