Advanced Integration Techniques: Parts and Substitution — A-Level数学进阶积分技巧：分部积分法与换元积分法详解

中文：在A-Level数学（Edexcel考纲）中，积分是纯数学部分最重要的技能之一。掌握了基本的幂函数积分和标准积分公式之后，学生需要进一步学习两种强大的积分技巧：分部积分法（Integration by Parts）和换元积分法（Integration by Substitution）。本文将从原理到应用，逐步解析这两种方法。

English: In A-Level Mathematics (Edexcel specification), integration is one of the most important skills in Pure Mathematics. After mastering basic power rule integration and standard integral formulas, students need to learn two powerful integration techniques: Integration by Parts and Integration by Substitution. This article will explain both methods step by step, from principles to applications.

一、分部积分法 / 1. Integration by Parts

1.1 基本原理 / Basic Principle

中文：分部积分法源自微分的乘法法则（Product Rule）。如果我们将乘法法则 d/dx(uv) = u·dv/dx + v·du/dx 两边同时积分并重新排列，就得到了分部积分公式：

English: Integration by Parts is derived from the Product Rule for differentiation. If we start with the product rule d/dx(uv) = u·dv/dx + v·du/dx and integrate both sides, rearranging yields the integration by parts formula:

中文：∫ u·(dv/dx) dx = uv − ∫ v·(du/dx) dx

English: ∫ u·(dv/dx) dx = uv − ∫ v·(du/dx) dx

中文：更简洁的表述是：∫ u dv = uv − ∫ v du。这个公式的核心思想是，将原本难以直接积分的被积函数，分解为 u 和 dv 两部分，使得右侧的积分 ∫ v du 比原积分更容易求解。

English: A more compact form is: ∫ u dv = uv − ∫ v du. The core idea is to decompose the original integrand (which is difficult to integrate directly) into two parts – u and dv – such that the integral on the right-hand side, ∫ v du, becomes easier to solve than the original.

1.2 选择 u 和 dv 的策略 — LIATE法则 / Strategy for Choosing u and dv — The LIATE Rule

中文：选择哪一部分作为 u 是分部积分法最关键的一步。错误的分配可能导致右侧积分比原积分更复杂。一般遵循 LIATE 优先级原则：

English: Choosing which part to assign as u is the most critical step in integration by parts. A poor choice can make the right-hand integral even more complex than the original. We generally follow the LIATE priority rule:

• L – Logarithmic（对数函数）：ln(x), log(x) 等 – 优先级最高 / highest priority
• I – Inverse trigonometric（反三角函数）：arcsin(x), arctan(x) 等
• A – Algebraic（代数函数）：x^n, 多项式等 / polynomials, etc.
• T – Trigonometric（三角函数）：sin(x), cos(x) 等
• E – Exponential（指数函数）：e^x, a^x 等 – 优先级最低 / lowest priority

中文：LIATE列表中越靠前的函数类型，越应该被选为 u。例如，∫ x e^x dx 中，x 是代数函数（A），e^x 是指数函数（E），因此选择 u = x, dv = e^x dx。

English: Functions appearing earlier in the LIATE list should be chosen as u. For example, in ∫ x e^x dx, x is algebraic (A) and e^x is exponential (E), so we choose u = x and dv = e^x dx.

1.3 典型例题 / Worked Examples

中文：例1 – 基础型：计算 ∫ x e^(2x) dx

English: Example 1 – Basic: Evaluate ∫ x e^(2x) dx

中文：根据 LIATE 规则，令 u = x, dv = e^(2x) dx。则 du = dx, v = (1/2)e^(2x)。代入公式：

English: Following the LIATE rule, let u = x, dv = e^(2x) dx. Then du = dx, v = (1/2)e^(2x). Substituting into the formula:

中文：∫ x e^(2x) dx = x · (1/2)e^(2x) − ∫ (1/2)e^(2x) dx = (1/2)xe^(2x) − (1/4)e^(2x) + C

English: ∫ x e^(2x) dx = x · (1/2)e^(2x) − ∫ (1/2)e^(2x) dx = (1/2)xe^(2x) − (1/4)e^(2x) + C

中文：例2 – 循环型（三角×指数）：计算 ∫ e^x sin x dx

English: Example 2 – Cyclic (Trig × Exponential): Evaluate ∫ e^x sin x dx

中文：令 u = sin x, dv = e^x dx。则 du = cos x dx, v = e^x。

English: Let u = sin x, dv = e^x dx. Then du = cos x dx, v = e^x.

