不定积分-工具

发表: 4/28/2026 更新: 6/1/2026 字数: 0 字时长: 0 分钟

(一) 基本公式

常数
$\begin{array}{r} \int k d x = k x + C \end{array}$
幂函数不定积分公式
① $\begin{array}{r} a \neq - 1 : \int x^{a} d x = \frac{1}{a + 1} x^{a + 1} + C; \end{array}$
② $\begin{array}{r} a = - 1 : \int x^{- 1} d x = \ln | x | + C; \end{array}$

简单证明

$∵ x < 0 时, [\ln (- x)]^{'} = \frac{1}{x}, ∴ \int \frac{1}{x} d x = \ln (- x) + C .$
$x > 0 时, (\ln x)^{'} = \frac{1}{x}, ∴ \int \frac{1}{x} d x = \ln x + C .$
$∴ \int \frac{1}{x} d x = \ln | x | + C .$

指数函数
① $\begin{array}{r} a \neq e : \int a^{x} d x = \frac{a^{x}}{\ln a} + C; \end{array}$
② $\begin{array}{r} a = e : \int e^{x} d x = e^{x} + C; \end{array}$
三角函数
① $\begin{array}{r} \int \sin x d x = - \cos x + C; \end{array}$
② $\begin{array}{r} \int \cos x d x = \sin x + C; \end{array}$
③ $\begin{array}{r} \int \tan x d x = \ln | \sec x | + C; \end{array}$
④ $\begin{array}{r} \int \cot x d x = \ln | \sin x | + C; \end{array}$
⑤ ⭐ $\begin{array}{r} \int \sec x d x = \ln | \sec x + \tan x | + C; \end{array}$
⑥ ⭐ $\begin{array}{r} \int \csc x d x = \ln | \csc x - \cot x | + C; \end{array}$
⑦ $\begin{array}{r} \int \sec^{2} x d x = \tan x + C; \end{array}$
⑧ $\begin{array}{r} \int \csc^{2} x d x = - \cot x + C; \end{array}$
⑨ $\begin{array}{r} \int \sec x \tan x d x = \sec x + C; \end{array}$
⑩ $\begin{array}{r} \int \csc x \cot x d x = - \csc x + C; \end{array}$

推导

$\begin{aligned} \int \tan x d x & = \int \frac{\sin x}{\cos x} d x = \int - \frac{d (\cos x)}{\cos x} = \int d (- \ln | \cos x |) \\ = \int d (\ln | \sec x |) = \ln | \sec x | + C . \end{aligned}$

$\begin{array}{l} \begin{array}{r} \int \sec x d x = \int \frac{1}{\cos x} d x = \int \frac{\cos x}{\cos^{2} x} d x = \int \frac{d \sin x}{\cos^{2} x} = \int \frac{d \sin x}{1 - \sin^{2} x} \end{array} \\ \begin{aligned} \overset{令 t = \sin x}{=} \int \frac{1}{1 - t^{2}} d t & = \frac{1}{2} \int (\frac{1}{1 + t} + \frac{1}{1 - t}) d t \\ = \frac{1}{2} (\int \frac{1}{1 + t} d (1 + t) - \int \frac{1}{1 - t} d (1 - t)) \end{aligned} \\ \begin{aligned} \int \frac{d \sin x}{1 - \sin^{2} x} & = \frac{1}{2} (\ln | 1 + t | - \ln | 1 - t |) + C = \frac{1}{2} \ln | \frac{1 + t}{1 - t} | + C \\ = \frac{1}{2} \ln | \frac{1 + \sin x}{1 - \sin x} | + C = \frac{1}{2} \ln | \frac{(1 + \sin x)^{2}}{1 - \sin^{2} x} | + C \\ = \ln | \frac{1 + \sin x}{\cos x} | + C \\ = \ln | \sec x + \tan x | + C . \end{aligned} \end{array}$

递推公式

$I_{n} = \int \sec^{n} x d x = \frac{\sec^{n - 2} x \tan x}{n - 1} + \frac{n - 2}{n - 1} I_{n - 2} (n \geq 2)$
$I_{n} = \int \sin^{n} x d x = - \frac{1}{n} \sin^{n - 1} x \cos x + \frac{n - 1}{n} I_{n - 2} (n \geq 2)$
$I_{n} = \int \cos^{n} x d x = \frac{1}{n} \cos^{n - 1} x \sin x + \frac{n - 1}{n} I_{n - 2} (n \geq 2)$

