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dx-x-2-x-1-3-




Question Number 20383 by tammi last updated on 26/Aug/17
∫(dx/(x(√(2+(x)^(1/3) ))))
$$\int\frac{{dx}}{{x}\sqrt{\mathrm{2}+\left({x}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} }} \\ $$
Answered by $@ty@m last updated on 26/Aug/17
let 2+x^(1/3) =t^2 ⇒x=(t^2 −2)^3   ⇒dx=6(t^2 −2)^2 tdt  ∴ the given integral∫((6(t^2 −2)^2 tdt)/((t^2 −2)^3 t))  =6∫(dt/(t^2 −2))  =(6/(2(√2)))ln∣((t−(√2))/(t+(√2)))∣+C
$${let}\:\mathrm{2}+{x}^{\frac{\mathrm{1}}{\mathrm{3}}} ={t}^{\mathrm{2}} \Rightarrow{x}=\left({t}^{\mathrm{2}} −\mathrm{2}\right)^{\mathrm{3}} \\ $$$$\Rightarrow{dx}=\mathrm{6}\left({t}^{\mathrm{2}} −\mathrm{2}\right)^{\mathrm{2}} {tdt} \\ $$$$\therefore\:{the}\:{given}\:{integral}\int\frac{\mathrm{6}\left({t}^{\mathrm{2}} −\mathrm{2}\right)^{\mathrm{2}} {tdt}}{\left({t}^{\mathrm{2}} −\mathrm{2}\right)^{\mathrm{3}} {t}} \\ $$$$=\mathrm{6}\int\frac{{dt}}{{t}^{\mathrm{2}} −\mathrm{2}} \\ $$$$=\frac{\mathrm{6}}{\mathrm{2}\sqrt{\mathrm{2}}}{ln}\mid\frac{{t}−\sqrt{\mathrm{2}}}{{t}+\sqrt{\mathrm{2}}}\mid+{C} \\ $$

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