710 lines
17 KiB
Markdown
710 lines
17 KiB
Markdown
## 笔记记录
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### 要点 01 - 积分与极限求和式的转化
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根据公式
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$$
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\int_a^b f(x)dx = \lim_{n\to\infty}\sum_{i=1}^n f(x_i)\Delta x
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$$
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对于均匀矩形分割的情况,实际上只用分离出 $\frac{1}{n}$
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$$
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\int_a^b f(x)dx = \lim_{n\to\infty}\sum_{i=1}^n f\left(a + \frac{(b-a) i}{n}\right) \frac{b-a}{n}
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$$
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---
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### 要点 02 - 分式型积分
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#### 基本公式
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$$
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\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \arctan \frac{x}{a} + C \quad (a > 0)
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$$
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**推导**(换元法):令 $x = a \tan t$,则 $dx = a \sec^2 t \, dt$
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$$
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\int \frac{dx}{x^2 + a^2} = \int \frac{a \sec^2 t}{a^2 + a^2 \tan^2 t} \, dt
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$$
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$$
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= \int \frac{a \sec^2 t}{a^2 \sec^2 t} \, dt
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$$
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$$
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= \int \frac{dt}{a}
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$$
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$$
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= \frac{t}{a} + C
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$$
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$$
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= \frac{1}{a} \arctan \frac{x}{a} + C
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$$
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#### 推广形式
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$$
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\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \arctan \frac{x}{a} + C
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$$
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$$
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\int \frac{dx}{b^2 + (x + c)^2} = \frac{1}{b} \arctan \frac{x + c}{b} + C
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$$
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$$
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\int \frac{x \, dx}{x^2 + a^2} = \frac{1}{2} \ln(x^2 + a^2) + C
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$$
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$$
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\int \frac{dx}{(x^2 + a^2)^2} = \frac{x}{2a^2(x^2 + a^2)} + \frac{1}{2a^3} \arctan \frac{x}{a} + C
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$$
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---
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### 要点 03 - 根号分式型积分
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#### 基本公式
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$$
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\int \frac{dx}{\sqrt{x^2 + a^2}} = \ln\left|x + \sqrt{x^2 + a^2}\right| + C
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$$
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$$
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\int \frac{dx}{\sqrt{x^2 + a^2}} = \operatorname{arsinh} \frac{x}{a} + C
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$$
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$$
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\int \frac{dx}{\sqrt{x^2 - a^2}} = \ln\left|x + \sqrt{x^2 - a^2}\right| + C \quad (|x| > |a|)
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$$
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$$
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\int \frac{dx}{\sqrt{x^2 - a^2}} = \operatorname{arcosh} \frac{x}{a} + C \quad (|x| > |a|)
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$$
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$$