有理分式
① $\begin{array}{r} \int \frac{d x}{\sqrt{1 - x^{2}}} = \arcsin x + C; \end{array}$
② $\begin{array}{r} \int \frac{d x}{\sqrt{a^{2} - x^{2}}} = \int \frac{d (\frac{x}{a})}{\sqrt{1 - (\frac{x}{a})^{2}}} = \arcsin \frac{x}{a} + C; \end{array}$
③ $\begin{array}{r} \int \frac{d x}{1 + x^{2}} = \arctan x + C; \end{array}$
④ $\begin{array}{r} \int \frac{d x}{a^{2} + x^{2}} = \frac{1}{a} \int \frac{d (\frac{x}{a})}{1 + (\frac{x}{a})^{2}} = \frac{1}{a} \arctan \frac{x}{a} + C; \end{array}$
⑤ $\begin{array}{r} \int \frac{d x}{x^{2} - a^{2}} = \frac{1}{2 a} [\int \frac{d (x - a)}{x - a} - \int \frac{d (x + a)}{x + a}] = \frac{1}{2 a} \ln | \frac{x - a}{x + a} | + C; \end{array}$
⑥ ⭐ $\begin{array}{r} \int \frac{d x}{\sqrt{x^{2} - a^{2}}} = \ln | x + \sqrt{x^{2} - a^{2}} | + C; \end{array}$
⑦ ⭐ $\begin{array}{r} \int \frac{d x}{\sqrt{x^{2} + a^{2}}} = \ln | x + \sqrt{x^{2} + a^{2}} | + C; \end{array}$
⑧ ⭐ $\begin{array}{r} \int \sqrt{a^{2} - x^{2}} d x = \frac{a^{2}}{2} \arcsin \frac{x}{a} + \frac{x}{2} \sqrt{a^{2} - x^{2}} + C; \end{array}$
⑨ $\begin{array}{r} \int \sqrt{x^{2} - a^{2}} d x = \frac{1}{2} x \sqrt{x^{2} - a^{2}} - \frac{a^{2}}{2} \ln | x + \sqrt{x^{2} - a^{2}} | + C; \end{array}$
⑩ $\begin{array}{r} \int \sqrt{x^{2} + a^{2}} d x = \frac{1}{2} x \sqrt{x^{2} + a^{2}} + \frac{a^{2}}{2} \ln | x + \sqrt{x^{2} + a^{2}} | + C . \end{array}$

推导

$\begin{array}{l} ⑥ \int \frac{d x}{\sqrt{x^{2} - a^{2}}} \\ 令 \tan ϑ = \frac{\sqrt{x^{2} - a^{2}}}{a}, \cos ϑ = \frac{a}{x} . \\ \Rightarrow \sqrt{x^{2} - a^{2}} = a \tan ϑ, x = a \sec ϑ . \\ \begin{aligned} 原式 & = \int \frac{1}{a \tan ϑ} d (a \sec ϑ) = \int \cot ϑ d \sec ϑ = \int \cot ϑ \tan ϑ \sec ϑ d ϑ = \int \sec ϑ d ϑ \\ = \ln | \sec ϑ + \tan ϑ | + C = \ln | \frac{x}{a} + \frac{\sqrt{x^{2} - a^{2}}}{a} | + C = \ln | x + \sqrt{x^{2} - a^{2}} | + C . \end{aligned} \end{array}$

$三角函数辅助图1$ 图一三角函数辅助图1

$\begin{array}{l} ⑦ \int \frac{d x}{\sqrt{x^{2} + a^{2}}} \\ 令 \cos ϑ = \frac{a}{\sqrt{x^{2} + a^{2}}}, \tan ϑ = \frac{x}{a} . \\ \Rightarrow \sqrt{x^{2} + a^{2}} = a \sec ϑ, x = a \tan ϑ . \\ \begin{aligned} 原式 & = \int \frac{1}{a \sec ϑ} d (a \tan ϑ) = \int \cos ϑ d \tan ϑ = \int \cos ϑ \sec^{2} ϑ d ϑ = \int \sec ϑ d ϑ \\ = \ln | \sec ϑ + \tan ϑ | + C = \ln | \frac{\sqrt{x^{2} + a^{2}}}{a} + \frac{x}{a} | + C = \ln | x + \sqrt{x^{2} + a^{2}} | + C . \end{aligned} \end{array}$

