Let 0.

Question

Let $\alpha>0$. If $\int\limits_0^\alpha \frac{x}{\sqrt{x+\alpha}-\sqrt{x}} \mathrm{~d} x=\frac{16+20 \sqrt{2}}{15}$, then $\alpha$ is equal to :

Accepted Answer

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

$$
\begin{aligned}
& =\frac{1}{\alpha} \int\limits_{0}^{\alpha} x(\sqrt{x+\alpha}+\sqrt{x}) d x
\end{aligned}
$$

$$\eqalign{
  \&  = {1 \over \alpha }\int\limits_0^\alpha  {\left( {x\sqrt {x + \alpha }  + x\sqrt x } \right)} dx  \cr
  \&  = {1 \over \alpha }\int\limits_0^\alpha  {\left[ {\left( {x + \alpha  - \alpha } \right)\sqrt {x + \alpha }  + {x^{{3 \over 2}}}} \right]} dx \cr} $$

$$
\begin{aligned}
& =\frac{1}{\alpha}\int\limits_0^\alpha \left[(x+\alpha)^{3 / 2}-\alpha(x+\alpha)^{1 / 2}+x^{3 / 2}\right] d x \\
& =\frac{1}{\alpha}\left[\frac{2}{5}(x+\alpha)^{5 / 2}-\alpha \frac{2}{3}(x+\alpha)^{3 / 2}+\frac{2}{5} x^{5 / 2}\right]_0^\alpha \\
& =\frac{1}{\alpha}\left(\frac{2}{5}(2 \alpha)^{5 / 2}-\frac{2 \alpha}{3}(2 \alpha)^{3 / 2}+\frac{2}{5} \alpha^{5 / 2}-\frac{2}{5} \alpha^{5 / 2}+\frac{2}{3} \alpha^{5 / 2}\right) \\
& =\frac{1}{\alpha}\left(\frac{2^{7 / 2} \alpha^{5 / 2}}{5}-\frac{2^{5 / 2} \alpha^{5 / 2}}{3}+\frac{2}{3} \alpha^{5 / 2}\right)=\alpha^{3 / 2}\left(\frac{2^{7 / 2}}{5}-\frac{2^{5 / 2}}{3}+\frac{2}{3}\right) \\
& =\frac{\alpha^{3 / 2}}{15}(24 \sqrt{2}-20 \sqrt{2}+10)=\frac{\alpha^{3 / 2}}{15}(4 \sqrt{2}+10)
\end{aligned}
$$

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