3. [4] Let $a$ be a positive real number. Find the value of $a$ such that the definite integral
$$
\int_{a}^{a^{2}} \frac{\mathrm{d} x}{x+\sqrt{x}}
$$
achieves its smallest possible value.
Answer: $3-2 \sqrt{2}$. Let $F(a)$ denote the given definite integral. Then
$$
F^{\prime}(a)=\frac{\mathrm{d}}{\mathrm{d} a} \int_{a}^{a^{2}} \frac{\mathrm{d} x}{x+\sqrt{x}}=2 a \cdot \frac{1}{a^{2}+\sqrt{a^{2}}}-\frac{1}{a+\sqrt{a}} .
$$

Setting $F^{\prime}(a)=0$, we find that $2 a+2 \sqrt{a}=a+1$ or $(\sqrt{a}+1)^{2}=2$. We find $\sqrt{a}= \pm \sqrt{2}-1$, and because $\sqrt{a}>0, a=(\sqrt{2}-1)^{2}=3-2 \sqrt{2}$.