$三角函数辅助图2$ 图二三角函数辅助图2

$\begin{array}{l} ⑧ \int \sqrt{a^{2} - x^{2}} d x \\ 令 \cos ϑ = \frac{\sqrt{a^{2} - x^{2}}}{a}, \sin ϑ = \frac{x}{a} . \\ \Rightarrow \sqrt{a^{2} - x^{2}} = a \cos ϑ, x = a \sin ϑ . \\ \begin{aligned} 原式 & = a^{2} \int \cos ϑ d (\sin ϑ) = a^{2} \int \cos^{2} ϑ d ϑ = a^{2} \int \frac{\cos 2 ϑ + 1}{2} d ϑ = \frac{a^{2}}{2} (\int \cos 2 ϑ d ϑ + \int d ϑ) \\ = \frac{a^{2}}{4} \sin 2 ϑ + \frac{a^{2}}{2} ϑ + C = \frac{a^{2}}{2} \sin ϑ \cos ϑ + \frac{a^{2}}{2} \arcsin \frac{x}{a} + C \\ = \frac{a^{2}}{2} \arcsin \frac{x}{a} + \frac{x}{2} \sqrt{a^{2} - x^{2}} + C . \end{aligned} \end{array}$

$三角函数辅助图3$ 图三三角函数辅助图3

$\begin{array}{l} ⑨ \int \sqrt{x^{2} - a^{2}} d x \\ 由 ⑥ 得 : \int \sqrt{x^{2} - a^{2}} d x = a^{2} \int \tan ϑ d (\sec ϑ) = a^{2} \int \tan^{2} ϑ \sec ϑ d ϑ \\ 令 t = \int \tan^{2} ϑ \sec ϑ d ϑ \\ \begin{aligned} t & = \tan ϑ \sec ϑ - \int \sec ϑ d (\tan ϑ) \\ = \tan ϑ \sec ϑ - \int \sec ϑ \cdot \sec^{2} ϑ d ϑ \\ = \tan ϑ \sec ϑ - \int \sec ϑ (1 + \tan^{2} ϑ) d ϑ \\ = \tan ϑ \sec ϑ - \int \sec ϑ d ϑ - t . \\ \Rightarrow t & = \frac{1}{2} \tan ϑ \sec ϑ - \frac{1}{2} \ln | \sec ϑ + \tan ϑ | + C \\ = \frac{1}{2 a^{2}} x \sqrt{x^{2} - a^{2}} - \frac{1}{2} \ln | x + \sqrt{x^{2} - a^{2}} | + C . \end{aligned} \\ ∴ \int \sqrt{x^{2} - a^{2}} d x = a^{2} t = \frac{1}{2} x \sqrt{x^{2} - a^{2}} - \frac{a^{2}}{2} \ln | x + \sqrt{x^{2} - a^{2}} | + C . \end{array}$

$\begin{array}{l} ⑩ \int \sqrt{x^{2} + a^{2}} d x \\ 由 ⑦ 得 : \int \sqrt{x^{2} + a^{2}} d x = a^{2} \int \sec ϑ d (\tan ϑ) = a^{2} \int \sec^{3} ϑ d ϑ \\ 令 t = \int \sec^{3} ϑ d ϑ \\ \begin{aligned} t & = \sec ϑ \tan ϑ - \int \tan ϑ d (\sec ϑ) \\ = \sec ϑ \tan ϑ - \int \tan^{2} ϑ \cdot \sec ϑ d ϑ \\ = \sec ϑ \tan ϑ - \int (\sec^{2} ϑ - 1) \sec ϑ d ϑ \\ = \sec ϑ \tan ϑ - t + \int \sec ϑ d ϑ . \\ \Rightarrow t & = \frac{1}{2} \sec ϑ \tan ϑ + \frac{1}{2} \ln | \sec ϑ + \tan ϑ | + C \\ = \frac{1}{2 a^{2}} x \sqrt{x^{2} + a^{2}} + \frac{1}{2} \ln | x + \sqrt{x^{2} + a^{2}} | + C . \end{aligned} \\ ∴ \int \sqrt{x^{2} + a^{2}} d x = a^{2} t = \frac{1}{2} x \sqrt{x^{2} + a^{2}} + \frac{a^{2}}{2} \ln | x + \sqrt{x^{2} + a^{2}} | + C . \end{array}$

