Exploit the Symmetry

$\begin{aligned} I & = \int_0^2 \left(\sqrt{\frac {4-x}x} - \sqrt{\frac x{4-x}}\right) dx & \small \color{#3D99F6} \text{Using identity: }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx \\ & = \int_0^2 \left(\sqrt{\frac {2+x}{2-x}} - \sqrt{\frac {2-x}{2+x}}\right) dx \\ & = \int_0^2 \frac {2x}{\sqrt{4-x^2}} dx \\ & = 2\sqrt{4-x^2} \ \bigg|_2^0 \\ & = \boxed{4} \end{aligned}$

Evaluated with the help of inverse functions

• Relevant Wiki: Integral of Inverse Functions

The inverse functions of $\displaystyle \sqrt{\frac{4-x}{x}}$ and $\displaystyle\sqrt{\frac{x}{4-x}}$ are $\displaystyle\frac{4}{x^2+1}$ and $\displaystyle\frac{4x^2}{x^2+1}$ ,respectively.

Using Integral of Inverse functions formula: $\int_{a}^{b}f(x)\text{d}x+\int_{f(a)}^{f(b)}f^{-1}(x)\text{d}x=bf(b)-af(a)$ $\int_{a}^{b}f(x)\text{d}x=bf(b)-af(a)-\int_{f(a)}^{f(b)}f^{-1}(x)\text{d}x ~~~ \cdot\cdot\cdot\cdot\cdot\cdot\cdot~(1)$

Returning into our problem: $\large\ \int _{ 0 }^{ 2 }{ \left( \sqrt { \frac { 4 - x }{ x } } - \sqrt { \frac { x }{ 4 - x } } \right) \text{d}x }$ Split them up into two integrals: $\large\ \int _{ 0 }^{ 2 }{ \sqrt { \frac { 4 - x }{ x } }\text{d}x - \int_{0}^{2}\sqrt { \frac { x }{ 4 - x } } \text{d}x }$

Apply the inverse functions formula (1) into the equation (Notice there's a $0\cdot\infty$ situation so a limit must be considered.) : $\left(\ 2(1) - \lim_{x\rightarrow 0}x\cdot\sqrt{\frac{4-x}{x}} - \int_{\infty}^{1}\frac{4}{x^2+1}\text{d}x \right) - \left( 2(1)-0(0) - \int_{0}^{1}\frac{4x^2}{x^2+1} \right)$ $2 - (-\pi) - ( 2 - [4 - \pi] )$ $\boxed{4}$

First of all, we note the many symmetries of the given expression. Specifically, we have $\sqrt { \frac { 4 - x }{ x } }$ and we subtract its reciprocal. We also recall that square roots, when we take their derivative, give us their reciprocal. This inspires the guess that the function $f\left( x \right) = \sqrt { x } \sqrt { 4 - x }$ . Indeed, we find that $f^{ ' }\left( x \right) = \left( \sqrt { \frac { 4 - x }{ x } } - \sqrt { \frac { x }{ 4 - x } } \right)$ so that $\int _{ 0 }^{ 2 }{ \left( \sqrt { \frac { 4 - x }{ x } } - \sqrt { \frac { x }{ 4 - x } } \right) } = 4 dx$ .