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\int \frac{dx}{\sqrt{a^2 - x^2}} = \arcsin \frac{x}{a} + C \quad (|x| < |a|)
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$$
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$$
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\int \frac{dx}{\sqrt{a^2 - x^2}} = -\arccos \frac{x}{a} + C \quad (|x| < |a|)
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$$
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#### 推导方法
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$$
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\begin{align}
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\int \frac{dx}{\sqrt{x^2 + a^2}}
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&= \int \frac{a \cosh t}{a \cosh t} \, dt && (x = a \sinh t) \\
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&= \int dt \\
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&= t + C \\
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&= \ln\left|x + \sqrt{x^2 + a^2}\right| + C
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\end{align}
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$$
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$$
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\begin{align}
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\int \frac{dx}{\sqrt{x^2 - a^2}}
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&= \int \frac{a \sinh t}{a \sinh t} \, dt && (x = a \cosh t) \\
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&= \int dt \\
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&= t + C \\
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&= \ln\left|x + \sqrt{x^2 - a^2}\right| + C
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\end{align}
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$$
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$$
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\begin{align}
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\int \frac{dx}{\sqrt{a^2 - x^2}}
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&= \int \frac{a \cos t}{a \cos t} \, dt && (x = a \sin t) \\
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&= \int dt \\
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&= t + C \\
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&= \arcsin \frac{x}{a} + C
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\end{align}
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$$
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#### 等效形式
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$$
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\int \frac{dx}{\sqrt{x^2 - a^2}} = \ln\left|\frac{x}{a} + \sqrt{\frac{x^2}{a^2} - 1}\right| + C
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$$
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#### 推广形式
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$$
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\int \frac{dx}{\sqrt{(x + b)^2 + a^2}} = \ln\left|x + b + \sqrt{(x + b)^2 + a^2}\right| + C
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$$
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$$
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\int \frac{dx}{\sqrt{(x + b)^2 - a^2}} = \ln\left|x + b + \sqrt{(x + b)^2 - a^2}\right| + C \quad (|x + b| > |a|)
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$$
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$$
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\int \frac{x \, dx}{\sqrt{x^2 + a^2}} = \sqrt{x^2 + a^2} + C
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$$
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$$
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\int \frac{x \, dx}{\sqrt{x^2 - a^2}} = \sqrt{x^2 - a^2} + C
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$$
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$$
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\int \sqrt{x^2 + a^2} \, dx = \frac{x}{2}\sqrt{x^2 + a^2} + \frac{a^2}{2}\ln\left|x + \sqrt{x^2 + a^2}\right| + C
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$$
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$$
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\int \sqrt{x^2 - a^2} \, dx = \frac{x}{2}\sqrt{x^2 - a^2} - \frac{a^2}{2}\ln\left|x + \sqrt{x^2 - a^2}\right| + C
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$$
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---
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### 要点 04 - 根号二次型积分
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#### 基本公式
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$$
