How can I evaluate this given improper integral? How can I evaluate this integral:
$$\int _{ 0 }^{3}{ \frac { x }{ (3-x)^{\frac{1}{3}}} dx} \ ?$$
 A: Hint: Try the substitution $u=3-x$.
A: 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?
A: 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}
$.
A: 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
$$ \mathrm{\beta}(x,y) = \int_0^1t^{x-1}(1-t)^{y-1}\,dt=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}\,, \quad  \!\textrm{Re}(x), \textrm{Re}(y) > 0.\, $$
