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

Toolbox:
• Method of substitution :
• Given f(x)dx can be transformed into another form by changing independent variable x to t by substituting x=g(t).
• Consider $I=\int f(x)dx.$
• Put x=g(t) so that $\frac{dx}{dt}=g'(t).$
• dx=g'(t)dt.
• Thus $I=\int f(g(t).g'(t))dt.$
Given $I=\int\frac{x}{\sqrt{x+4}}dx$.

Let x+4=t [dx=dt]

$\Rightarrow x=t-4$.

Now substituting for x and dx we get,

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

$\;\;\;\qquad\quad\;=\int \sqrt tdt-\int\frac{4}{\sqrt t}dt.$

On integrating we get,

$\;\;\;\qquad\quad\;=\begin{bmatrix}\frac{t^\frac{3}{2}}{\frac{3}{2}}\end{bmatrix}-4\begin{bmatrix}\frac{t^\frac{1}{2}}{\frac{1}{2}}\end{bmatrix}+c.$

$\;\;\;\qquad\quad\;=\frac{2}{3}[t^\frac{3}{2}]-8(t^\frac{1}{2})+c.$

Substituting back for t we get,
$\;\;\;\qquad\quad\;=\frac{2}{3}\begin{bmatrix}(x+4)^\frac{3}{2}\end{bmatrix}-4\begin{bmatrix}(x+4)^\frac{1}{2}\end{bmatrix}+c.$

On further simplifying,
$\;\;\;\qquad\quad\;=\frac{2}{3}[(x+4)^\frac{1}{2}](x+4-12)+c.$

$\;\;\;\qquad\quad\;=\frac{2}{3}\sqrt{(x+4)}(x-8)+c.$

Hence $\int\frac{x}{\sqrt{x+4}}dx=\frac{2}{3}\sqrt{(x+4)}(x-8)+c.$