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\int \sqrt{x^2 + a^2} \, dx = \frac{x}{2}\sqrt{x^2 + a^2} + \frac{a^2}{2}\ln\left|x + \sqrt{x^2 + a^2}\right| + C
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$$
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$$
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\int \sqrt{x^2 - a^2} \, dx = \frac{x}{2}\sqrt{x^2 - a^2} - \frac{a^2}{2}\ln\left|x + \sqrt{x^2 - a^2}\right| + C \quad (|x| > |a|)
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$$
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$$
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\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\arcsin\frac{x}{a} + C \quad (|x| < |a|)
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$$
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#### 推导方法
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$$
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\begin{align}
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\int \sqrt{x^2 + a^2} \, dx
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&= \int a \cosh t \cdot a \cosh t \, dt = a^2 \int \cosh^2 t \, dt && (x = a \sinh t) \\
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&= a^2 \int \frac{\cosh 2t + 1}{2} \, dt = \frac{a^2}{2}\left(\frac{\sinh 2t}{2} + t\right) + C \\
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&= \frac{a^2}{2}(\sinh t \cosh t + t) + C \\
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&= \frac{x}{2}\sqrt{x^2 + a^2} + \frac{a^2}{2}\ln\left|x + \sqrt{x^2 + a^2}\right| + C
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\end{align}
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$$
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$$
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\begin{align}
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\int \sqrt{x^2 - a^2} \, dx
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&= \int a \sinh t \cdot a \sinh t \, dt = a^2 \int \sinh^2 t \, dt && (x = a \cosh t) \\
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&= a^2 \int \frac{\cosh 2t - 1}{2} \, dt = \frac{a^2}{2}\left(\frac{\sinh 2t}{2} - t\right) + C \\
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&= \frac{a^2}{2}(\sinh t \cosh t - t) + C \\
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&= \frac{x}{2}\sqrt{x^2 - a^2} - \frac{a^2}{2}\ln\left|x + \sqrt{x^2 - a^2}\right| + C
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\end{align}
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$$
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$$
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\begin{align}
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\int \sqrt{a^2 - x^2} \, dx
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&= \int a \cos t \cdot a \cos t \, dt = a^2 \int \cos^2 t \, dt && (x = a \sin t) \\
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&= \frac{a^2}{2}\left(t + \frac{\sin 2t}{2}\right) + C \\
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&= \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\arcsin\frac{x}{a} + C
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\end{align}
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$$
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#### 推广形式
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$$
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\int (x + b)\sqrt{x^2 + a^2} \, dx = \frac{1}{3}(x + b)(x^2 + a^2)^{3/2} - \frac{b}{2}x\sqrt{x^2 + a^2} - \frac{ab^2}{2}\ln\left|x + \sqrt{x^2 + a^2}\right| + C
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$$
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$$
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\int x\sqrt{x^2 + a^2} \, dx = \frac{1}{3}(x^2 + a^2)^{3/2} + C
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$$
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$$
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\int x\sqrt{x^2 - a^2} \, dx = \frac{1}{3}(x^2 - a^2)^{3/2} + C
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$$
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$$
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\int x\sqrt{a^2 - x^2} \, dx = -\frac{1}{3}(a^2 - x^2)^{3/2} + C
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$$
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---
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### 要点 05 - 三角函数积分
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#### 降幂公式
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$$
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\sin^2 x = \frac{1 - \cos 2x}{2}
