Question

Solve:

$\int x\sqrt{x+2}dx$

Answer 1

$I=\int x\sqrt{x+2}dx$

Let x + 2 = t

dx = dt

$I=\int (t-2)\sqrt{t}dt$

$=\int t\sqrt{t}dt-\int 2\sqrt{t}dt$

$=\int t^{3/2}dt-2\int t^{1/2}dt$

$=\frac{t^{3/2+1}}{3/2+1}-2\frac{t^{1/2+1}}{1/2+1}+c$

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

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

$\frac{2}{5}(x+2{)}^{5/2}-\frac{4}{3}(x+2)\sqrt{x+2}+c$

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

$=\frac{2}{1}(x+2)\sqrt{x+2}(\frac{(x+2)}{5}-\frac{2}{3})+c$

$=2(x+2)\sqrt{x+2}\left(\frac{3x+6-10}{15}\right)+c$

$=2(x+2)\sqrt{x+2}\left(\frac{3x-4}{15}\right)+c$

$=\frac{2}{15}(x+2)(3x-4)\sqrt{x+2}+c$