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第一类换元法

## 第一类换元法(凑微分法)
 **定理1** 设 $f(u)$ 具有原函数 $F(x), u=\varphi(x)$ 可导,则 $F[\varphi(x)]$ 是 $f[\varphi(x)] \varphi^{\prime}(x)$ 的原函数, 有换元公式

$$
\int f(\varphi(x)) \varphi^{\prime}(x) \mathrm{d} x=\left[\int f(u) \mathrm{d} u\right]_{u=\varphi(x)} .
$$

证明 因为 $F(x)$ 是 $f(u)$ 的原函数, 故 $\int f(u) \mathrm{d} u=F(u)+C$. 又 $u=\varphi(x)$ 可微, 故由复合函数微分法, 有 $\mathrm{d} u=\varphi^{\prime}(x) \mathrm{d} x=f[\varphi(x)] \varphi^{\prime}(x) d x$, 从而由不定积分定义得

$$
\int f[\varphi(x)] \varphi^{\prime}(x) \mathrm{d} x=F[\varphi(x)]+C=\left[\int f(u) d u\right]_{u=\varphi(x)} .
$$

> 注意：第一类换元法，通常俗称为**凑积分法**。这要求学生对能积分表**倒背如流**。 当拿到一个积分时，脑子里要第一时间感觉这个积分和哪个积分表相似，然后往上凑。

`例` 求 $\int(3 x+2)^5 \mathrm{~d} x$.
解：拿到这个积分，脑子里首先想到应该想到和基本积分表 $ \int x^5 \mathrm{~d} x  = \frac{1}{6} x^6+C $ 相似，所以尽可能往这个上面凑

①因为 $3 {~d} x={d}(3 x)={d}(3 x+2)$ (**正数常数项可以随意往$d$里放，因为常数求导为零，所以这里在d里额外增加了一个2**),

所以

$$
\int(3 x+2)^5 \mathrm{~d} x=\frac{1}{3} \int(3 x+2)^5 \cdot 3 \mathrm{~d} x=\frac{1}{3} \int(3 x+2)^5 \mathrm{~d}(3 x+2) \text {; }
$$

②设变量代换:令 $u=3 x+2$, 得 $\mathrm{d} u=\mathrm{d}(3 x+2)=3 \mathrm{~d} x$, 即有

$$
\int(3 x+2)^5 \mathrm{~d} x=\frac{1}{3} \int u^5 \mathrm{~d} u
$$

③利用基本积分公式求上式右端积分: $\frac{1}{3} \int u^5 \mathrm{~d} u=\frac{1}{3} \cdot \frac{1}{6} u^6+C$, 从而 $\int(3 x+2)^5 \mathrm{~d} x=\frac{1}{18} u^6+C$;

④回到原来变量, 将 $u=3 x+2$ 代入上式, 得所求积分为

$$
\int(3 x+2)^5 \mathrm{~d} x=\frac{1}{18}(3 x+2)^6+C .
$$

**以后我们就不用写出中间变量 $u$, 从求积分的第一步的式子中,将 $(3 x+2)$ 当作 中间变量 $u$, 直接得到第四步的结果. 这样, 本例求积分过程可简化为**

$$
\int(3 x+2)^5 \mathrm{~d} x=\frac{1}{3} \int(3 x+2)^5 \mathrm{~d}(3 x+2)=\frac{1}{18}(3 x+2)^6+C .
$$

一般地, 当被积函数形式为 $f(a x+b)$ 时, 可作变换 $u=a x+b$, 即若 $f(x)$ 有原 函数 $F(x)$, 则

$$
\int f(a x+b) \mathrm{d} x=\frac{1}{a} F(a x+b)+C .
$$

`例`求不定积分 $\int \frac{1}{1+2 x} d x$.
解： 拿到这个积分，首先想到的是他和 $\int \frac{1}{x} d x= ln |x|$ 很像，所以往这个公式上凑。
$$
  \begin{aligned}
\int \frac{1}{1+2 x} \mathrm{~d} x & =\frac{1}{2} \int \frac{1}{1+2 x} \cdot 2 \mathrm{~d} x=\frac{1}{2} \int \frac{1}{1+2 x} \cdot(1+2 x)^{\prime} \mathrm{d} x \\
& =\frac{1}{2} \int \frac{1}{1+2 x} \mathrm{~d}(1+2 x) \underline{\underline{1+2 x=u}} \frac{1}{2} \int \frac{1}{u} \mathrm{~d} u \\
& =\frac{1}{2} \ln |u|+C \underline{\underline{u=1+2 x}} \frac{1}{2} \ln |1+2 x|+C .
\end{aligned}
$$

