A-Level 数学：分部积分法完全指南 | Integration by Parts: The Complete Guide

在 A-Level 数学（尤其是 Edexcel、AQA 和 OCR 考试局）中，分部积分法（Integration by Parts） 是必考的核心积分技巧。它不仅出现在纯数学（Pure Mathematics）卷中，还在力学和统计学的应用题中频繁使用。本文将带你从基础公式到高阶技巧，全面掌握分部积分法。

In A-Level Mathematics — particularly for Edexcel, AQA, and OCR exam boards — Integration by Parts is an essential integration technique that appears frequently. It features not only in Pure Mathematics papers but also in applied contexts across Mechanics and Statistics. This guide takes you from the fundamental formula to advanced strategies for complete mastery.

1. 分部积分公式 | The Integration by Parts Formula

分部积分法源于微分的乘积法则（Product Rule）。回顾一下：如果 \( u \) 和 \( v \) 都是 \( x \) 的函数，那么：

Integration by Parts is derived from the Product Rule for differentiation. Recall: if \( u \) and \( v \) are both functions of \( x \), then:

\[ \frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx} \]

两边积分并移项，得到分部积分公式：

Integrating both sides and rearranging gives the Integration by Parts formula:

\[ \int u \, \frac{dv}{dx} \, dx = uv – \int v \, \frac{du}{dx} \, dx \]

更简洁的写法（A-Level 考试中常用）：

A more compact form (commonly used in A-Level exams):

\[ \int u \, dv = uv – \int v \, du \]

💡 关键思路：我们把一个复杂的积分 \( \int u\,dv \) 转化为一个（希望）更简单的积分 \( \int v\,du \)。

💡 Core Idea: We transform a complex integral \( \int u\,dv \) into a (hopefully) simpler integral \( \int v\,du \).

2. 如何选择 \( u \) 和 \( dv \) | How to Choose \( u \) and \( dv \)

选择正确的 \( u \) 是分部积分法成功的关键。使用 LIATE 法则（优先级从高到低）：

Choosing the right \( u \) is the key to success with Integration by Parts. Use the LIATE Rule (priority from highest to lowest):

• L — Logarithmic functions 对数函数：\( \ln x, \ln(ax+b) \)
• I — Inverse trigonometric functions 反三角函数：\( \arcsin x, \arctan x \)
• A — Algebraic functions 代数函数：\( x^n, (ax+b)^n \)
• T — Trigonometric functions 三角函数：\( \sin x, \cos x, \tan x \)
• E — Exponential functions 指数函数：\( e^x, a^x \)

选择 LIATE 中较高的类别作为 \( u \)，其余部分（包括 \( dx \)）作为 \( dv \)。

Choose the higher-ranked category in LIATE as \( u \), and the rest (including \( dx \)) as \( dv \).

💡 记忆口诀（中文）：”对数反代三角指 —— 谁在前面谁当 u”

3. 经典例题 | Classic Worked Examples

例题 1 | Example 1: \( \int x e^x \, dx \)

按照 LIATE：Algebraic (\( x \)) 优先于 Exponential (\( e^x \))，因此：

Following LIATE: Algebraic (\( x \)) ranks above Exponential (\( e^x \)), so:

\[ u = x, \quad dv = e^x \, dx \]

\[ du = dx, \quad v = e^x \]

代入公式：

Substituting into the formula:

\[ \int x e^x \, dx = x e^x – \int e^x \, dx = x e^x – e^x + C = e^x(x – 1) + C \]

✅ 验证：对 \( e^x(x-1) \) 求导，使用乘积法则，得到 \( e^x(x-1) + e^x \cdot 1 = xe^x \)。正确！

✅ Check: Differentiate \( e^x(x-1) \) using the Product Rule: \( e^x(x-1) + e^x \cdot 1 = xe^x \). Correct!

