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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

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

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