`例` $\int \frac{\ln x}{x} \mathrm{~d} x$;
解 因为 $\frac{1}{x} \mathrm{~d} x=\mathrm{d} \ln x$, 故

$$
\int \frac{\ln x}{x} \mathrm{~d} x=\int \ln x \mathrm{~d}(\ln x) \underline{\underline{\ln x=u}} \int u \mathrm{~d} u=\frac{1}{2} u^2+C \underline{\underline{u=\ln x}} \frac{1}{2} \ln ^2 x+C .
$$

注 $f(\ln x) \frac{1}{x} \mathrm{~d} x=f(\ln x) \mathrm{d}(\ln x)$.

`例` $\int \frac{1}{x \ln x \ln \ln x} d x$.

$$
\int \frac{1}{x \ln x \ln \ln x} \mathrm{~d} x=\int \frac{1}{\ln x \ln \ln x} \mathrm{~d} \ln x=\int \frac{1}{\ln \ln x} \mathrm{~d} \ln \ln x=\ln |\ln \ln x|+C .
$$

`例` $\int 2 x \sin \left(x^2\right) \mathrm{d} x$;

$$
\int 2 x \sin \left(x^2\right) \mathrm{d} x=\int \sin \left(x^2\right) \mathrm{d}\left(x^2\right)=-\cos \left(x^2\right)+C
$$

`例`$\int x \sqrt{x^2+3} \mathrm{~d} x$.
解 因为 $\left.x \mathrm{~d} x=\left(\frac{1}{2} x^2\right)\right)^{\prime} \mathrm{d} x=\frac{1}{2}\left(x^2\right)^{\prime} \mathrm{d} x=\frac{1}{2} \mathrm{~d}\left(x^2\right)=\frac{1}{2} \mathrm{~d}\left(x^2+3\right)$, 故

$$
\int x \sqrt{x^2+3} \mathrm{~d} x=\frac{1}{2} \int \sqrt{x^2+3} \mathrm{~d}\left(x^2+3\right)=\frac{1}{3}\left(x^2+3\right)^{\frac{3}{2}}+C \text {. }
$$

注 若 $f(x)$ 有原函数 $F(x)$, 则

$$
\int f\left(x^2\right) 2 x \mathrm{~d} x=F\left(x^2\right)+C .
$$

`例` 求下列不定积分：(1) $\int \frac{1}{1+\mathrm{e}^x} \mathrm{~d} x$;
解

$$
\begin{aligned}
\text { 原式 } & =\int \frac{1+\mathrm{e}^x-\mathrm{e}^x}{1+\mathrm{e}^x} \mathrm{~d} x=\int\left(1-\frac{\mathrm{e}^x}{1+\mathrm{e}^x}\right) \mathrm{d} x \\
& =\int \mathrm{d} x-\int \frac{\mathrm{e}^x}{1+\mathrm{e}^x} \mathrm{~d} x=\int \mathrm{d} x-\int \frac{1}{1+\mathrm{e}^x} \mathrm{~d}\left(1+\mathrm{e}^x\right) \\
& =x-\ln \left(1+\mathrm{e}^x\right)+C .
\end{aligned}
$$

`例` $\int \frac{\sin \frac{1}{x}}{x^2} d x$.

$$
\begin{aligned}
& \int \frac{\sin \frac{1}{x}}{x^2} \mathrm{~d} x=\int \sin \left(\frac{1}{x}\right) \cdot\left(-\frac{1}{x}\right) \mathrm{d} x=-\int \sin \left(\frac{1}{x}\right) \cdot \mathrm{d}\left(\frac{1}{x}\right)=\cos \left(\frac{1}{x}\right)+C . \\
& \text { 注 } \int f\left(\mathrm{e}^x\right) \mathrm{e}^x \mathrm{~d} x=\int f\left(\mathrm{e}^x\right) \mathrm{d}\left(\mathrm{e}^x\right) ; \int f\left(\frac{1}{x}\right) \frac{1}{x^2} \mathrm{~d} x=-\int f\left(\frac{1}{x}\right) \mathrm{d}\left(\frac{1}{x}\right) .
\end{aligned}
$$

