dx-x-3-4-x-3-1-3-

Question Number 84170 by jagoll last updated on 10/Mar/20

\int \frac{dx}{x^{3} \sqrt[3]{4 - x^{3}}} ?

Answered by john santu last updated on 10/Mar/20

\int \frac{dx}{x^{4} \sqrt[3]{\frac{4}{x^{3}} - 1}} =

let u = \frac{4}{x^{3}} - 1 \Rightarrow du = - \frac{12}{x^{4}} dx

\frac{dx}{x^{4}} = - \frac{1}{12} du .

\Rightarrow \int - \frac{d u}{12 u^{\frac{1}{3}}} = - \frac{1}{12} \times \frac{3}{2} \sqrt[3]{u^{2}} + c

= - \frac{1}{8} \sqrt[3]{{(\frac{4 - x^{3}}{x^{3}})}^{2}} + c

- \frac{1}{8 x^{2}} \sqrt[3]{{(4 - x^{3})}^{2}} + c

Answered by MJS last updated on 10/Mar/20

\int \frac{d x}{x^{3} \sqrt[3]{4 - x^{3}}} = - \int \frac{d x}{x^{3} \sqrt[3]{x^{3} - 4}} =

[t = \frac{\sqrt[3]{x^{3} - 4}}{x} \to d x = \frac{x^{2} {(x^{3} - 4)}^{2 / 3}}{4}]

= - \frac{1}{4} \int t d t = - \frac{1}{8} t^{2} = - \frac{{(x^{3} - 4)}^{2 / 3}}{8 x^{2}} + C

Commented by john santu last updated on 10/Mar/20

oo thank you sir

Commented by john santu last updated on 10/Mar/20

how to get t = \frac{\sqrt[3]{x^{3} - 4}}{x} sir ?

by feeling ?

Commented by MJS last updated on 10/Mar/20

I tried t = \sqrt[3]{x^{3} - 4} \to d x = \frac{{(x^{3} - 4)}^{2 / 3}}{3 x^{2}} first and

because of {(\frac{u}{v})}^{^{'}} = \frac{u^{'} v - {u v}^{'}}{v^{2}} with u = \sqrt[3]{w} is

\frac{\frac{1}{3} w^{- 2 / 3} v - w^{1 / 3} v^{'}}{v^{2}} = \frac{v - {w v}^{'}}{3 v^{2} w^{2 / 3}} I tried v = x

maybe {it}^{'} s just experience mixed with good luck

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