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

$$\int_0^{\alpha} \frac{x}{\sqrt{x+\alpha}-\sqrt{x}} dx = \frac{16+20\sqrt{2}}{15}$$

$$\frac{1}{\alpha} \int_0^{\alpha} x(\sqrt{x+\alpha}+\sqrt{x}) dx = \frac{16+20\sqrt{2}}{15}$$

$$\frac{1}{\alpha} \left[ \int_0^{\alpha} x\sqrt{x+\alpha} \, dx + \int_0^{\alpha} x^{3/2} \, dx \right] = \frac{16+20\sqrt{2}}{15}$$

Using substitution $$x+\alpha = t$$ for the first integral:

$$\int_0^{\alpha} x\sqrt{x+\alpha} \, dx = \int_{\alpha}^{2\alpha} (t-\alpha)\sqrt{t} \, dt = \left[ \frac{2}{5}t^{5/2} - \frac{2}{3}\alpha t^{3/2} \right]_{\alpha}^{2\alpha} = \alpha^{5/2} \left( \frac{4\sqrt{2}}{15} + \frac{4}{15} \right)$$

Evaluating the second integral: $$\int_0^{\alpha} x^{3/2} \, dx = \left[ \frac{2}{5}x^{5/2} \right]_0^{\alpha} = \frac{6}{15}\alpha^{5/2}$$

$$\frac{1}{\alpha} \cdot \alpha^{5/2} \left( \frac{10 + 4\sqrt{2}}{15} \right) = \frac{16+20\sqrt{2}}{15}$$

$$\alpha^{3/2}(10 + 4\sqrt{2}) = 16 + 20\sqrt{2}$$

$$\alpha^{3/2}(10 + 4\sqrt{2}) = 2\sqrt{2}(10 + 4\sqrt{2}) \implies \alpha^{3/2} = 2^{3/2} \implies \alpha = 2$$