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How can I evaluate this integral:

$$\int _{ 0 }^{3}{ \frac { x }{ (3-x)^{\frac{1}{3}}} dx} \ ?$$

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Can you show us what you have tried or explain what is bothering you with attempting the problem? –  5space May 9 '13 at 4:41

4 Answers 4

Hint: Try the substitution $u=3-x$.

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Ok, I'll try, thank you. –  Lia Denisse May 9 '13 at 4:46

Letting $u = 3-x$, then we see $du = -dx$, and our integral becomes:

$$\int _{ 0 }^{3}{ \frac { x }{ (3-x)^{\frac{1}{3}}} dx} \ = \int_{0}^{3} \frac{3-u}{u^{\frac{1}{3}}}du = \int_{0}^{3} 3u^{-\frac{1}{3}} - u^{\frac{2}{3}}du$$

Can you take it from here?

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Try $y = (3-x)^{1/3}$. $y^3 = 3-x$, so $x = 3-y^3$ and $dx = -3 y^2 dy$.

Putting this in, since $y$ goes from $3^{1/3}$ to $0$ as $x$ goes from $0$ to $3$,

$\begin{align} \int_{ 0 }^{3}{ \frac { x }{ (3-x)^{\frac{1}{3}}} dx} &=\int_{3^{1/3}}^0 \frac{3-y^3}{y}(-3 y^2)dy\\ &=\int_0^{3^{1/3}} \frac{3-y^3}{y}(3 y^2)dy\\ &=3 \int_0^{3^{1/3}} y(3-y^3)dy\\ &=3 \int_0^{3^{1/3}} (3y-y^4)dy\\ &=3 (3y^2/2-y^5/5)\big|_0^{3^{1/3}}\\ &=3(3\cdot3^{2/3}/2 - 3^{5/3}/5)\\ &=3(3^{5/3}/2 - 3^{5/3}/5)\\ &=3\cdot 3^{5/3}(1/2-1/5)\\ &=3\cdot 3^{5/3}(3/10)\\ &=(9/10) 3^{5/3}\\ &=(27/10) 3^{2/3}\\ \end{align} $.

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Using the change of variables $x=3y$ casts the integral in terms of the beta function $$\int _{ 0 }^{3}{ \frac { x }{ (3-x)^{\frac{1}{3}}} dx} = 3^{5/3}\int _{ 0 }^{1}{ { y }{ (1-y)^{-\frac{1}{3}}} dy} = 3^{5/3}\beta(2,2/3)$$

$$ =3^{5/3}\frac{\Gamma(2)\Gamma(2/3)}{\Gamma(2+2/3)}=3^{2/3}\frac{27}{10}.$$

Beta function is defined by

$$ \mathrm{\beta}(x,y) = \int_0^1t^{x-1}(1-t)^{y-1}\,dt=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}\,, \quad \!\textrm{Re}(x), \textrm{Re}(y) > 0.\, $$

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