1. Simplify the integrand

We need to evaluate

$$I(\alpha)=\int_0^\alpha \frac{x}{\sqrt{x+\alpha}-\sqrt{x}}\,dx,$$

where $\alpha>0$.

Rationalize the denominator:

$$\frac{x}{\sqrt{x+\alpha}-\sqrt{x}}\cdot \frac{\sqrt{x+\alpha}+\sqrt{x}}{\sqrt{x+\alpha}+\sqrt{x}} =\frac{x(\sqrt{x+\alpha}+\sqrt{x})}{(x+\alpha)-x} =\frac{x(\sqrt{x+\alpha}+\sqrt{x})}{\alpha}.$$

So

$$I(\alpha)=\frac{1}{\alpha}\int_0^\alpha x\sqrt{x+\alpha}\,dx+\frac{1}{\alpha}\int_0^\alpha x\sqrt{x}\,dx.$$

2. Evaluate the second integral

$$\int_0^\alpha x\sqrt{x}\,dx=\int_0^\alpha x^{3/2}dx =\left[\frac{2}{5}x^{5/2}\right]_0^\alpha =\frac{2}{5}\alpha^{5/2}.$$

Thus its contribution is

$$\frac{1}{\alpha}\cdot \frac{2}{5}\alpha^{5/2}=\frac{2}{5}\alpha^{3/2}.$$

3. Evaluate the first integral

Let

$$u=x+\alpha \implies x=u-\alpha,$$

so when $x=0$, $u=\alpha$, and when $x=\alpha$, $u=2\alpha$.

Then

$$\int_0^\alpha x\sqrt{x+\alpha}\,dx =\int_\alpha^{2\alpha}(u-\alpha)u^{1/2}du =\int_\alpha^{2\alpha}\left(u^{3/2}-\alpha u^{1/2}\right)du.$$

Now,

$$\int u^{3/2}du=\frac{2}{5}u^{5/2}, \qquad \int u^{1/2}du=\frac{2}{3}u^{3/2}.$$

Hence

$$\int_0^\alpha x\sqrt{x+\alpha}\,dx =\left[\frac{2}{5}u^{5/2}-\alpha\cdot \frac{2}{3}u^{3/2}\right]_\alpha^{2\alpha}.$$

Factor out $\alpha^{5/2}$:

$$(2\alpha)^{5/2}=2^{5/2}\alpha^{5/2}=4\sqrt{2}\,\alpha^{5/2},$$ $$(2\alpha)^{3/2}=2^{3/2}\alpha^{3/2}=2\sqrt{2}\,\alpha^{3/2}.$$

So

$$\int_0^\alpha x\sqrt{x+\alpha}\,dx =\alpha^{5/2}\left[\frac{2}{5}(4\sqrt2-1)-\frac{2}{3}(2\sqrt2-1)\right].$$

Simplify the bracket:

$$\frac{8\sqrt2-2}{5}-\frac{4\sqrt2-2}{3} =\frac{3(8\sqrt2-2)-5(4\sqrt2-2)}{15} =\frac{24\sqrt2-6-20\sqrt2+10}{15} =\frac{4\sqrt2+4}{15}.$$

Thus

$$\int_0^\alpha x\sqrt{x+\alpha}\,dx =\frac{4(\sqrt2+1)}{15}\alpha^{5/2}.$$

Therefore its contribution is

$$\frac{1}{\alpha}\cdot \frac{4(\sqrt2+1)}{15}\alpha^{5/2} =\frac{4(\sqrt2+1)}{15}\alpha^{3/2}.$$

4. Combine both parts

So

$$I(\alpha)=\alpha^{3/2}\left(\frac{4(\sqrt2+1)}{15}+\frac{2}{5}\right).$$

Since

$$\frac{2}{5}=\frac{6}{15},$$

we get

$$I(\alpha)=\alpha^{3/2}\cdot \frac{4\sqrt2+4+6}{15} =\alpha^{3/2}\cdot \frac{10+4\sqrt2}{15}.$$

So

$$I(\alpha)=\frac{(10+4\sqrt2)\alpha^{3/2}}{15}.$$

Given that

$$I(\alpha)=\frac{16+20\sqrt2}{15},$$

we equate:

$$(10+4\sqrt2)\alpha^{3/2}=16+20\sqrt2.$$

Factor both sides:

$$10+4\sqrt2=2(5+2\sqrt2),$$ $$16+20\sqrt2=4(4+5\sqrt2).$$

Instead of factoring further, test the options.

5. Check options

• If $\alpha=4$, then $\alpha^{3/2}=8$:

  $$I=\frac{(10+4\sqrt2)8}{15}=\frac{80+32\sqrt2}{15}\neq \frac{16+20\sqrt2}{15}.$$

• If $\alpha=2$, then $\alpha^{3/2}=2\sqrt2$:

  $$I=\frac{(10+4\sqrt2)(2\sqrt2)}{15} =\frac{20\sqrt2+16}{15} =\frac{16+20\sqrt2}{15}.$$

  This matches.

Hence,

$$\boxed{\alpha=2}.$$

6. Comparison with stored answer

Stored correct answer is B, i.e. $2$.

Our derived answer is also B.