$\int_9^4 \frac{1}{1+\sqrt{x}} \mathrm{~d} x=$
【答案】 $2 \ln \frac{4}{3}-2$

【解析】
【参考解析】令 $u=\sqrt{x}$ ,则 $x=u^2(u > 0), \mathrm{d} x=2 u \mathrm{~d} u$ , 当 $x=9$ 时, $u=3$ ;当 $x=4$ 时, $u=2$ ,于是
$$
\begin{aligned}
&\int_9^4 \frac{1}{1+\sqrt{x}} \mathrm{~d} x=\int_3^2 \frac{1}{1+u} \cdot 2 u \mathrm{~d} u=2 \int_3^2 \frac{u}{1+u} \mathrm{~d} u \\
&=2 \int_3^2\left(1-\frac{1}{1+u}\right) \mathrm{d} u=2\left[1 \cdot(2-3)-\left.\ln (1+u)\right|_3 ^2\right] \\
&=-2-2 \ln \frac{3}{4}=2 \ln \frac{4}{3}-2
\end{aligned}
$$
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