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Question Number 101747 by Dwaipayan Shikari last updated on 04/Jul/20

∫(x^((−1)/2) /(1+x^(1/3) ))dx

$$\int\frac{{x}^{\frac{−\mathrm{1}}{\mathrm{2}}} }{\mathrm{1}+{x}^{\frac{\mathrm{1}}{\mathrm{3}}} }{dx} \\ $$

Answered by bemath last updated on 04/Jul/20

set x = t^6 ⇒dx = 6t^5 dt ∫ (t^(−3) /(1+t^2 )) .6t^5 dt = ∫ ((6t^2 )/(1+t^2 )) dt set t = tan θ ∫ ((6tan^2 θ sec^2 θ dθ)/(sec^2 θ)) = 6tan θ−6tan^(−1) (θ)+C = 6t−6tan^(−1) (t) + c =6 (x)^(1/(6 )) − 6tan ((x)^(1/(6 )) ) + c ★

$${set}\:{x}\:=\:{t}^{\mathrm{6}} \:\Rightarrow\mathrm{dx}\:=\:\mathrm{6t}^{\mathrm{5}} \:\mathrm{dt}\: \\ $$$$\int\:\frac{\mathrm{t}^{−\mathrm{3}} }{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\:.\mathrm{6t}^{\mathrm{5}} \:\mathrm{dt}\:=\:\int\:\frac{\mathrm{6t}^{\mathrm{2}} }{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\:\mathrm{dt} \\ $$$$\mathrm{set}\:\mathrm{t}\:=\:\mathrm{tan}\:\theta \\ $$$$\int\:\frac{\mathrm{6tan}\:^{\mathrm{2}} \theta\:\mathrm{sec}\:^{\mathrm{2}} \:\theta\:\mathrm{d}\theta}{\mathrm{sec}\:^{\mathrm{2}} \theta}\:=\:\mathrm{6tan}\:\theta−\mathrm{6tan}^{−\mathrm{1}} \left(\theta\right)+\mathrm{C} \\ $$$$=\:\mathrm{6t}−\mathrm{6tan}^{−\mathrm{1}} \left(\mathrm{t}\right)\:+\:\mathrm{c} \\ $$$$=\mathrm{6}\:\sqrt[{\mathrm{6}\:}]{{x}}\:−\:\mathrm{6tan}\:\left(\sqrt[{\mathrm{6}\:}]{{x}}\right)\:+\:{c}\:\bigstar \\ $$

Commented by Dwaipayan Shikari last updated on 04/Jul/20

Great!

$$\boldsymbol{{Great}}! \\ $$

Commented by bemath last updated on 04/Jul/20

thank you

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

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