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Question:

$\int_{0}^{2} x \sqrt{x+2}\left(\right.$ Put $\left.x+2=t^{2}\right)$

Solution:

$\int_{0}^{2} x \sqrt{x+2} d x$

Let $x+2=t^{2} \Rightarrow d x=2 t d t$

When $x=0, t=\sqrt{2}$ and when $x=2, t=2$

$\therefore \int_{0}^{2} x \sqrt{x+2} d x=\int_{\sqrt{2}}^{2}\left(t^{2}-2\right) \sqrt{t^{2}} 2 t d t$

$=2 \int_{\sqrt{2}}^{2}\left(t^{2}-2\right) t^{2} d t$

$=2 \int_{\sqrt{2}}^{2}\left(t^{4}-2 t^{2}\right) d t$

$=2\left[\frac{t^{5}}{5}-\frac{2 t^{3}}{3}\right]_{\sqrt{2}}^{2}$

$=2\left[\frac{32}{5}-\frac{16}{3}-\frac{4 \sqrt{2}}{5}+\frac{4 \sqrt{2}}{3}\right]$

$=2\left[\frac{96-80-12 \sqrt{2}+20 \sqrt{2}}{15}\right]$

$=2\left[\frac{16+8 \sqrt{2}}{15}\right]$

$=\frac{16(2+\sqrt{2})}{15}$

$=\frac{16 \sqrt{2}(\sqrt{2}+1)}{15}$

 

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