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Question

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 :

Accepted Answer

$I=\int _{0}^{\alpha }\frac{x}{\sqrt{x+\alpha }-\sqrt{x}}dx,\alpha >0$

$\begin{gathered} =\frac{1}{\alpha }\int _{0}^{\alpha }x(\sqrt{x+\alpha }+\sqrt{x})dx \end{gathered}$

$\begin{gathered} =\frac{1}{\alpha }\int _{0}^{\alpha }(x\sqrt{x+\alpha }+x\sqrt{x})dx \ =\frac{1}{\alpha }\int _{0}^{\alpha }[(x+\alpha -\alpha )\sqrt{x+\alpha }+{x}^{\frac{3}{2}}]dx \end{gathered}$

$\begin{gathered} =\frac{1}{\alpha }\int _{0}^{\alpha }[(x+\alpha {)}^{3/2}-\alpha (x+\alpha {)}^{1/2}+{x}^{3/2}]dx \ =\frac{1}{\alpha }{[\frac{2}{5}(x+\alpha {)}^{5/2}-\alpha \frac{2}{3}(x+\alpha {)}^{3/2}+\frac{2}{5}{x}^{5/2}]}_{\alpha }^{0} \ =\frac{1}{\alpha }(\frac{2}{5}(2\alpha {)}^{5/2}-\frac{2\alpha }{3}(2\alpha {)}^{3/2}+\frac{2}{5}{\alpha }^{5/2}-\frac{2}{5}{\alpha }^{5/2}+\frac{2}{3}{\alpha }^{5/2}) \ =\frac{1}{\alpha }(\frac{{2}^{7/2}{\alpha }^{5/2}}{5}-\frac{{2}^{5/2}{\alpha }^{5/2}}{3}+\frac{2}{3}{\alpha }^{5/2})={\alpha }^{3/2}(\frac{{2}^{7/2}}{5}-\frac{{2}^{5/2}}{3}+\frac{2}{3}) \ =\frac{{\alpha }^{3/2}}{15}(24\sqrt{2}-20\sqrt{2}+10)=\frac{{\alpha }^{3/2}}{15}(4\sqrt{2}+10) \end{gathered}$

Now, $\frac{{\alpha }^{3/2}}{15}(4\sqrt{2}+10)=\frac{16+20\sqrt{2}}{15}=\frac{2\sqrt{2}(4\sqrt{2}+10)}{15}$

$\Rightarrow {\alpha }^{3/2}=2\sqrt{2}={2}^{3/2}$

$\Rightarrow \alpha =2$

Solutions · 1