(二) 积分法

1.换元积分法

Case1 第一类换元积分法

$例 1. \int \frac{x}{(2 x + 1)^{2}} d x$
$例 2. \int \frac{d x}{x^{2} + 2 x + 5}$
$例 3. \int \frac{d x}{\sqrt{2 x - x^{2}}}$
$例 4. \int \frac{d x}{\sqrt{x} (4 + x)}$
$例 5. \int \frac{d x}{\sqrt{x^{2} + x}}$
$例 6. \int \frac{e^{x}}{4 + e^{2 x}} d x$
$例 7. \int \frac{d x}{x \ln^{2} x}$

形如 $\int f [φ (x)] φ^{'} (x) d x = \int f [φ (x)] d (φ (x)) \overset{令 t = φ (x)}{=} \int f (t) d t = F (t) + C = F [φ (x)] + C .$
$\begin{aligned} 如 : \int \frac{\sin^{2} \sqrt{x}}{\sqrt{x}} d x & = 2 \int \sin^{2} \sqrt{x} d (\sqrt{x}) \overset{令 t = \sqrt{x}}{=} 2 \int \sin^{2} t d t \\ = \int (1 - \cos 2 t) d t = t - \int \cos 2 t d t \\ = t - \frac{1}{2} \sin 2 t + C = \sqrt{x} - \frac{1}{2} \sin 2 \sqrt{x} + C . \end{aligned}$

Case2 第二类换元积分法

$例 1. \int \frac{t}{\sqrt{t + 1}} d t$
$例 2. \int \frac{\sqrt{t + 1}}{t} d t$
$例 3. \int \frac{d t}{t \sqrt{t + 1}}$
$例 4. \int t \sqrt{t + 1} d t$
$例 5. \int \frac{t}{\sqrt{t^{2} + 1}} d t$
$例 6. \int \frac{d t}{\sqrt{(x^{2} + 1)^{3}}}$

形如 $\int f (x) d x \overset{令 x = ψ (t)}{=} \int f [ψ (t)] ψ^{'} (t) d t = G (t) = G [ψ^{- 1} (x)] + C .$

2.分部积分法

已知 $(u v)^{'} = u^{'} v + u v^{'}$
$d (u v) = d (u) v + u d (v)$
两边同时积分: $u v = \int v d u + \int u d v$
$\Rightarrow \int u d v = u v - \int v d u$

Case1.∫ 幂×指 dx

例1. $\int x^{2} e^{x} d x$

Case2.∫ 幂×对 dx

例2. $\int x \ln^{2} x d x$

Case3.∫ 幂×三角 dx

例3. $\int x \sin^{2} x d x$
例4. $\int x \tan^{2} x d x$

Case4.∫ 幂×反三角 dx

例5. $\int x \arctan x d x$
例6. $\int \arcsin x d x$

Case5. $e^{a x} \times {_{\sin b x}^{\cos a x} d x$

$⭐ 例 7. \int e^{2 x} \cos x d x$

$\begin{array}{l} \begin{aligned} 解 : I & = \int e^{2 x} \cos x d x = \int e^{2 x} d (\sin x) \\ = e^{2 x} \sin x - 2 \int \sin x e^{2 x} d x = e^{2 x} \sin x + 2 \int e^{2 x} d (\cos x) \\ = e^{2 x} \sin x + 2 (e^{2 x} \cos x - 2 \int \cos x e^{2 x} d x) \\ = e^{2 x} \sin x + 2 e^{2 x} \cos x - 4 I \end{aligned} \\ \Rightarrow I = \frac{1}{5} e^{2 x} (\sin x + 2 \cos x) + C . \end{array}$

Case6. $\int \sec^{n} x (或 \csc^{n} x) d x (n 为奇数)$

例8. $\int \sec^{4} x d x$
例9. $\int \sec^{3} x d x$

错题集

其他重要的公式定理

不定积分-工具

(一) 基本公式

(二) 积分法

1.换元积分法

2.分部积分法

$⭐ 例 7. \int e^{2 x} \cos x d x$

不定积分-工具 ​

(一) 基本公式 ​

(二) 积分法 ​

1.换元积分法 ​

2.分部积分法 ​

⭐例⭐例7.∫e2xcos⁡xdx

不定积分-工具

(一) 基本公式

(二) 积分法

1.换元积分法

2.分部积分法

$⭐ 例 7. \int e^{2 x} \cos x d x$