这道题你去几次根号可以解出来？

一、题目

难度评级：

二、解析

§1. 解法一

由于：

$$
\begin{aligned}
& \int \frac{1}{\sqrt{x ^{3} (1-x)}} \mathrm{~d} x \\ \\
= & \int \frac{1}{\sqrt{x ^{4} \left( \frac{1-x}{x} \right) }} \mathrm{~d} x \\ \\
= & \int \frac{1}{ \textcolor{magenta}{x ^{2} } \sqrt{ \frac{1-x}{x} }} \mathrm{~d} x \\ \\
= & \int \frac{1}{ \sqrt{ \frac{1-x}{x} }} \cdot \textcolor{magenta}{ \frac{1}{x ^{2}} } \mathrm{~d} x \\ \\
= & \int \frac{1}{ \sqrt{ \frac{1-x}{x} }} \mathrm{~d} \left( \textcolor{magenta}{ \frac{-1}{x} } \right) \\ \\
= & – \int \frac{1}{ \sqrt{ \frac{1}{x} – 1 }} \mathrm{~d} \left( \frac{1}{x} \right) \\ \\
\xlongequal{k = 1/x} & – \int \frac{1}{ \sqrt{ k – 1 }} \mathrm{~d} \left( k \right) \\ \\
= & – \int \left( k – 1 \right) ^{\frac{-1}{2}} \mathrm{~d} \left( k – 1 \right) \\ \\
= & -2 \sqrt{k-1} + C_{0} \\ \\
= & \textcolor{magenta}{-2 \sqrt{\frac{1}{x} – 1}} + C_{0}
\end{aligned}
$$

其中，$C_{0}$ 为任意常数。

也就是说：

$$
\left( \textcolor{magenta}{\sqrt{\frac{1}{x} – 1}} \right) ^{\prime} _{x} = \textcolor{orangered}{-2} \cdot \textcolor{springgreen}{\frac{1}{\sqrt{x ^{3} (1-x)}}}
$$

于是：

$$
\begin{aligned}
& I \\ \\
= & \int \frac{\ln x}{\textcolor{springgreen}{\sqrt{x ^{3} (1-x)}} } \mathrm{~d} x \\ \\
= & \textcolor{orangered}{- 2} \int \ln x \mathrm { ~ d } \left( \textcolor{magenta}{ \sqrt { \frac { 1 – x } { x } } } \right) \\ \\
\xlongequal{\text{分部积分}} & – 2 \sqrt { \frac { 1 – x } { x } } \ln x + 2 \int \sqrt { \frac { 1 – x } { x } } \mathrm { ~ d } \left( \ln x \right) \\ \\
= & – 2 \sqrt { \frac { 1 – x } { x } } \ln x + \textcolor{#ffec00}{2 \int \sqrt { \frac { 1 – x } { x } } \cdot \frac { 1 } { x } \mathrm { ~ d } x} \\ \\
\end{aligned}
$$

又因为：

$$
\begin{aligned}
& \textcolor{#ffec00}{ 2 \int \sqrt { \frac { 1 – x } { x } } \cdot \frac { 1 } { x } \mathrm { ~ d } x} \\ \\
= & 2 \int \frac{1}{x} \cdot \frac{\sqrt{1-x}}{\sqrt{x}} \mathrm{~d} x \\ \\
= & 2 \int \frac{1}{x} \cdot \frac{(\sqrt{1-x}) (\sqrt{1-x})}{(\sqrt{x}) (\sqrt{1-x})} \mathrm{~d} x \\ \\
= & 2 \int \frac{1}{x} \cdot \frac{1-x}{\sqrt{x (1-x) } } \mathrm{~d} x \\ \\
= & 2 \int \frac { \textcolor{springgreen}{1} – \textcolor{orangered}{x} } { x \sqrt { x ( 1 – x ) } } \mathrm { ~ d } x \\ \\
= & 2 \textcolor{springgreen}{\int \frac { \mathrm { d } x } { x \sqrt { x ( 1 – x ) } } } – 2 \textcolor{orangered}{\int \frac { \mathrm { d } x } { \sqrt { x ( 1 – x ) } } } \\ \\
= & 4 \int \frac { \textcolor{tan}{ \mathrm { d } \left( \frac { 1 – x } { x } \right) } } { \textcolor{tan}{-2} \sqrt { \frac { 1 – x } { x } } } – 2 \int \frac { \textcolor{pink}{1} } { \textcolor{pink}{\sqrt{x}} \cdot \sqrt { 1 – x } } \mathrm { ~ d } x \\ \\
= & 4 \int \frac { \textcolor{tan}{ \mathrm { d } \left( \frac { 1 – x } { x } \right) } } { \textcolor{tan}{-2} \sqrt { \frac { 1 – x } { x } } } – 2 \int \frac { \textcolor{pink}{\frac{1}{\sqrt{x}}} } { \sqrt { 1 – x } } \mathrm { ~ d } x \\ \\
\xlongequal{\textcolor{pink}{ (\sqrt{x}) ^{\prime} = \frac{1}{2} \frac{1}{\sqrt{x}}}} & 4 \int \frac { \textcolor{tan}{ \mathrm { d } \left( \frac { 1 – x } { x } \right) } } { \textcolor{tan}{-2} \sqrt { \frac { 1 – x } { x } } } – 2 \cdot \textcolor{pink}{2} \int \frac { 1 } { \sqrt { 1 – x } } \mathrm { ~ d } \left( \textcolor{pink}{\sqrt{x}} \right) \\ \\
= & 4 \int \frac { \mathrm { d } \left( \frac { 1 – x } { x } \right) } { -2 \sqrt { \frac { 1 – x } { x } } } – 4 \int \frac { 1 } { \sqrt { 1 – ( \sqrt { x } ) ^ { 2 } } } \mathrm { ~ d } (\sqrt{x}) \\ \\
= & – 4 \sqrt { \frac { 1 – x } { x } } – 4 \arcsin \sqrt { x } + C_{1}
\end{aligned}
$$