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$$
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$$
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\cos^2 x = \frac{1 + \cos 2x}{2}
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$$
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$$
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\sin^3 x = \frac{3\sin x - \sin 3x}{4}
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$$
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$$
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\cos^3 x = \frac{3\cos x + \cos 3x}{4}
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$$
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#### 基本积分
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$$
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\int \sin x \, dx = -\cos x + C
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$$
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$$
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\int \cos x \, dx = \sin x + C
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$$
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$$
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\int \tan x \, dx = -\ln|\cos x| + C
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$$
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$$
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\int \cot x \, dx = \ln|\sin x| + C
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$$
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$$
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\int \sec x \, dx = \ln|\sec x + \tan x| + C
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$$
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$$
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\int \csc x \, dx = -\ln|\csc x + \cot x| + C
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$$
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$$
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\int \sec^2 x \, dx = \tan x + C
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$$
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$$
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\int \csc^2 x \, dx = -\cot x + C
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$$
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$$
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\int \sec x \tan x \, dx = \sec x + C
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$$
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$$
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\int \csc x \cot x \, dx = -\csc x + C
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$$
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#### 万能代换
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令 $t = \tan\frac{x}{2}$,则:
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$$
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\sin x = \frac{2t}{1 + t^2}, \quad \cos x = \frac{1 - t^2}{1 + t^2}, \quad dx = \frac{2 \, dt}{1 + t^2}
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$$
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适用类型:$R(\sin x, \cos x)$(有理函数形式)
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#### 常用结论
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$$
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\int \sin^n x \, dx = -\frac{\sin^{n-1} x \cos x}{n} + \frac{n-1}{n}\int \sin^{n-2} x \, dx
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$$
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$$
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\int \cos^n x \, dx = \frac{\cos^{n-1} x \sin x}{n} + \frac{n-1}{n}\int \cos^{n-2} x \, dx
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$$
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$$
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\int \sec^n x \, dx = \frac{\sec^{n-2} x \tan x}{n-1} + \frac{n-2}{n-1}\int \sec^{n-2} x \, dx \quad (n \neq 1)
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$$
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$$
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\int \csc^n x \, dx = -\frac{\csc^{n-2} x \cot x}{n-1} + \frac{n-2}{n-1}\int \csc^{n-2} x \, dx \quad (n \neq 1)
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$$
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**secⁿ 递推公式推导**(分部积分法):
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设 $I_n = \int \sec^n x \, dx$,改写为 $\int \sec^{n-2} x \cdot \sec^2 x \, dx$。
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令 $u = \sec^{n-2} x$,$dv = \sec^2 x \, dx$,则:
