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Question Number 156422 by SANOGO last updated on 11/Oct/21

Commented by cortano last updated on 11/Oct/21

let x=t^6 →dx=6t^5 dt I=∫ (t^3 /(1−t^2 )) (6t^5 dt) = ∫ ((6t^8 )/(1−t^2 )) dt partial fraction

$$\:\mathrm{let}\:\mathrm{x}=\mathrm{t}^{\mathrm{6}} \:\rightarrow\mathrm{dx}=\mathrm{6t}^{\mathrm{5}} \:\mathrm{dt} \\ $$$$\mathrm{I}=\int\:\frac{\mathrm{t}^{\mathrm{3}} }{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }\:\left(\mathrm{6t}^{\mathrm{5}} \:\mathrm{dt}\right) \\ $$$$=\:\int\:\frac{\mathrm{6t}^{\mathrm{8}} }{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }\:\mathrm{dt}\: \\ $$$$\mathrm{partial}\:\mathrm{fraction} \\ $$

Commented by SANOGO last updated on 11/Oct/21

merci bien

$${merci}\:{bien} \\ $$

Answered by puissant last updated on 11/Oct/21

K=∫(1/( (√x)−(x)^(1/3) ))dx ; x=u^6 →dx=6u^5 du ⇒ K=∫((6u^5 )/(u^3 −u^2 ))du = 6∫(u^3 /(u−1))du =6∫((u^3 −1+1)/(u−1))du = 6∫((u^3 −1)/(u−1))du+6∫(1/(u−1))du =6∫{u^2 +u+1}du+6ln∣u∣ ⇒ K=2u^3 +3u^2 +6u+6ln∣u∣+C ⇒ K=2(x^(1/6) )^3 +3(x^(1/6) )^2 +6x^(1/6) +6ln∣x^(1/6) ∣+C ∴∵ K=2(√x)+3(x)^(1/3) +6(x)^(1/6) +ln∣x∣+C.. ..........Le puissant...........

$${K}=\int\frac{\mathrm{1}}{\:\sqrt{{x}}−\sqrt[{\mathrm{3}}]{{x}}}{dx}\:;\:{x}={u}^{\mathrm{6}} \rightarrow{dx}=\mathrm{6}{u}^{\mathrm{5}} {du} \\ $$$$\Rightarrow\:{K}=\int\frac{\mathrm{6}{u}^{\mathrm{5}} }{{u}^{\mathrm{3}} −{u}^{\mathrm{2}} }{du}\:=\:\mathrm{6}\int\frac{{u}^{\mathrm{3}} }{{u}−\mathrm{1}}{du} \\ $$$$=\mathrm{6}\int\frac{{u}^{\mathrm{3}} −\mathrm{1}+\mathrm{1}}{{u}−\mathrm{1}}{du}\:=\:\mathrm{6}\int\frac{{u}^{\mathrm{3}} −\mathrm{1}}{{u}−\mathrm{1}}{du}+\mathrm{6}\int\frac{\mathrm{1}}{{u}−\mathrm{1}}{du} \\ $$$$=\mathrm{6}\int\left\{{u}^{\mathrm{2}} +{u}+\mathrm{1}\right\}{du}+\mathrm{6}{ln}\mid{u}\mid \\ $$$$\Rightarrow\:{K}=\mathrm{2}{u}^{\mathrm{3}} +\mathrm{3}{u}^{\mathrm{2}} +\mathrm{6}{u}+\mathrm{6}{ln}\mid{u}\mid+{C} \\ $$$$\Rightarrow\:{K}=\mathrm{2}\left({x}^{\frac{\mathrm{1}}{\mathrm{6}}} \right)^{\mathrm{3}} +\mathrm{3}\left({x}^{\frac{\mathrm{1}}{\mathrm{6}}} \right)^{\mathrm{2}} +\mathrm{6}{x}^{\frac{\mathrm{1}}{\mathrm{6}}} +\mathrm{6}{ln}\mid{x}^{\frac{\mathrm{1}}{\mathrm{6}}} \mid+{C} \\ $$$$ \\ $$$$\therefore\because\:{K}=\mathrm{2}\sqrt{{x}}+\mathrm{3}\sqrt[{\mathrm{3}}]{{x}}+\mathrm{6}\sqrt[{\mathrm{6}}]{{x}}+{ln}\mid{x}\mid+{C}.. \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:..........\mathscr{L}{e}\:{puissant}........... \\ $$

Commented by SANOGO last updated on 11/Oct/21

merci bien

$${merci}\:{bien} \\ $$

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