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

$\begin{array}{l}{\int }_0^{\mathrm{\alpha }} \frac{\mathrm{x}}{\sqrt{\mathrm{x}+\mathrm{\alpha }}-\sqrt{\mathrm{x}}}\mathrm{dx}=\frac{16+20\sqrt{2}}{15}\\ \frac{1}{\mathrm{\alpha }}{\int }_0^{\mathrm{\alpha }}\mathrm{x} (\sqrt{\mathrm{x}+\mathrm{\alpha }}+\sqrt{\mathrm{x}})\mathrm{dx}\\ \frac{1}{\mathrm{\alpha }}\cdot {\int }_0^{\mathrm{\alpha }} \left((\mathrm{x}+\mathrm{\alpha }-\mathrm{\alpha })\sqrt{\mathrm{x}+\mathrm{\alpha }}+{\mathrm{x}}^{\frac{3}{2}}\right)\\ \frac{1}{\mathrm{\alpha }}\cdot {\int }_0^{\mathrm{\alpha }} \left((\mathrm{x}+\mathrm{\alpha }{)}^{\frac{3}{2}}-\mathrm{\alpha }(\mathrm{x}+\mathrm{\alpha }{)}^{\frac{1}{2}}+{\mathrm{x}}^{\frac{3}{2}}\right)\mathrm{dx}\\ =\frac{1}{\mathrm{\alpha }}\cdot {\left((\mathrm{x}+\mathrm{\alpha }{)}^{\frac{5}{2}}\cdot \frac{2}{5}-\mathrm{\alpha }\cdot \frac{2}{3}(\mathrm{x}+\mathrm{\alpha }{)}^{\frac{3}{2}}+{\mathrm{x}}^{\frac{5}{2}}\cdot \frac{2}{5}\right)}_0^{\mathrm{\alpha }}\\ \frac{1}{\alpha }\left(\frac{2}{5}(2\alpha {)}^{5/2}-\frac{2\alpha }{3}(2\alpha {)}^{3/2}+\frac{2}{5}{\alpha }^{5/2}-\frac{2}{5}{\alpha }^{5/2}+\frac{2}{3}{\alpha }^{5/2}\right)\\ =\frac{1}{\alpha }\left(\frac{2^{7/2}{\alpha }^{5/2}}{5}\frac{2^{5/2}{\alpha }^{5/2}}{3}+\frac{2}{3}{\alpha }^{5/2}\right)\\ ={\alpha }^{3/2}\left(\frac{2^{7/2}}{5}-\frac{2^{5/2}}{3}+\frac{2}{3}\right)\\ ==\frac{{\alpha }^{3/2}}{15}(24\sqrt{2}-20\sqrt{2}+10)=\frac{{\alpha }^{3/2}}{15}(4\sqrt{2}+10)\\ =\mathrm{\alpha }\sqrt{\mathrm{\alpha }}\frac{(10+4\sqrt{2})}{15}=\frac{16+20\sqrt{2}}{15}\Rightarrow \mathrm{\alpha }=2\\ \mathrm{\alpha }=2\end{array}$