拨开云雾，直抵核心：不要被这个积分中的三个 “$\ln$” 函数迷惑了

一、题目

$$
I = \int \frac{1}{x \ln x \ln \ln x} \mathrm{~d} x = ?
$$

难度评级：

二、解析

在这道题目中，我们首先要根据对数函数“只对右侧变量生效”的定理，对原式加括号，以明确每个 “$\ln$” 的作用范围：

$$
\begin{aligned}
I = & \ \int \frac{1}{x \ln x \ln \ln x} \mathrm{~d} x \\ \\
= & \ \int \frac{1}{(x) \ln x \ln \ln x} \mathrm{~d} x \\ \\
= & \ \int \frac{1}{(x) (\ln x) \ln \ln x} \mathrm{~d} x \\ \\
= & \ \textcolor{yellow}{ \int \frac{1}{(x) (\ln x) (\ln \ln x)} \mathrm{~d} x } \\ \\
\end{aligned}
$$

于是：

$$
\begin{aligned}
I \\ \\
= & \ \int \frac{1}{(\textcolor{orangered}{x}) (\ln x) (\ln \ln x)} \mathrm{~d} x \\ \\
= & \ \int \frac{1}{(\ln x) (\ln \ln x)} \mathrm{~d} (\textcolor{orangered}{ \ln x }) \\ \\
\Leftrightarrow & \ \textcolor{gray}{ k_{1} = \ln x } \\ \\
= & \ \int \frac{1}{(\textcolor{pink}{ k_{1} }) (\ln k_{1})} \mathrm{~d} (k_{1}) \\ \\
= & \ \int \frac{1}{\ln k_{1}} \mathrm{~d} (\textcolor{pink}{\ln k_{1}}) \\ \\
\Leftrightarrow & \ \textcolor{gray}{k_{2} = \ln k_{1}} \\ \\
= & \ \int \frac{1}{k_{2}} \mathrm{~d} (k_{2}) \\ \\
= & \ \textcolor{yellow}{\ln k_{2} + C} \\ \\
= & \ \ln (\ln k_{1}) + C \\ \\
= & \ \textcolor{yellow}{ \ln [\ln (\ln x)] + C } \\ \\
= & \ \textcolor{springgreen}{\boldsymbol{ \ln \ln \ln x + C }}
\end{aligned}
$$

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