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 :

Given,

${\int }_0^{\alpha }\frac{x}{\sqrt{x+\alpha }-\sqrt{x}}dx=\frac{16+20\sqrt{2}}{15}....\left(1\right)$

Rationalise the denominator in L.H.S we get,

${\int }_0^{\alpha }\frac{x(\sqrt{\alpha +x}+\sqrt{x})}{\alpha }dx=\frac{1}{\alpha }\left[{\int }_0^{\alpha }x\sqrt{\alpha +x}dx+{\int }_0^{\alpha }{x}^{3/2}dx\right]$

Put $\alpha +x=t^2\Rightarrow dx=2\mathrm{tdt}$

Limit when $x=0\to t=\sqrt{\alpha }, x=\alpha \to t=\sqrt{2\alpha }$

$\frac{1}{\alpha }{\int }_{\sqrt{\alpha }}^{\sqrt{2\alpha }}\left({\mathrm{t}}^2-\alpha \right)2{\mathrm{t}}^2\mathrm{dt}+\frac{2{\alpha }^{3/2}}{5}$

$=\frac{1}{\alpha }{\left[\frac{2t^5}{5}-\frac{2\alpha t^3}{3}\right]}_{\sqrt{\alpha }}^{\sqrt{2\alpha }}+\frac{2{\alpha }^{3/2}}{5}$

$=\frac{4\alpha \sqrt{2\alpha }+10\alpha \sqrt{\alpha }}{15}$

From equation $\left(1\right)$

${\int }_0^{\alpha }\frac{x}{\sqrt{x+\alpha }-\sqrt{x}}dx=\frac{16+20\sqrt{2}}{15}$

$\Rightarrow \frac{4\alpha \sqrt{2\alpha }+10\alpha \sqrt{\alpha }}{15}=\frac{16+20\sqrt{2}}{15}$

On comparing both sides we get,

$\alpha =2$