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Question Number 121962 by bemath last updated on 13/Nov/20

$$\underset{0}{\stackrel{1}{\int }}\frac{dx}{\sqrt[3]{x^2-x^3}}?$$

Answered by MJS_new last updated on 13/Nov/20

$$\underset{0}{\stackrel{1}{\int }}\frac{dx}{{(x^2-x^3)}^{1/3}}=\underset{0}{\stackrel{1}{\int }}\frac{dx}{x^{2/3}{(1-x)}^{1/3}}=$$$$[t=\frac{x^{1/3}}{{(1-x)}^{1/3}}\to dx=3x^{2/3}{(1-x)}^{4/3}dt]$$$$=3\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{dt}{t^3+1}=\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{dt}{t+1}-\frac{1}{2}\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{2t-1}{t^2-t+1}+\frac{3}{2}\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{dt}{t^2-t+1}=$$$$={[\ln \mid t+1\mid -\frac{1}{2}\ln (t^2-t+1)+\sqrt{3}\arctan \frac{2t-1}{\sqrt{3}}]}_0^{\mathrm{\infty }}=$$$$=\frac{2\pi \sqrt{3}}{3}$$

Commented by bemath last updated on 13/Nov/20

$$howgetupperlimit\mathrm{\infty }andlower$$$$limit0sir$$

Commented by MJS_new last updated on 13/Nov/20

$$t=\frac{x^{1/3}}{{(1-x)}^{1/3}}$$$$x=0\Rightarrow t=0$$$$x\to 1^{-}\Rightarrow t\to \mathrm{\infty }$$

Commented by MJS_new last updated on 13/Nov/20

$$\mathrm{corrected}\mathrm{a}\mathrm{typo}...$$

Commented by bemath last updated on 13/Nov/20

$$ooyesthankyousir$$

Answered by Dwaipayan Shikari last updated on 13/Nov/20

$${\int }_0^1\frac{dx}{x^{\frac{2}{3}}{(1-x)}^{\frac{1}{3}}}$$$$={\int }_0^1{x}^{-\frac{2}{3}}{(1-x)}^{-\frac{1}{3}}$$$$=\beta (\frac{1}{3},\frac{2}{3})=\frac{\mathrm{\Gamma }\left(\frac{1}{3}\right)\mathrm{\Gamma }\left(\frac{2}{3}\right)}{\mathrm{\Gamma }\left(1\right)}=\frac{\pi }{\left(sin\frac{\pi }{3}\right)}=\frac{2\pi }{\sqrt{3}}$$

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