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# Integrate the function $x\;\sqrt{x+2}$

Toolbox:
• Method of substitution:
• Given $\int f(x)dx$ 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 $\frac{dx}{dx}=g'(t).$
• dx=g'(t)dt.
• Thus $I=\int f(g(t).g'(t))dt.$
Given $I=x\sqrt {x+2}.$

Let t=x+2 $\Rightarrow x=t-2.$

Hence dt=dx.

Substituting for t and dt we get,

$I=\int (t-2)\sqrt tdt.$

$\;\;\;=\int(t\sqrt t-2\sqrt t)dt.$

$\;\;\;=\int t^\frac{3}{2}dt-2\int t^\frac{1}{2}dt.$

On integrating we get,
$\;\;\;=\frac{t^\frac{5}{2}}{\frac{5}{2}}-2\frac{t^\frac{3}{2}}{\frac{3}{2}}+c.$

$\;\;\;=\frac{2}{5}t^\frac{5}{2}-\frac{4}{3}t^\frac{3}{2}+c.$

Substituting back for t we get,

$\;\;\;=\frac{2}{5}(x+2)^\frac{5}{2}-\frac{4}{3}(x+2)^\frac{3}{2}+c.$

edited Jan 28, 2013