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Question Number 101808 by Ar Brandon last updated on 04/Jul/20

$${\int }_{\frac{1}{3}}^1\frac{{(\mathrm{x}-{\mathrm{x}}^3)}^{1/3}}{{\mathrm{x}}^4}\mathrm{dx}$$

Answered by Dwaipayan Shikari last updated on 04/Jul/20

$${\int }_{\frac{1}{3}}^1\frac{x{(\frac{1}{x^2}-1)}^{\frac{1}{3}}dx}{x^4}=-\frac{1}{2}{\int }_{\frac{1}{3}}^1\frac{-2}{x^3}{(\frac{1}{x^2}-1)}^{\frac{1}{3}}dx=\frac{1}{2}{\int }_0^8{t}^{\frac{1}{3}}dt=\frac{3}{8}{\left[t^{\frac{4}{3}}\right]}_0^8=6$$$$\{Assuming(\frac{1}{x^2}-1)=t$$

Commented by Ar Brandon last updated on 04/Jul/20

Cool ��

Commented by Dwaipayan Shikari last updated on 04/Jul/20

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Answered by bobhans last updated on 05/Jul/20

$$\int \sqrt[3]{\frac{x-x^3}{x^{12}}}dx=\int \sqrt[3]{\frac{1}{x^{11}}-\frac{1}{x^9}}dx$$$$=\int \frac{1}{x^3}\sqrt[3]{\frac{1}{x^2}-1}dx=-\frac{1}{3}\int \sqrt[3]{\frac{1}{x^2}-1}d(\frac{1}{x^2}-1)$$$$=-\frac{1}{3}\times \frac{3}{4}\sqrt[3]{{(\frac{1}{x^2}-1)}^4}+C\left(Bob\right)$$

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