$\cfrac{x}{{\sqrt {x + 4} }},x > 0$

Let $I = \int {\cfrac{x}{{\sqrt {x + 4} }}dx}$

Put $x + 4 = t$ $\Rightarrow$ $dx = dt$
$\therefore$ $I = \int {\cfrac{x}{{\sqrt {x + 4} }}} dx = \int {\cfrac{{t - 4}}{{\sqrt t }}} dt$

$= \int {\left( {{t^{1/2}} - 4{t^{ - 1/2}}} \right)dt} = \cfrac{2}{3}{t^{3/2}} - 4 \times 2{t^{1/2}} + C$
$\Rightarrow$

$I = \cfrac{2}{3}{\left( {x + 4} \right)^{3/2}} - 8{\left( {x + 4} \right)^{1/2}} + C = \cfrac{2}{3}{\left( {x + 4} \right)^{1/2}}\left[ {x + 4 - 12} \right] + C$

$= \cfrac{2}{3}{\left( {x + 4} \right)^{1/2}}\left( {x - 8} \right) + C$