于是：

$$
\textcolor{springgreen}{
\boldsymbol{
I = – 2 \sqrt { \frac { 1 – x } { x } } \ln x – 4 \sqrt { \frac { 1 – x } { x } } – 4 \arcsin \sqrt { x } + C_{1}
}
}
$$

其中，$C_{1}$ 为任意常数。

§2. 解法二

首先，将被积函数分母根号中的 “$x ^{3}$” 与 “$(1 – x)$” 分离开：

$$
\begin{aligned}
& I \\ \\
= & \int \frac{\ln x}{\sqrt{x ^{3} (1-x)}} \mathrm{~d} x \\ \\
= & \textcolor{orange}{\int \frac{\ln x}{\sqrt{(x ^{3})} \cdot \sqrt{(1-x)}} \mathrm{~d} x}
\end{aligned}
$$

之后，通过令 $t$ $=$ $\sqrt{1-x}$ 对上面式子中的 $\sqrt{(1-x)}$ 做整体代换，因为这样可以去掉一个根号。此时 $x$ $=$ $1 – t ^{2}$:

$$
\begin{aligned}
& I \\ \\
= & \textcolor{orange}{\int \frac{\ln x}{\sqrt{(x ^{3})} \cdot \sqrt{(1-x)}} \mathrm{~d} x} \\ \\
\xlongequal[x = 1 – t ^{2}]{t = \sqrt{1 – x}} & \int \frac{\ln (1 – t ^{2})}{\sqrt{(1 – t^{2}) ^{3}} \times t} \mathrm{~d} (1 – t ^{2}) \\ \\
= (-2) \cdot & \int \frac{\ln (1 – t ^{2}) \times \textcolor{black}{\colorbox{orange}{t}} }{\sqrt{(1 – t^{2}) ^{3}} \times \textcolor{black}{\colorbox{orange}{t}} } \mathrm{~d} t \\ \\
= \textcolor{orange}{(-2) \cdot} & \textcolor{orange}{\int \frac{\ln (1 – t ^{2})}{\sqrt{(1 – t^{2}) ^{3}}} \mathrm{~d} t} \\ \\
\end{aligned}
$$

