Question

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

Answer 1

Let x+4 = t

∴ dx = dt

$\int \frac{x}{\sqrt x+4} dx =\int \frac{(t-4)}{\sqrt t}dt$

= $\int \bigg(\sqrt t-\frac{4}{\sqrt t}\bigg)dt$

= $\frac{t^{\frac{3}{2}}}{\frac{3}{2}}-4\bigg(\frac{t^{\frac{1}{2}}}{\frac{1}{2}}\bigg)+C$

=$\frac{2}{3}(t)^{\frac{3}{2}}-8(t)^{\frac{1}{2}}+C$

=$\frac{2}{3}t.t^{\frac{1}{2}}-8t^{\frac{1}{2}}+C$

=$\frac{2}{3}t^{\frac{1}{2}}(t-12)+C$

= $\frac{2}{3}(x+4)^{\frac{1}{2}}(x+4-12)+C$

= $\frac{2}{3}\sqrt {x+4}(x-8)+C$