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$$
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du = (n-2)\sec^{n-2} x \tan x \, dx,\quad v = \tan x
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$$
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代入分部积分公式 $\int u \, dv = uv - \int v \, du$:
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$$
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\begin{aligned}
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I_n &= \sec^{n-2} x \tan x - \int \tan x \cdot (n-2)\sec^{n-2} x \tan x \, dx \\
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&= \sec^{n-2} x \tan x - (n-2)\int \sec^{n-2} x \tan^2 x \, dx
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\end{aligned}
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$$
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利用 $\tan^2 x = \sec^2 x - 1$:
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$$
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I_n = \sec^{n-2} x \tan x - (n-2)\int \sec^{n-2} x (\sec^2 x - 1) \, dx
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$$
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$$
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I_n = \sec^{n-2} x \tan x - (n-2) I_n + (n-2) I_{n-2}
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$$
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移项合并 $I_n$ 项:
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$$
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(n-1) I_n = \sec^{n-2} x \tan x + (n-2) I_{n-2}
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$$
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$$
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\boxed{I_n = \frac{\sec^{n-2} x \tan x}{n-1} + \frac{n-2}{n-1} I_{n-2}}
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$$
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需要两个初始条件:
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- $I_1 = \int \sec x \, dx = \ln|\sec x + \tan x| + C$
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- $I_2 = \int \sec^2 x \, dx = \tan x + C$
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**cscⁿ 递推公式推导**类似,利用 $\cot^2 x = \csc^2 x - 1$。
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#### 积化和差
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$$
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\sin A \cos B = \frac{1}{2}\sin(A+B) + \frac{1}{2}\sin(A-B)
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$$
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$$
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\cos A \cos B = \frac{1}{2}\cos(A+B) + \frac{1}{2}\cos(A-B)
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$$
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$$
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\sin A \sin B = \frac{1}{2}\cos(A-B) - \frac{1}{2}\cos(A+B)
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$$
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---
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### 要点 06 - 换元积分法
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#### 第一类换元法(凑微分法)
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若 $\int f(u) \, du = F(u) + C$,$u = \varphi(x)$ 可微,则:
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$$
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\int f[\varphi(x)] \, \varphi'(x) \, dx = \int f(u) \, du = F(u) + C = F[\varphi(x)] + C
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$$
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**核心思想**:将被积函数中一部分"凑"成某个函数的导数,令其为一个新变量。
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**常见凑微分形式**:
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| 类型 | 凑微分 | 令 $u$ |
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|------|--------|--------|
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| $\int f(ax+b) \, dx$ | $\displaystyle\frac{1}{a} \int f(ax+b) \, d(ax+b)$ | $u = ax+b$ |
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| $\int f(x^n) \cdot x^{n-1} \, dx$ | $\displaystyle\frac{1}{n} \int f(x^n) \, d(x^n)$ | $u = x^n$ |
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| $\int f(\sin x) \cos x \, dx$ | $\displaystyle\int f(\sin x) \, d(\sin x)$ | $u = \sin x$ |
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| $\int f(\cos x) \sin x \, dx$ | $\displaystyle-\int f(\cos x) \, d(\cos x)$ | $u = \cos x$ |
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| $\int f(\tan x) \sec^2 x \, dx$ | $\displaystyle\int f(\tan x) \, d(\tan x)$ | $u = \tan x$ |
