Let alpha 0. If int limits _0^alpha frac{ x | vedclass

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(A) Rationalizing the denominator of the integrand:$\frac{x}{\sqrt{x+\alpha}-\sqrt{x}} = \frac{x(\sqrt{x+\alpha}+\sqrt{x})}{(x+\alpha)-x} = \frac{x(\s

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$2$

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(A) Rationalizing the denominator of the integrand:$\frac{x}{\sqrt{x+\alpha}-\sqrt{x}} = \frac{x(\sqrt{x+\alpha}+\sqrt{x})}{(x+\alpha)-x} = \frac{x(\sqrt{x+\alpha}+\sqrt{x})}{\alpha} = \frac{1}{\alpha}(x(x+\alpha)^{1/2} + x^{3/2})$We can rewrite $x(x+\alpha)^{1/2}$ as $((x+\alpha)-\alpha)(x+\alpha)^{1/2} = (x+\alpha)^{3/2} - \alpha(x+\alpha)^{1/2}$.So,the integral becomes:$\frac{1}{\alpha} \int_0^{\alpha} ((x+\alpha)^{3/2} - \alpha(x+\alpha)^{1/2} + x^{3/2}) dx$Integrating term by term:$= \frac{1}{\alpha} \left[ \frac{2}{5}(x+\alpha)^{5/2} - \alpha \cdot \frac{2}{3}(x+\alpha)^{3/2} + \frac{2}{5}x^{5/2} \right]_0^{\alpha}$$= \frac{1}{\alpha} \left( \left( \frac{2}{5}(2\alpha)^{5/2} - \frac{2\alpha}{3}(2\alpha)^{3/2} + \frac{2}{5}\alpha^{5/2} \right) - \left( \frac{2}{5}\alpha^{5/2} - \frac{2\alpha}{3}\alpha^{3/2} + 0 \right) \right)$$= \frac{1}{\alpha} \left( \frac{2}{5} \cdot 4\sqrt{2} \alpha^{5/2} - \frac{2}{3} \cdot 2\sqrt{2} \alpha^{5/2} + \frac{2}{5}\alpha^{5/2} - \frac{2}{5}\alpha^{5/2} + \frac{2}{3}\alpha^{5/2} \right)$$= \alpha^{3/2} \left( \frac{8\sqrt{2}}{5} - \frac{4\sqrt{2}}{3} + \frac{2}{3} \right) = \alpha^{3/2} \left( \frac{24\sqrt{2} - 20\sqrt{2} + 10}{15} \right) = \alpha^{3/2} \left( \frac{4\sqrt{2} + 10}{15} \right)$Given $\alpha^{3/2} \left( \frac{10 + 4\sqrt{2}}{15} \right) = \frac{16 + 20\sqrt{2}}{15}$.Comparing terms,if $\alpha = 2$,then $\alpha^{3/2} = 2\sqrt{2}$.$2\sqrt{2} \cdot \frac{10 + 4\sqrt{2}}{15} = \frac{20\sqrt{2} + 8(2)}{15} = \frac{20\sqrt{2} + 16}{15}$.Thus,$\alpha = 2$.

Let $\alpha > 0$. If $\int \limits _0^\alpha \frac{ x }{\sqrt{ x +\alpha}-\sqrt{ x }} dx =\frac{16+20 \sqrt{2}}{15}$,then $\alpha$ is equal to :

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