例题 2 | Example 2: \( \int x^2 \sin x \, dx \)

按照 LIATE：Algebraic (\( x^2 \)) 优先于 Trigonometric (\( \sin x \))：

Following LIATE: Algebraic (\( x^2 \)) ranks above Trigonometric (\( \sin x \)):

\[ u = x^2, \quad dv = \sin x \, dx \]

\[ du = 2x\,dx, \quad v = -\cos x \]

\[ \int x^2 \sin x \, dx = -x^2 \cos x + \int 2x \cos x \, dx \]

新的积分 \( \int 2x\cos x \, dx \) 仍需要分部积分法（再次使用！）：

The new integral \( \int 2x\cos x \, dx \) still requires Integration by Parts (apply again!):

\[ u = 2x, \quad dv = \cos x \, dx; \quad du = 2\,dx, \quad v = \sin x \]

\[ \int 2x \cos x \, dx = 2x\sin x – \int 2\sin x \, dx = 2x\sin x + 2\cos x + C \]

最终答案：

Final answer:

\[ \int x^2 \sin x \, dx = -x^2\cos x + 2x\sin x + 2\cos x + C \]

📌 这个例题展示了重复使用分部积分法的技巧——每次积分都会降低 \( x \) 的次数。

📌 This example demonstrates the repeated Integration by Parts technique — each integration reduces the power of \( x \).

例题 3 | Example 3: \( \int \ln x \, dx \) (A-Level 经典题)

初看，这似乎只有一个函数。技巧是：设 \( u = \ln x \)，\( dv = 1 \cdot dx \)：

At first glance, this looks like a single function. The trick: set \( u = \ln x \), \( dv = 1 \cdot dx \):

\[ u = \ln x, \quad dv = dx \]

\[ du = \frac{1}{x}dx, \quad v = x \]

\[ \int \ln x \, dx = x\ln x – \int x \cdot \frac{1}{x} \, dx = x\ln x – \int 1 \, dx = x\ln x – x + C \]

🎯 这是 A-Level 考试的高频题，必须熟记！

🎯 This is a high-frequency exam question in A-Level — memorize it!

4. 高级技巧 | Advanced Techniques

技巧 1：循环积分 | Technique 1: Cyclic Integration

当两次分部积分后，原积分重新出现时，可以像解方程一样求出结果。

When the original integral reappears after two rounds of Integration by Parts, you can solve for it like an equation.

例题 | Example: \( \int e^x \sin x \, dx \)

设 \( I = \int e^x \sin x \, dx \)

第一次分部积分（选 \( u = \sin x, dv = e^x dx \)）：

First Integration by Parts (choose \( u = \sin x, dv = e^x dx \)):

\[ I = e^x\sin x – \int e^x \cos x \, dx \]

对剩余积分再次使用分部积分法（\( u = \cos x, dv = e^x dx \)）：

Apply Integration by Parts again to the remaining integral (\( u = \cos x, dv = e^x dx \)):

\[ I = e^x\sin x – \left( e^x\cos x + \int e^x \sin x \, dx \right) \]

\[ I = e^x\sin x – e^x\cos x – I \]

\[ 2I = e^x(\sin x – \cos x) \]

\[ I = \frac{e^x(\sin x – \cos x)}{2} + C \]

⚠️ 常见陷阱：如果你两次都选同类型的函数作为 \( u \)（比如两次都选三角函数），循环不会闭合。第一次选三角函数，第二次也要选三角函数（不能用指数函数代替），这样原积分才会精确重现。

⚠️ Common Pitfall: If you switch which type of function you choose as \( u \) in the second round, the original integral won’t reappear. Keep the same type (trig in both rounds, or exponential in both) for the cycle to close.