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| $\int f(e^x) e^x \, dx$ | $\displaystyle\int f(e^x) \, d(e^x)$ | $u = e^x$ |
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| $\int f(\ln x) \frac{1}{x} \, dx$ | $\displaystyle\int f(\ln x) \, d(\ln x)$ | $u = \ln x$ |
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| $\int f(\arcsin x) \frac{dx}{\sqrt{1-x^2}}$ | $\displaystyle\int f(\arcsin x) \, d(\arcsin x)$ | $u = \arcsin x$ |
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| $\int f(\arctan x) \frac{dx}{1+x^2}$ | $\displaystyle\int f(\arctan x) \, d(\arctan x)$ | $u = \arctan x$ |
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**示例**:
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$$
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\int \frac{dx}{x \ln x} = \int \frac{d(\ln x)}{\ln x} = \ln|\ln x| + C
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$$
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---
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#### 第二类换元法(变量代换法)
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令 $x = \psi(t)$,其中 $\psi(t)$ 单调可导且 $\psi'(t) \neq 0$,则:
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$$
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\int f(x) \, dx = \int f[\psi(t)] \, \psi'(t) \, dt
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$$
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**常用代换类型**:
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##### 1. 三角代换
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| 被积函数含 | 代换 | 适用区间 | 微元 |
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|-----------|------|---------|------|
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| $\sqrt{a^2 - x^2}$ | $x = a \sin t$ | $\displaystyle[-\frac{\pi}{2}, \frac{\pi}{2}]$ | $dx = a \cos t \, dt$ |
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| $\sqrt{a^2 + x^2}$ | $x = a \tan t$ | $\displaystyle(-\frac{\pi}{2}, \frac{\pi}{2})$ | $dx = a \sec^2 t \, dt$ |
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| $\sqrt{x^2 - a^2}$ | $x = a \sec t$ | $\displaystyle[0, \frac{\pi}{2}) \cup (\frac{\pi}{2}, \pi]$ | $dx = a \sec t \tan t \, dt$ |
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##### 2. 双曲函数代换
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| 被积函数含 | 代换 | 微元 |
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|-----------|------|------|
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| $\sqrt{a^2 + x^2}$ | $x = a \sinh t$ | $dx = a \cosh t \, dt$ |
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| $\sqrt{x^2 - a^2}$ | $x = a \cosh t$ | $dx = a \sinh t \, dt$ |
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双曲函数代换优势:无需分类讨论符号,计算更简洁。
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##### 3. 根式代换
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令 $t = \sqrt[n]{ax + b}$,则 $x = \dfrac{t^n - b}{a}$,$dx = \dfrac{n t^{n-1}}{a} \, dt$
|
||
|
||
适用类型:$\displaystyle\int R(x, \sqrt[n]{ax+b}) \, dx$
|
||
|
||
##### 4. 倒代换
|
||
|
||
令 $x = \dfrac{1}{t}$,则 $dx = -\dfrac{1}{t^2} \, dt$
|
||
|
||
适用类型:分母次数比分子次数高较多时(通常差 $2$ 次以上)
|
||
|
||
##### 5. 指数代换
|
||
|
||
令 $t = e^x$,则 $x = \ln t$,$dx = \dfrac{dt}{t}$
|
||
|
||
适用类型:$\displaystyle\int R(e^x) \, dx$
|
||
|
||
##### 6. 万能代换
|
||
|
||
令 $t = \tan\dfrac{x}{2}$,则:
|
||
|
||
$$
|
||
\sin x = \frac{2t}{1+t^2}, \quad \cos x = \frac{1-t^2}{1+t^2}, \quad dx = \frac{2}{1+t^2} \, dt
|
||
$$
|
||
|
||
适用类型:$R(\sin x, \cos x)$ 有理函数形式(已在要点 05 中列出)
|
||
|
||
---
|
||
|
||
#### 两类换元法对比
|
||
|
||
| 对比项 | 第一类换元法(凑微分) | 第二类换元法(变量代换) |
|
||
|-------|----------------------|----------------------|
|
||
| 本质 | $u = \varphi(x)$,从 $x$ 到 $u$ | $x = \psi(t)$,从 $x$ 到 $t$ |
|
||
| 适用场景 | 被积函数中有"导数因子" | 被积函数含根式、复杂表达式 |
|
||
| 操作难度 | 较简单,需观察导数关系 | 较复杂,需选择合适的代换 |
|
||
| 常见类型 | 凑微分表 | 三角/双曲/根式/倒代换 |
|
||
|
||
---
|
||
|
||
### 要点 07 - 分部积分法
|
||
|
||
#### 基本公式
|
||
|
||
由乘法求导法则 $(uv)' = u'v + uv'$ 两边积分得:
|
||
|
||
$$
|
||
\int u \, dv = uv - \int v \, du
|
||
$$
|
||
|
||
或写作:
|
||
|
||
$$
|
||
\int u(x) v'(x) \, dx = u(x) v(x) - \int v(x) u'(x) \, dx
|
||
$$
|
||
|
||
**核心思想**:将被积函数分为两部分 $u$ 和 $dv$,通过公式将不易直接积分的部分转化为更易积分的形式。
|
||
|
||
---
|
||
|
||
#### 选择 $u$ 和 $dv$ 的原则
|
||
|
||
**关键**:$u$ 应使导数变简单,$dv$ 应易于积分。
|
||
|
||
##### LIATE 优先序(反-对-幂-三-指)
|
||
|
||
按以下顺序选择 $u$(优先级从高到低):
|
||
|
||
| 类别 | 英文 | 示例 |
|
||
|-----|------|------|
|
||
| **L** - 反三角函数 | **L**ogarithmic inverse | $\arcsin x, \arctan x$ |
|
||
| **I** - 对数函数 | **I**nverse trigonometric | $\ln x, \log_a x$ |
|
||
| **A** - 幂函数 | **A**lgebraic | $x^n, ax+b$ |
|
||
| **T** - 三角函数 | **T**rigonometric | $\sin x, \cos x, \sec^2 x$ |
|
||
| **E** - 指数函数 | **E**xponential | $e^x, a^x$ |
