Question

Integrate the function $\frac{x}{\sqrt{x+4}},x>0$

Answer 1

Let $x+4=t\Rightarrow dx=dt$
$\therefore \int \frac{x}{\sqrt{x+4}}dx=\int \frac{(t-4)}{\sqrt{t}}dt$
$=\int (\sqrt{t}-\frac{4}{\sqrt{t}})dt$
$=\frac{t^{\frac{3}{2}}}{\frac{3}{2}}-4\left(\frac{t^{\frac{1}{2}}}{\frac{1}{2}}\right)+C$
$=\frac{2}{3}(t{)}^{\frac{3}{2}}-8(t{)}^{\frac{1}{2}}+C$
$=\frac{2}{3}t\cdot t^{\frac{1}{2}}-8t^{\frac{1}{2}}+C$
$=\frac{2}{3}{t}^{\frac{1}{2}}(t-12)+C$
$=\frac{2}{3}(x+4{)}^{\frac{1}{2}}(x+4-12)+C$
$=\frac{2}{3}\sqrt{x+4}(x-8)+C$