If int_{0}^{sqrt{3}} frac{15 x^{3}}{sqrt{1+x^ | vedclass

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(A) Let $1 + x^{2} = t^{2}$. Then $2x dx = 2t dt$,so $x dx = t dt$.When $x = 0$,$t = 1$. When $x = \sqrt{3}$,$t = 2$.The integral becomes $\int_{1}^{2

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$10$

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(A) Let $1 + x^{2} = t^{2}$. Then $2x dx = 2t dt$,so $x dx = t dt$.When $x = 0$,$t = 1$. When $x = \sqrt{3}$,$t = 2$.The integral becomes $\int_{1}^{2} \frac{15(t^{2}-1) t dt}{\sqrt{t^{2} + t^{3}}} = 15 \int_{1}^{2} \frac{t(t^{2}-1)}{t \sqrt{1+t}} dt = 15 \int_{1}^{2} \frac{t^{2}-1}{\sqrt{1+t}} dt$.Let $1 + t = u^{2}$,so $t = u^{2} - 1$ and $dt = 2u du$.When $t = 1$,$u = \sqrt{2}$. When $t = 2$,$u = \sqrt{3}$.The integral becomes $15 \int_{\sqrt{2}}^{\sqrt{3}} \frac{(u^{2}-1)^{2}-1}{u} (2u du) = 30 \int_{\sqrt{2}}^{\sqrt{3}} (u^{4} - 2u^{2}) du$.Evaluating the integral: $30 \left[ \frac{u^{5}}{5} - \frac{2u^{3}}{3} \right]_{\sqrt{2}}^{\sqrt{3}} = 30 \left[ \left( \frac{9\sqrt{3}}{5} - \frac{6\sqrt{3}}{3} \right) - \left( \frac{4\sqrt{2}}{5} - \frac{4\sqrt{2}}{3} \right) \right]$.$= 30 \left[ \left( \frac{9\sqrt{3} - 10\sqrt{3}}{5} \right) - \left( \frac{12\sqrt{2} - 20\sqrt{2}}{15} \right) \right] = 30 \left[ -\frac{\sqrt{3}}{5} + \frac{8\sqrt{2}}{15} \right] = -6\sqrt{3} + 16\sqrt{2}$.Comparing with $\alpha \sqrt{2} + \beta \sqrt{3}$,we get $\alpha = 16$ and $\beta = -6$.Thus,$\alpha + \beta = 16 - 6 = 10$.

If $\int_{0}^{\sqrt{3}} \frac{15 x^{3}}{\sqrt{1+x^{2}+\sqrt{(1+x^{2})^{3}}}} dx = \alpha \sqrt{2} + \beta \sqrt{3}$,where $\alpha, \beta$ are integers,then $\alpha + \beta$ is equal to.

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