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dx-x-1-3-2-




Question Number 103846 by Dwaipayan Shikari last updated on 17/Jul/20
∫(dx/(x^(1/3) +2))
$$\int\frac{{dx}}{{x}^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{2}} \\ $$
Commented by Dwaipayan Shikari last updated on 17/Jul/20
∫((3u^2 du)/(u+2))=∫((3u^2 −12)/(u+2))+((12)/(u+2))=∫3u−6+12log(x^(1/3) +2)=(3/2)x^(2/3) −6x^(1/3) +12log(x^(1/3) +2)+C
$$\int\frac{\mathrm{3}{u}^{\mathrm{2}} {du}}{{u}+\mathrm{2}}=\int\frac{\mathrm{3}{u}^{\mathrm{2}} −\mathrm{12}}{{u}+\mathrm{2}}+\frac{\mathrm{12}}{{u}+\mathrm{2}}=\int\mathrm{3}{u}−\mathrm{6}+\mathrm{12}{log}\left({x}^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{2}\right)=\frac{\mathrm{3}}{\mathrm{2}}{x}^{\frac{\mathrm{2}}{\mathrm{3}}} −\mathrm{6}{x}^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{12}{log}\left({x}^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{2}\right)+{C} \\ $$
Commented by PRITHWISH SEN 2 last updated on 17/Jul/20
x=t^3 ⇒ dx=3t^2 dt  ∫((3t^2 dt)/(t+2))=3 ∫(((t+2)^2 −4(t+2)+4)/((t+2)))dt  =3{∫(t+2)dt−4∫dt+4∫(dt/(t+2))
$$\mathrm{x}=\mathrm{t}^{\mathrm{3}} \Rightarrow\:\mathrm{dx}=\mathrm{3t}^{\mathrm{2}} \mathrm{dt} \\ $$$$\int\frac{\mathrm{3t}^{\mathrm{2}} \mathrm{dt}}{\mathrm{t}+\mathrm{2}}=\mathrm{3}\:\int\frac{\left(\mathrm{t}+\mathrm{2}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{t}+\mathrm{2}\right)+\mathrm{4}}{\left(\mathrm{t}+\mathrm{2}\right)}\mathrm{dt} \\ $$$$=\mathrm{3}\left\{\int\left(\mathrm{t}+\mathrm{2}\right)\mathrm{dt}−\mathrm{4}\int\mathrm{dt}+\mathrm{4}\int\frac{\mathrm{dt}}{\mathrm{t}+\mathrm{2}}\right. \\ $$

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