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




Question Number 20384 by tammi last updated on 26/Aug/17
∫((x^3 dx)/((2+3x)^2 ))
$$\int\frac{{x}^{\mathrm{3}} {dx}}{\left(\mathrm{2}+\mathrm{3}{x}\right)^{\mathrm{2}} } \\ $$
Answered by $@ty@m last updated on 26/Aug/17
let 2+3x=t ⇒3dx=dt & x=((t−2)/3)  ∫(((t−2)/3))^3 (dt/(3t^2 ))  =(1/(81))∫((t^3 −8−6t^2 +12t)/t^2 )dt  =(1/(81))∫(t−(8/t^2 )−6+((12)/t))dt  =(1/(81))((t^2 /2)+(8/t)−6t+12lnt)+C
$${let}\:\mathrm{2}+\mathrm{3}{x}={t}\:\Rightarrow\mathrm{3}{dx}={dt}\:\&\:{x}=\frac{{t}−\mathrm{2}}{\mathrm{3}} \\ $$$$\int\left(\frac{{t}−\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{3}} \frac{{dt}}{\mathrm{3}{t}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{81}}\int\frac{{t}^{\mathrm{3}} −\mathrm{8}−\mathrm{6}{t}^{\mathrm{2}} +\mathrm{12}{t}}{{t}^{\mathrm{2}} }{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{81}}\int\left({t}−\frac{\mathrm{8}}{{t}^{\mathrm{2}} }−\mathrm{6}+\frac{\mathrm{12}}{{t}}\right){dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{81}}\left(\frac{{t}^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{8}}{{t}}−\mathrm{6}{t}+\mathrm{12}{lnt}\right)+{C} \\ $$

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