Question-146131

Question Number 146131 by mathlove last updated on 11/Jul/21

Commented by iloveisrael last updated on 11/Jul/21

$$\mathrm{great}\:\mathrm{ustad} \\ $$

Answered by EDWIN88 last updated on 11/Jul/21

Solve ∫ (dx/( (x)^(1/3) . (((1−x)^8 ))^(1/3) )) . Solution : G= ∫ (dx/( (x)^(1/3) ((x^8 (x^(−1) −1)^8 ))^(1/3) )) G = ∫ (dx/(x^(1/3) .x^2 ((x^2 (x^(−1) −1)^8 ))^(1/3) )) G = ∫ (dx/(x.x^2 (((x^(−1) −1)^8 ))^(1/3) )) let u=x^(−1) −1 ⇒(dx/x^2 ) =−du then ⇒x =(1/(u+1)) G=∫ ((−du)/(((1/(u+1)))(u^8 )^(1/3) )) =−∫ ((u+1)/u^(8/3) ) du G =−∫u^(−5/3) du−∫u^(−8/3) du G =−(−(3/2))u^(−2/3) −(−(3/5))u^(−5/3) +c G =(3/(2 (u^2 )^(1/3) )) + (3/(5 (u^5 )^(1/3) )) + c G = (3/(2 (((x^(−1) −1)^2 ))^(1/3) )) +(3/(5 (((x^(−1) −1)^5 ))^(1/3) )) +c G = (3/(2 (((((1−x)/x))^2 ))^(1/3) )) +(3/(5 (((((1−x)/x))^5 ))^(1/3) )) + c G =(3/2)((x^2 /((1−x)^2 )))^(1/(3 )) +(3/5)((x^5 /((1−x)^5 )))^(1/(3 )) + c ∗Edwin−88∗

$$\:{Solve}\:\int\:\frac{{dx}}{\:\sqrt[{\mathrm{3}}]{{x}}.\:\sqrt[{\mathrm{3}}]{\left(\mathrm{1}−{x}\right)^{\mathrm{8}} }}\:. \\ $$$${Solution}\:: \\ $$$${G}=\:\int\:\frac{{dx}}{\:\sqrt[{\mathrm{3}}]{{x}}\:\sqrt[{\mathrm{3}}]{{x}^{\mathrm{8}} \left({x}^{−\mathrm{1}} −\mathrm{1}\right)^{\mathrm{8}} }}\: \\ $$$${G}\:=\:\int\:\frac{{dx}}{{x}^{\mathrm{1}/\mathrm{3}} .{x}^{\mathrm{2}} \:\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} \left({x}^{−\mathrm{1}} −\mathrm{1}\right)^{\mathrm{8}} }} \\ $$$${G}\:=\:\int\:\frac{{dx}}{{x}.{x}^{\mathrm{2}} \:\sqrt[{\mathrm{3}}]{\left({x}^{−\mathrm{1}} −\mathrm{1}\right)^{\mathrm{8}} }}\: \\ $$$$\:{let}\:{u}={x}^{−\mathrm{1}} −\mathrm{1}\:\Rightarrow\frac{{dx}}{{x}^{\mathrm{2}} }\:=−{du}\: \\ $$$${then}\:\:\Rightarrow{x}\:=\frac{\mathrm{1}}{{u}+\mathrm{1}}\: \\ $$$${G}=\int\:\frac{−{du}}{\left(\frac{\mathrm{1}}{{u}+\mathrm{1}}\right)\sqrt[{\mathrm{3}}]{{u}^{\mathrm{8}} }}\:=−\int\:\frac{{u}+\mathrm{1}}{{u}^{\mathrm{8}/\mathrm{3}} }\:{du}\: \\ $$$${G}\:=−\int{u}^{−\mathrm{5}/\mathrm{3}} {du}−\int{u}^{−\mathrm{8}/\mathrm{3}} \:{du}\: \\ $$$${G}\:=−\left(−\frac{\mathrm{3}}{\mathrm{2}}\right){u}^{−\mathrm{2}/\mathrm{3}} −\left(−\frac{\mathrm{3}}{\mathrm{5}}\right){u}^{−\mathrm{5}/\mathrm{3}} +{c} \\ $$$${G}\:=\frac{\mathrm{3}}{\mathrm{2}\:\sqrt[{\mathrm{3}}]{{u}^{\mathrm{2}} }}\:+\:\frac{\mathrm{3}}{\mathrm{5}\:\sqrt[{\mathrm{3}}]{{u}^{\mathrm{5}} }}\:+\:{c}\: \\ $$$${G}\:=\:\frac{\mathrm{3}}{\mathrm{2}\:\sqrt[{\mathrm{3}}]{\left({x}^{−\mathrm{1}} −\mathrm{1}\right)^{\mathrm{2}} }}\:+\frac{\mathrm{3}}{\mathrm{5}\:\sqrt[{\mathrm{3}}]{\left({x}^{−\mathrm{1}} −\mathrm{1}\right)^{\mathrm{5}} }}\:+{c}\: \\ $$$${G}\:=\:\frac{\mathrm{3}}{\mathrm{2}\:\sqrt[{\mathrm{3}}]{\left(\frac{\mathrm{1}−{x}}{{x}}\right)^{\mathrm{2}} }}\:+\frac{\mathrm{3}}{\mathrm{5}\:\sqrt[{\mathrm{3}}]{\left(\frac{\mathrm{1}−{x}}{{x}}\right)^{\mathrm{5}} }}\:+\:{c} \\ $$$${G}\:=\frac{\mathrm{3}}{\mathrm{2}}\sqrt[{\mathrm{3}\:}]{\frac{{x}^{\mathrm{2}} }{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }}\:+\frac{\mathrm{3}}{\mathrm{5}}\sqrt[{\mathrm{3}\:}]{\frac{{x}^{\mathrm{5}} }{\left(\mathrm{1}−{x}\right)^{\mathrm{5}} }}\:+\:{c}\: \\ $$$$\:\ast{Edwin}−\mathrm{88}\ast\: \\ $$

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