|
||
|
||
**规则**:排名靠前的选为 $u$,靠后的选为 $dv$。
|
||
|
||
**示例**:
|
||
|
||
$$
|
||
\int x e^x \, dx \quad \Longrightarrow \quad u = x,\; dv = e^x \, dx
|
||
$$
|
||
|
||
$$
|
||
\int x \ln x \, dx \quad \Longrightarrow \quad u = \ln x,\; dv = x \, dx
|
||
$$
|
||
|
||
$$
|
||
\int e^x \sin x \, dx \quad \Longrightarrow \quad u = \sin x,\; dv = e^x \, dx \text{(指数和三角函数任选其一)}
|
||
$$
|
||
|
||
---
|
||
|
||
#### 常见类型与技巧
|
||
|
||
##### 类型 1:幂函数 $\times$ 指数/三角函数($u$ 取幂函数)
|
||
|
||
$$
|
||
\int x^n e^{ax} \, dx,\quad \int x^n \sin(ax) \, dx,\quad \int x^n \cos(ax) \, dx
|
||
$$
|
||
|
||
令 $u = x^n$,$dv = e^{ax} \, dx$(或 $\sin(ax) \, dx$、$\cos(ax) \, dx$),**需多次分部**直至幂次降为 $0$。
|
||
|
||
##### 类型 2:幂函数 $\times$ 对数/反三角($u$ 取对数/反三角)
|
||
|
||
$$
|
||
\int x^n \ln x \, dx,\quad \int x^n \arcsin x \, dx,\quad \int x^n \arctan x \, dx
|
||
$$
|
||
|
||
令 $u = \ln x$(或 $\arcsin x$、$\arctan x$),$dv = x^n \, dx$,一次分部即可消去对数/反三角。
|
||
|
||
##### 类型 3:指数 $\times$ 三角函数(循环分部)
|
||
|
||
$$
|
||
\int e^{ax} \sin(bx) \, dx,\quad \int e^{ax} \cos(bx) \, dx
|
||
$$
|
||
|
||
任选其一为 $u$,两次分部后出现原积分,**移项求解**。
|
||
|
||
**示例**:
|
||
|
||
$$
|
||
\begin{align}
|
||
I &= \int e^{ax} \sin(bx) \, dx \\
|
||
&= -\frac{e^{ax} \cos(bx)}{b} + \frac{a}{b} \int e^{ax} \cos(bx) \, dx \\
|
||
&= -\frac{e^{ax} \cos(bx)}{b} + \frac{a}{b}\left(\frac{e^{ax} \sin(bx)}{b} - \frac{a}{b} I\right) \\
|
||
&\Rightarrow I = \frac{e^{ax}(a \sin(bx) - b \cos(bx))}{a^2 + b^2} + C
|
||
\end{align}
|
||
$$
|
||
|
||
##### 类型 4:单独一个函数
|
||
|
||
$$
|
||
\int \ln x \, dx,\quad \int \arcsin x \, dx,\quad \int \arctan x \, dx
|
||
$$
|
||
|
||
令 $u = \ln x$(或 $\arcsin x$、$\arctan x$),**$dv = dx$**(凑出 $1$ 作为 $dv$)。
|
||
|
||
**示例**:
|
||
|
||
$$
|
||
\int \ln x \, dx = x \ln x - \int x \cdot \frac{1}{x} \, dx = x \ln x - x + C
|
||
$$
|
||
|
||
##### 类型 5:分部与换元结合
|
||
|
||
先换元化简,再分部积分。常见于被积函数含复合结构时。
|
||
|
||
---
|
||
|
||
#### 分部积分法推广公式
|
||
|
||
反复应用分部积分法则可得:
|
||
|
||
$$
|
||
\int u v^{(n+1)} \, dx = u v^{(n)} - u' v^{(n-1)} + u'' v^{(n-2)} - \cdots + (-1)^{n+1} \int u^{(n+1)} v \, dx
|
||
$$
|
||
|
||
---
|
||
|
||
### 要点 08 - 有理分式积分
|
||
|
||
#### 基本概念
|
||
|
||
**有理分式**:两个多项式的比 $\dfrac{P(x)}{Q(x)}$
|
||
|
||
**真分式**:分子次数 $<$ 分母次数
|
||
|
||
**假分式**:分子次数 $\geq$ 分母次数,需先化为多项式 + 真分式
|
||
|
||
#### 部分分式分解法
|
||
|
||
将真分式分解为若干简单分式之和:
|
||
|
||
##### 1. 分母仅有线性因子
|
||
|
||
若分母可分解为 $(x-a_1)^{m_1}(x-a_2)^{m_2}\cdots$,则:
|
||
|
||
$$
|
||
\frac{P(x)}{Q(x)} = \frac{A_1}{x-a_1} + \frac{A_2}{(x-a_1)^2} + \cdots + \frac{A_{m_1}}{(x-a_1)^{m_1}} + \frac{B_1}{x-a_2} + \cdots
|
||
$$
|
||
|
||
##### 2. 分母含二次因子
|
||
|
||
若分母含不可约二次因子 $x^2+px+q$,则对应项为:
|
||
|
||
$$
|
||
\frac{Ax+B}{x^2+px+q}, \quad \frac{Ax+B}{(x^2+px+q)^n}
|
||
$$
|
||
|
||
#### 常见积分类型
|
||
|
||
##### 类型 1:一次因子
|
||
|
||
$$
|
||
\int \frac{dx}{x-a} = \ln|x-a| + C
|
||
$$
|
||
|
||
$$
|
||
\int \frac{dx}{(x-a)^n} = -\frac{1}{(n-1)(x-a)^{n-1}} + C \quad (n \neq 1)
|
||
$$
|
||
|
||
##### 类型 2:二次质因子
|
||
|
||
配方后分项积分:
|
||
|
||
$$
|
||
\int \frac{x}{x^2+px+q} \, dx = \frac{1}{2}\ln(x^2+px+q) - \frac{p}{2}\int \frac{dx}{x^2+px+q}
|
||
$$
|
||
|
||
对于 $\displaystyle\int \frac{dx}{x^2+px+q}$,配方:
|
||
|
||
$$
|
||
x^2+px+q = \left(x+\frac{p}{2}\right)^2 + \left(q-\frac{p^2}{4}\right)
|
||
$$
|
||
|
||
则:
|
||
|
||
$$
|
||
\int \frac{dx}{x^2+px+q} = \frac{2}{\sqrt{4q-p^2}} \arctan \frac{2x+p}{\sqrt{4q-p^2}} + C \quad (4q > p^2)
|
||
$$
|
||
|
||
##### 类型 3:二次因子幂次
|
||
|
||
$$
|
||
\int \frac{dx}{(x^2+a^2)^n} = \frac{x}{2a^2(n-1)(x^2+a^2)^{n-1}} + \frac{2n-3}{2a^2(n-1)}\int \frac{dx}{(x^2+a^2)^{n-1}}
|
||
$$
|
||
|
||
特别地,当 $n=2$ 时:
|
||
|
||
$$
|
||
\int \frac{dx}{(x^2+a^2)^2} = \frac{x}{2a^2(x^2+a^2)} + \frac{1}{2a^3}\arctan\frac{x}{a} + C
|
||
$$
|
||
|
||
#### 积分步骤总结
|
||
|
||
1. **化简**:假分式化为多项式 + 真分式
|
||
2. **分解**:对分母因式分解,写出部分分式形式
|
||
3. **待定系数**:比较系数或代值法求系数
|
||
4. **积分**:逐项积分
|
||
|
||
#### 示例
|
||
|
||
**例**:求 $\displaystyle\int \frac{x+3}{x^2-5x+6} \, dx$
|
||
|
||
解:分母因式分解 $x^2-5x+6 = (x-2)(x-3)$
|
||
|
||
设 $\displaystyle\frac{x+3}{(x-2)(x-3)} = \frac{A}{x-2} + \frac{B}{x-3}$
|
||
|
||
则 $x+3 = A(x-3) + B(x-2) = (A+B)x - (3A+2B)$
|
||
|
||
比较系数:$\begin{cases} A+B=1 \\ 3A+2B=-3 \end{cases} \Rightarrow A=-5, B=6$
|
||
|
||
故 $\displaystyle\int \frac{x+3}{x^2-5x+6} \, dx = \int\left(\frac{-5}{x-2} + \frac{6}{x-3}\right)dx = -5\ln|x-2| + 6\ln|x-3| + C$
|
||
|
||
---
|
||
|
||
### 知识点
|
||
- 定积分的定义
|
||
- 黎曼和与积分的关系
|
||
- 均匀分割技巧
|
||
- $\frac{1}{x^2 + a^2}$ 型积分公式
|
||
- $\frac{1}{\sqrt{x^2 \pm a^2}}$ 型积分公式
|
||
- $\sqrt{x^2 \pm a^2}$ 型积分公式
|
||
- 三角函数积分(降幂、万能代换、积化和差)
|
||
- 第一类换元法(凑微分法)
|
||
- 第二类换元法(变量代换法:三角/双曲/根式/倒代换)
|
||
- 分部积分法(LIATE 优先序)
|
||
- 循环分部与移项求解
|
||
- 有理分式积分(部分分式分解法)
|