技巧 2：分部积分与定积分 | Technique 2: Integration by Parts with Definite Integrals

对于定积分，公式变为：

For definite integrals, the formula becomes:

\[ \int_a^b u \, dv = \Big[uv\Big]_a^b – \int_a^b v \, du \]

例题 | Example: \( \int_0^1 x e^{2x} \, dx \)

\[ u = x, \quad dv = e^{2x}dx; \quad du = dx, \quad v = \frac{1}{2}e^{2x} \]

\[ \int_0^1 x e^{2x} dx = \Big[\frac{x}{2}e^{2x}\Big]_0^1 – \int_0^1 \frac{1}{2}e^{2x} dx \]

\[ = \frac{1}{2}e^2 – 0 – \Big[\frac{1}{4}e^{2x}\Big]_0^1 = \frac{e^2}{2} – \left(\frac{e^2}{4} – \frac{1}{4}\right) = \frac{e^2 + 1}{4} \]

5. 考试技巧与常见错误 | Exam Tips & Common Mistakes

• ❌ 错误：忘了 “+C” —— 不定积分必须加积分常数！考试中每个遗漏扣1分。
• ❌ Mistake: Forgetting “+C” — Always add the constant of integration for indefinite integrals! One mark lost per omission.
• ❌ 错误：选反了 u —— 如果用 LIATE 选了错误的 u，积分会越做越复杂。例如在 \( \int xe^x dx \) 中如果选 \( u=e^x \)，会陷入死循环。
• ❌ Mistake: Wrong choice of u — Picking the wrong u (against LIATE) makes the integral more complex. In \( \int xe^x dx \), choosing \( u = e^x \) leads to a dead end.
• ❌ 错误：\( dv \) 必须包含 \( dx \) —— 许多学生忘记 \( dv \) 是导数形式，必须包含 \( dx \)。
• ❌ Mistake: dv must include dx — Many students forget that dv is in derivative form and must include dx.
• ✅ 技巧：先积分再代入 —— 对于定积分的分部积分，先做不定积分得到原函数，再代入上下限，通常更不容易出错。
• ✅ Tip: Integrate first, then substitute — For definite integrals with Integration by Parts, find the antiderivative first, then apply limits — usually less error-prone.
• ✅ 技巧：写出完整的 u, dv, du, v 表格 —— 考试中，清晰标注这四个量可以帮助你避免代数错误，并得到方法分（Method Marks）。
• ✅ Tip: Write a full u, dv, du, v table — In exams, clearly labeling these four quantities helps avoid algebraic errors and secures Method Marks.

6. 练习题 | Practice Questions

尝试以下题目检验你的理解（答案见文末）：

Try these questions to test your understanding (answers at the end):

1. \( \displaystyle \int x \cos x \, dx \)
2. \( \displaystyle \int x^3 \ln x \, dx \)
3. \( \displaystyle \int e^{2x} \cos x \, dx \)（提示：循环积分）
4. \( \displaystyle \int_0^{\pi/2} x \sin x \, dx \)（定积分）
5. \( \displaystyle \int (\ln x)^2 \, dx \)（需两次分部积分）

7. 总结 | Summary

步骤 Step	中文	English
1	用 LIATE 选择 \( u \)	Choose \( u \) using LIATE
2	确定 \( dv \)（剩余部分 + dx）	Identify \( dv \) (remaining part + dx)
3	计算 \( du \)（对 \( u \) 求导）	Compute \( du \) (differentiate \( u \))
4	计算 \( v \)（对 \( dv \) 积分）	Compute \( v \) (integrate \( dv \))
5	代入公式 \( uv – \int v\,du \)	Substitute into \( uv – \int v\,du \)
6	计算剩余积分 + 加 C	Evaluate remaining integral + add C

分部积分法是 A-Level 数学工具箱中不可或缺的一部分。掌握 LIATE 法则、循环积分技巧和定积分形式后，你将能够自信应对任何考试题目。记住：多练习是王道！

Integration by Parts is an indispensable tool in your A-Level Mathematics toolkit. With LIATE, the cyclic technique, and the definite integral form under your belt, you’ll tackle any exam question with confidence. Remember: practice makes perfect!

答案 | Answers

1. \( x\sin x + \cos x + C \)
2. \( \frac{x^4}{4}\ln x – \frac{x^4}{16} + C \)
3. \( \frac{e^{2x}}{5}(2\cos x + \sin x) + C \)
4. \( 1 \)
5. \( x(\ln x)^2 – 2x\ln x + 2x + C \)