`例`$\int \frac{1}{a^2+x^2} \mathrm{~d} x(a>0)$;
解 $\int \frac{1}{a^2+x^2} \mathrm{~d} x=\frac{1}{a^2} \int \frac{1}{1+\left(\frac{x}{a}\right)^2} \mathrm{~d} x=\frac{1}{a} \int \frac{1}{1+\left(\frac{x}{a}\right)^2} \mathrm{~d}\left(\frac{x}{a}\right)=\frac{1}{a} \arctan \frac{x}{a}+C$.

`例` $\int \frac{1}{x^2-a^2} \mathrm{~d} x(a>0)$.
解 先作分解 $\frac{1}{x^2-a^2}=\frac{1}{2 a}\left(\frac{1}{x-a}-\frac{1}{x+a}\right)$,然后积分.

$$
\begin{aligned}
\int \frac{1}{x^2-a^2} \mathrm{~d} x & =\int \frac{1}{2 a}\left(\frac{1}{x-a}-\frac{1}{x+a}\right) \mathrm{d} x \\
& =\frac{1}{2 a}\left(\int \frac{1}{x-a} \mathrm{~d}(x-a)-\int \frac{1}{x+a} \mathrm{~d}(x+a)\right) \\
& =\frac{1}{2 a}(\ln |x-a|-\ln |x+a|)=\frac{1}{2 a} \ln \left|\frac{x-a}{x+a}\right|+C .
\end{aligned}
$$

`例` $\int \frac{1}{\sqrt{2 x+3}+\sqrt{2 x-1}} \mathrm{~d} x$.
解 将分母有理化:

$$
\begin{aligned}
\text { 原式 } & =\int \frac{\sqrt{2 x+3}-\sqrt{2 x-1}}{(\sqrt{2 x+3}+\sqrt{2 x-1})(\sqrt{2 x+3}-\sqrt{2 x-1})} \mathrm{d} x \\
& =\frac{1}{4} \int \sqrt{2 x+3} \mathrm{~d} x-\frac{1}{4} \int \sqrt{2 x-1} \mathrm{~d} x \\
& =\frac{1}{8} \int \sqrt{2 x+3} \mathrm{~d}(2 x+3)-\frac{1}{8} \int \sqrt{2 x-1} \mathrm{~d}(2 x-1) \\
& =\frac{1}{12}(\sqrt{2 x+3})^3-\frac{1}{12}(\sqrt{2 x-1})^3+C .
\end{aligned}
$$

> 注 利用平方差公式进行根式有理化是化简积分计算的常用手段之一.

`例` $\int \frac{\mathrm{e}^{3 \sqrt{x}}}{\sqrt{x}} \mathrm{~d} x$;
解 $\int \frac{\mathrm{e}^{3 \sqrt{x}}}{\sqrt{x}} \mathrm{~d} x=2 \int \mathrm{e}^{3 \sqrt{x}} \mathrm{~d}(\sqrt{x})=\frac{2}{3} \int \mathrm{e}^{3 \sqrt{x}} \mathrm{~d}(3 \sqrt{x})=\frac{2}{3} \mathrm{e}^{3 \sqrt{x}}+C$;

`例`$\int \frac{\tan \sqrt{x}}{\sqrt{x}} d x$.
解 $\begin{aligned} \int \frac{\tan \sqrt{x}}{\sqrt{x}} \mathrm{~d} x & =2 \int \tan \sqrt{x} \mathrm{~d}(\sqrt{x})=2 \int \frac{\sin \sqrt{x}}{\cos \sqrt{x}} \mathrm{~d}(\sqrt{x}) \\ & =-2 \int \frac{1}{\cos \sqrt{x}} \mathrm{~d}(\cos \sqrt{x})=-2 \ln |\cos \sqrt{x}|+C .\end{aligned}$

