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Integration Questions

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

$$\mathrm{thank}\:\mathrm{you} \\$$