之后，我们使用三角代换的方式去除上式中的根号，令 $t$ $=$ $\sin \theta$, 即：

$$
\begin{aligned}
& I \\ \\
\xlongequal{t = \sin \theta} \textcolor{orange}{(-2) } & \textcolor{orange}{\int \frac{\ln (1 – t ^{2})}{\sqrt{(1 – t^{2}) ^{3}}} \mathrm{~d} t} \\ \\
= (-2) & \int \frac{\ln (\cos ^{2} \theta)}{\cos ^{3} \theta} \mathrm{~d} (\sin \theta) \\ \\
= (-2) & \int \frac{\cos \theta \cdot \ln (\cos ^{2} \theta)}{\cos ^{3} \theta} \mathrm{~d} \theta \\ \\
= (-2) & \int \frac{\ln (\cos ^{\textcolor{magenta}{2}} \theta)}{\cos ^{2} \theta} \mathrm{~d} \theta \\ \\
= (-2 \cdot \textcolor{magenta}{2}) & \int \frac{\ln (\cos \theta)}{\cos ^{2} \theta} \mathrm{~d} \theta \\ \\
= (-4) & \int \ln (\cos \theta) \cdot \frac{1}{\cos ^{2} \theta} \mathrm{~d} \theta \\ \\
= (-4) & \int \ln (\cos \theta) \mathrm{~d} (\tan \theta) \\ \\
\xlongequal{\text{ 分部积分 }} (-4) & \left[ \ln(\cos \theta) \cdot \tan \theta – \int \tan \theta \mathrm{~d} [\ln(\cos \theta)] \right] \\ \\
= -4 & \ln (\cos \theta) \tan \theta + 4 \int \tan \theta \cdot \frac{- \sin \theta}{\cos \theta} \mathrm{~d} \theta \\ \\
= -4 & \tan \theta \ln (\cos \theta) – 4 \int \textcolor{pink}{\tan ^{2} \theta} \mathrm{~d} \theta \\ \\
= -4 & \tan \theta \ln (\cos \theta) – 4 \int \textcolor{pink}{\frac{\sin ^{2} \theta}{\cos ^{2} \theta}} \mathrm{~d} \theta \\ \\
= -4 & \tan \theta \ln (\cos \theta) – 4 \int \textcolor{pink}{\frac{1 – \cos ^{2} \theta}{\cos ^{2} \theta}} \mathrm{~d} \theta \\ \\
= -4 & \tan \theta \ln (\cos \theta) – 4 \int \left( \textcolor{pink}{\frac{1}{\cos ^{2} \theta} – 1} \right) \mathrm{~d} \theta \\ \\
= -4 & \tan \theta \ln (\cos \theta) – 4 \int \textcolor{pink}{\frac{1}{\cos ^{2} \theta}} \mathrm{~d} \theta + 4 \int \textcolor{pink}{1} \mathrm{~d} \theta \\ \\
= \textcolor{orange}{-4} & \textcolor{orange}{\tan \theta \ln (\cos \theta) – 4 \textcolor{pink}{\tan \theta} + 4 \textcolor{pink}{\theta} + C_{2}} \\ \\
\end{aligned}
$$

接着，根据 $t$ $=$ $\sin \theta$ $=$ $\sqrt{1-x}$, $\theta$ $=$ $\arcsin t$ $=$ $\arcsin (\sqrt{1-x})$ 进行回代：

$$
\begin{aligned}
& I \\ \\
= \textcolor{orange}{-4} & \textcolor{orange}{\tan \theta \ln (\cos \theta) – 4 \textcolor{pink}{\tan \theta} + 4 \textcolor{pink}{\theta} + C_{2}} \\ \\
= -4 & \tan (\arcsin t) \cdot \ln [\cos (\arcsin t)] \\
– 4 & \tan (\arcsin t) + 4 \arcsin t+ C_{2} \\ \\
= -4 & \frac{t}{ \sqrt{1 – t ^{2}} } \cdot \ln ( \sqrt{1 – t ^{2}} ) – \frac{4t}{\sqrt{1 – t ^{2}}} + 4 \arcsin t + C_{2} \\ \\
= & \frac{-4 \sqrt{1-x}}{\sqrt{x}} \cdot \ln \sqrt{x} + \frac{-4 \sqrt{1-x}}{\sqrt{x}} + 4 \arcsin (\sqrt{1-x}) + C_{2} \\ \\
= & \frac{-4 \sqrt{1-x}}{\sqrt{x}} \cdot \ln x^{\textcolor{orangered}{\frac{1}{2}}} + \frac{-4 \sqrt{1-x}}{\sqrt{x}} + 4 \arcsin (\sqrt{1-x}) + C_{2} \\ \\
= & \textcolor{orangered}{\frac{1}{2}} \cdot \frac{-4 \sqrt{1-x}}{\sqrt{x}} \cdot \ln x + \frac{-4 \sqrt{1-x}}{\sqrt{x}} + 4 \arcsin (\sqrt{1-x}) + C_{2} \\ \\
= & \textcolor{springgreen}{\boldsymbol{-2 \sqrt{ \frac{1-x}{x} } \cdot \ln x – 4 \sqrt{ \frac{1-x}{x} } + 4 \arcsin (\sqrt{1-x}) + C_{2} }} \\ \\
\end{aligned}
$$

Note

参考连接一：$\cos (\arcsin x)$ $=$ $\sqrt{1 – x ^{2}}$;
参考连接二：$\tan (\arcsin x)$ $=$ $\frac{x}{\sqrt{1 – x ^{2}}}$.

zhaokaifeng.com

在解法二中，我们得到的最终积分结果与前两种方法得到的积分结果中的 “$4 \arcsin (\sqrt{1-x})$ $+$ $C_{2}$” 这一部分不相同，但其实作用呢是一样的，因为 “$4 \arcsin (\sqrt{1-x})$ $+$ $C_{2}$” 的求导结果与 “$-4 \arcsin (\sqrt{x})$ $+$ $C_{1}$” 的求导结果相等。

此外，从函数图象上可以看出来（如图 01 所示），函数 $y$ = $\arcsin \left(\sqrt{1-x}\right)$ （绿色实线）和函数 $y$ $=$ $- \arcsin \left(\sqrt{x}\right)$ 就是只相差一个常数：

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