`例` 求 $\int \frac{\arctan \sqrt{x}}{\sqrt{x}(1+x)} \mathrm{d} x$.
解 注意到 $\mathrm{d} \sqrt{x}=\frac{1}{2 \sqrt{x}} \mathrm{~d} x$ 即可.

$$
\begin{aligned}
\int \frac{\arctan \sqrt{x}}{\sqrt{x}(1+x)} \mathrm{d} x & =\int \frac{2 \arctan \sqrt{x}}{1+(\sqrt{x})^2} \mathrm{~d} \sqrt{x}=\int 2 \arctan \sqrt{x} \mathrm{darctan} \sqrt{x} . \\
& =(\arctan \sqrt{x})^2+C .
\end{aligned}
$$

`例` 求不定积分 $\int \sin 2 x \mathrm{~d} x$.
解法一 原式 $=\frac{1}{2} \int \sin 2 x \mathrm{~d}(2 x)=-\frac{1}{2} \cos 2 x+C$;
解法二 原式 $=2 \int \sin x \cos x \mathrm{~d} x=2 \int \sin x \mathrm{~d}(\sin x)=(\sin x)^2+C$;
解法三 原式 $=2 \int \sin x \cos x \mathrm{~d} x=-2 \int \cos x \mathrm{~d}(\cos x)=-(\cos x)^2+C$.
注 $\quad f(\sin x) \cos x \mathrm{~d} x=f(\sin x) \mathrm{d}(\sin x)$;
$f(\cos x) \sin x \mathrm{~d} x=-f(\cos x) \mathrm{d}(\cos x)$.

`例`求 (1) $\int \tan x \mathrm{~d} x ;(2) \int \cot x \mathrm{~d} x$.
解 (1) $\int \tan x \mathrm{~d} x=\int \frac{\sin x}{\cos x} \mathrm{~d} x=-\int \frac{1}{\cos x} \mathrm{~d} \cos x=-\ln |\cos x|+C$.
(2) 同理可得 $\int \cot x \mathrm{~d} x=\ln |\sin x|+C$.

`例` $\int \sin ^3 x \mathrm{~d} x$;
解 因为 $\sin ^3 x=\sin ^2 x \cdot \sin x=\sin ^2 x \cdot(-\cos x)^{\prime}=\left(1-\cos ^2 x\right) \cdot(-\cos x)^{\prime}$,
所以

$$
\int \sin ^3 x \mathrm{~d} x=-\int\left(1-\cos ^2 x\right) \mathrm{d} \cos x=-\cos x+\frac{1}{3} \cos ^3 x+C .
$$

`例` $\int \sin ^2 x \cos ^3 x d x$;
解 因为 $\sin ^2 x \cos ^3 x=\sin ^2 x \cos ^2 x \cdot \cos x=\sin ^2 x\left(1-\sin ^2 x\right) \cdot \cos x$, 所以

$$
\begin{aligned}
\int \sin ^2 x \cos ^3 x \mathrm{~d} x & =\int \sin ^2 x\left(1-\sin ^2 x\right) \cdot \cos x \mathrm{~d} x=\int \sin ^2 x\left(1-\sin ^2 x\right) \mathrm{d} \sin x \\
& =\int\left(\sin ^2 x-\sin ^4 x\right) \mathrm{d} \sin x \\
& =\frac{1}{3} \sin ^3 x-\frac{1}{5} \sin ^5 x+C .
\end{aligned}
$$

`例` $\int \cos ^2 x \mathrm{~d} x$;

$$
\text { 解 } \begin{aligned}
\int \cos ^2 x \mathrm{~d} x & =\int \frac{1+\cos 2 x}{2} \mathrm{~d} x=\int \frac{1}{2}(1+\cos 2 x) \mathrm{d} x=\frac{1}{2} \int(1+\cos 2 x) \mathrm{d} x \\
& =\frac{1}{2}\left(x+\frac{1}{2} \sin 2 x\right)+C=\frac{1}{2} x+\frac{1}{4} \sin 2 x+C .
\end{aligned}
$$

`例` $\int \sec ^4 x \mathrm{~d} x$.

