int_0^2 frac{x^3}{(x^2 + 1)^{3/2}} , dx = | vedclass

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(D) Let $t = x^2 + 1$,then $dt = 2x \, dx$ or $x \, dx = \frac{1}{2} \, dt$.When $x = 0$,$t = 1$. When $x = 2$,$t = 5$.The integral becomes $\int_0^2

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(D) Let $t = x^2 + 1$,then $dt = 2x \, dx$ or $x \, dx = \frac{1}{2} \, dt$.When $x = 0$,$t = 1$. When $x = 2$,$t = 5$.The integral becomes $\int_0^2 \frac{x^2 \cdot x \, dx}{(x^2 + 1)^{3/2}} = \frac{1}{2} \int_1^5 \frac{t - 1}{t^{3/2}} \, dt$.$= \frac{1}{2} \int_1^5 (t^{-1/2} - t^{-3/2}) \, dt$.$= \frac{1}{2} [2t^{1/2} - \frac{t^{-1/2}}{-1/2}]_1^5 = \frac{1}{2} [2\sqrt{t} + \frac{2}{\sqrt{t}}]_1^5$.$= [\sqrt{t} + \frac{1}{\sqrt{t}}]_1^5 = (\sqrt{5} + \frac{1}{\sqrt{5}}) - (1 + 1) = \frac{5+1}{\sqrt{5}} - 2 = \frac{6}{\sqrt{5}} - 2 = \frac{6 - 2\sqrt{5}}{\sqrt{5}}$.Since this value is not among the options,the correct choice is $D$.

$\int_0^2 \frac{x^3}{(x^2 + 1)^{3/2}} \, dx = $

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