对式子的等价转换除了有先加后减，还有先开方再平方

一、题目

$$
I = \int_{2}^{+ \infty} \frac{1}{x \sqrt{x-1}} \mathrm{~d} x = ?
$$

难度评级：

二、解析

$$
\begin{aligned}
I & \\ \\
& = \ \int_{2}^{+ \infty} \frac{1}{x \sqrt{x-1}} \mathrm{~d} x \\ \\
& = \ \int_{2}^{+ \infty} \frac{1}{x} \cdot \textcolor{pink}{ \frac{1}{\sqrt{x-1}} } \mathrm{~d} x \\ \\
& \Rightarrow \ \textcolor{gray}{(\sqrt{x-1}) ^{\prime} = [(x-1) ^{\frac{1}{2}}] ^{\prime} = \frac{1}{2} \frac{1}{\sqrt{x-1}}} \\ \\
& = \ \int_{2}^{+ \infty} \frac{1}{x} \cdot \textcolor{pink}{2 \mathrm{~d} (\sqrt{x-1}) } \\ \\
& = \ \int_{2}^{+ \infty} \frac{1}{x + \textcolor{orangered}{ 1 – 1 }} \cdot 2 \mathrm{~d} (\sqrt{x-1}) \\ \\
& = \ \int_{2}^{+ \infty} \frac{1}{1 + \textcolor{springgreen}{ x – 1 }} \cdot 2 \mathrm{~d} (\sqrt{x-1}) \\ \\
& = \ \int_{2}^{+ \infty} \frac{1}{1 + \textcolor{springgreen}{ [(x – 1)^{\textcolor{yellow}{\frac{1}{2}}}]^{\textcolor{yellow}{2}} }} \cdot 2 \mathrm{~d} (\sqrt{x-1}) \\ \\
& = \ \int_{\textcolor{red}{2}}^{+ \infty} \frac{1}{1 + \textcolor{springgreen}{(\sqrt{x-1})^{2}}} \cdot \textcolor{magenta}{2} \mathrm{~d} (\sqrt{x-1}) \\ \\
& \Rightarrow \ \textcolor{gray}{\begin{cases}
t = \sqrt{x-1} \\
t \in (1, + \infty)
\end{cases}} \\ \\
& = \ \textcolor{magenta}{2} \int_{\textcolor{red}{1}}^{+ \infty} \frac{1}{1 + t^{2}} \mathrm{~d} t \\ \\
& = \ 2 \arctan t \Big|_{\textcolor{tan}{1}}^{\textcolor{orange}{+ \infty}} \\ \\
& = \ 2 \left( \textcolor{tan}{\frac{\pi}{2}} – \textcolor{orange}{\frac{\pi}{4}} \right) \\ \\
& = \ \textcolor{springgreen}{\boldsymbol{ \frac{\pi}{2} }}
\end{aligned}
$$

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