Integrals — Class 12 Maths Solution

.: Let$I = \int {x\sqrt {x + 2\,} d} x$

Put $x + 2 = t$ $\Rightarrow$ $dx = dt$

Also, $x = t - 2$

$\therefore$ $I = \int {\left( {t - 2} \right)\sqrt t dt} = \int {\left( {{t^{3/2}} - 2{t^{1/2}}} \right)} dt$

$= \cfrac{2}{5}{t^{5/2}} - 2 \times \cfrac{2}{3}{t^{3/2}} + C = \cfrac{2}{5}{\left( {x + 2} \right)^{5/2}} - \cfrac{4}{3}{\left( {x + 2} \right)^{3/2}} + C$