$$
\text { 解 } \begin{aligned}
\int \sec ^4 x \mathrm{~d} x & =\int \sec ^2 x \cdot \sec ^2 x \mathrm{~d} x=\int \sec ^2 x \mathrm{~d} \tan x=\int\left(1+\tan ^2 x\right) \mathrm{d} \tan x \\
& =\tan x+\frac{1}{3} \tan ^3 x+C .
\end{aligned}
$$

`例` $\int \csc x \mathrm{~d} x$;

$$
\text { 解 } \begin{aligned}
& \int \csc x \mathrm{~d} x=\int \frac{1}{\sin x} \mathrm{~d} x=\int \frac{1}{2 \sin \frac{x}{2} \cos \frac{x}{2}} \mathrm{~d} x=\int \frac{1}{2} \frac{\sec ^2 \frac{x}{2}}{\tan \frac{x}{2}} \mathrm{~d} x=\int \frac{1}{\tan \frac{x}{2}} \mathrm{~d} \tan \frac{x}{2} \\
&=\ln \left|\tan \frac{x}{2}\right|+C \\
& \text { 又 } \tan \frac{x}{2}=\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}=\frac{2 \sin ^2 \frac{x}{2}}{\sin x}=\frac{1-\cos x}{\sin x}=\csc x-\cot x \text {, 故 } \\
& \int \csc x \mathrm{~d} x=\ln |\csc x-\cot x|+C ;
\end{aligned}
$$

`例` $\int \sec x \mathrm{~d} x$.

$$
\text { 解 } \begin{aligned}
\int \sec x \mathrm{~d} x=\int \frac{1}{\cos x} \mathrm{~d} x & =\int \frac{\mathrm{d}\left(x+\frac{\pi}{2}\right)}{\sin \left(x+\frac{\pi}{2}\right)}=\ln \left|\csc \left(x+\frac{\pi}{2}\right)-\cot \left(x+\frac{\pi}{2}\right)\right|+C \\
& =\ln |\sec x+\tan x|+C .
\end{aligned}
$$

`例`$\int \sec x \mathrm{~d} x$.
也可以用下列方法解决：

$$
\begin{aligned}
\int \sec x \mathrm{~d} x & =\int \frac{1}{\cos x} \mathrm{~d} x=\int \frac{\cos x}{1-\sin ^2 x} \mathrm{~d} x \\
& =\int \frac{1}{2}\left[\frac{1}{1-\sin x}+\frac{1}{1+\sin x}\right] \mathrm{d} \sin x=\frac{1}{2} \ln \left|\frac{1+\sin x}{1-\sin x}\right|+C \\
& =\frac{1}{2} \ln \left|\frac{(1+\sin x)^2}{\cos ^2 x}\right|+C=\ln \left|\frac{1+\sin x}{\cos x}\right|+C \\
& =\ln |\sec x+\tan x|+C .
\end{aligned}
$$

`例`求 (1) $\int \cos 3 x \cos x \mathrm{~d} x$;
解 利用三角函数的积化和差公式

$$
\cos \alpha \cos \beta=\frac{1}{2}[\cos (\alpha+\beta)+\cos (\alpha-\beta)] \text {, }
$$

可得

$$
\begin{aligned}
\int \cos 3 x \cos x \mathrm{~d} x & =\int \frac{1}{2}(\cos 4 x+\cos 2 x) \mathrm{d} x=\frac{1}{2}\left(\frac{1}{4} \sin 4 x+\frac{\sin 2 x}{2}\right)+C \\
& =\frac{1}{8} \sin 4 x+\frac{1}{4} \sin 2 x+C .
\end{aligned}
$$

`例`求 (2) $\int \sin 3 x \cos x \mathrm{~d} x$;
解 利用三角函数的积化和差公式

$$
\sin \alpha \cos \beta=\frac{1}{2}[\sin (\alpha+\beta)+\sin (\alpha-\beta)],
$$

可得 $\int \sin 3 x \cos x \mathrm{~d} x=\int \frac{1}{2}(\sin 4 x+\sin 2 x) \mathrm{d} x=\frac{1}{2}\left(-\frac{\cos 4 x}{4}-\frac{\cos 2 x}{2}\right)+C$

$$
=-\frac{\cos 4 x}{8}-\frac{\cos 2 x}{4}+C \text {. }
$$

注
第一类换元法也称凑微分法, 有时也需要一定的技巧, 这需要读者进一步练习.

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