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Question

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


Solution

$$ I=\int x\sqrt{x+2}dx$$
Let x + 2 = t
dx = dt
$$ I=\int (t-2)\sqrt{t}dt$$
$$=\int t\sqrt{t}dt-\int 2\sqrt{t}dt$$
$$=\int t^{3/2}dt-2\int t^{1/2}dt$$
$$=\frac{t^{3/2+1}}{3/2+1}-2\frac{t^{1/2+1}}{1/2+1}+c$$
$$=\frac{t^{5/2}}{5/2}-2.\frac{t^{3/2}}{3/2}+c$$
$$=\frac{2}{5}t^{5/2}-\frac{4}{3}t^{3/2}+c$$
$$\frac{2}{5}(x+2)^{5/2}-\frac{4}{3}(x+2)\sqrt{x+2}+c$$
= $$\frac{2}{5} (x+2)^{2}\sqrt{(x+2)}-\frac{4}{3}(x+2)\sqrt{x+2}+c$$
$$=\frac{2}{1}(x+2)\sqrt{x+2}(\frac{(x+2)}{5}-\frac{2}{3})+c$$
$$=2(x+2)\sqrt{x+2}(\frac{3x+6-10}{15})+c$$
$$=2(x+2)\sqrt{x+2}(\frac{3x-4}{15})+c$$
$$=\frac{2}{15}(x+2)(3x-4)\sqrt{x+2}+c$$

1170892_1296348_ans_1bb7ee012e1341dd939b78bf6da903a0.JPG

Mathematics

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