中文：∫ e^x sin x dx = e^x sin x − ∫ e^x cos x dx

English: ∫ e^x sin x dx = e^x sin x − ∫ e^x cos x dx

中文：现在对 ∫ e^x cos x dx 再次使用分部积分：令 u = cos x, dv = e^x dx，得：

English: Now apply integration by parts again to ∫ e^x cos x dx: let u = cos x, dv = e^x dx, giving:

中文：∫ e^x cos x dx = e^x cos x + ∫ e^x sin x dx

English: ∫ e^x cos x dx = e^x cos x + ∫ e^x sin x dx

中文：代回原式：设 I = ∫ e^x sin x dx，则 I = e^x sin x − (e^x cos x + I)，即 2I = e^x(sin x − cos x)，最终：I = (1/2)e^x(sin x − cos x) + C

English: Substituting back: let I = ∫ e^x sin x dx, then I = e^x sin x − (e^x cos x + I), so 2I = e^x(sin x − cos x). Finally: I = (1/2)e^x(sin x − cos x) + C

1.4 特殊技巧：ln(x)的积分 / Special Technique: Integrating ln(x)

中文：ln x 看起来不像是两个函数的乘积，但我们可以写成 ln x × 1，然后令 u = ln x, dv = 1·dx。这样 du = (1/x)dx, v = x：

English: ln x doesn’t appear to be a product, but we can write it as ln x × 1, then let u = ln x, dv = 1·dx. This gives du = (1/x)dx, v = x:

中文：∫ ln x dx = x ln x − ∫ x · (1/x) dx = x ln x − x + C

English: ∫ ln x dx = x ln x − ∫ x · (1/x) dx = x ln x − x + C

二、换元积分法 / 2. Integration by Substitution

2.1 基本原理 / Basic Principle

中文：换元积分法本质上是链式法则（Chain Rule）的逆运算。当我们面对一个复合函数的积分时，通过引入一个新的变量来简化被积函数的结构。基本思想是：如果被积函数可以写成 f(g(x)) · g'(x) 的形式，那么令 u = g(x)，则积分变为 ∫ f(u) du。

English: Integration by Substitution is essentially the reverse of the Chain Rule. When faced with the integral of a composite function, we introduce a new variable to simplify the structure of the integrand. The basic idea is: if the integrand can be written in the form f(g(x)) · g'(x), then let u = g(x), and the integral becomes ∫ f(u) du.

2.2 标准换元步骤 / Standard Substitution Steps

中文：A-Level考试中，换元法通常有两种出题方式：题目直接给出替换式（如 “Use the substitution u = …”），或不给提示，要求你自己选择合适的替换。

English: In A-Level exams, substitution questions typically appear in two forms: the question directly provides the substitution (e.g., “Use the substitution u = …”), or no hint is given and you must select an appropriate substitution yourself.

中文：标准步骤：

English: Standard Procedure:

1. 选择替换 / Choose the substitution：令 u = 内部函数（通常是括号内的表达式、根号下的内容、或分母）/ Let u = the inner function (usually the expression inside brackets, under a square root, or in the denominator).
2. 求导并表达 dx / Differentiate and express dx：计算 du/dx，然后写成 dx = du/(du/dx) 的形式 / Compute du/dx, then write dx = du/(du/dx).
3. 替换所有 x / Replace all x：将被积函数中所有包含 x 的部分（包括 dx）替换为 u 的表达式 / Replace every part of the integrand involving x (including dx) with expressions in u.
4. 改变积分限（定积分）/ Change the limits (definite integrals)：如果是定积分，将 x 的上下限转换为 u 的对应值 / For definite integrals, convert the x-limits to the corresponding u-values.
5. 积分并回代（不定积分）/ Integrate and back-substitute (indefinite)：对 u 积分后，将 u 替换回 x 的表达式 / After integrating with respect to u, substitute back the expression for u in terms of x.

2.3 典型例题 / Worked Examples

中文：例3 – 直接换元：计算 ∫ x(x² + 3)⁵ dx

English: Example 3 – Direct Substitution: Evaluate ∫ x(x² + 3)⁵ dx

中文：令 u = x² + 3，则 du/dx = 2x，即 dx = du/(2x)。代入：

English: Let u = x² + 3, then du/dx = 2x, so dx = du/(2x). Substituting:

中文：∫ x(x² + 3)⁵ dx = ∫ x · u⁵ · du/(2x) = (1/2)∫ u⁵ du = (1/2) · (u⁶/6) + C = (x²+3)⁶/12 + C

English: ∫ x(x² + 3)⁵ dx = ∫ x · u⁵ · du/(2x) = (1/2)∫ u⁵ du = (1/2) · (u⁶/6) + C = (x²+3)⁶/12 + C

中文：例4 – 定积分换元：计算 ∫₀¹ (2x)/(√(x²+1)) dx

English: Example 4 – Definite Integral with Substitution: Evaluate ∫₀¹ (2x)/(√(x²+1)) dx

中文：令 u = x² + 1，则 du/dx = 2x，即 dx = du/(2x)。改变积分限：当 x=0 时 u=1；当 x=1 时 u=2。

English: Let u = x² + 1, then du/dx = 2x, so dx = du/(2x). Change limits: when x=0, u=1; when x=1, u=2.

中文：∫₀¹ (2x)/(√(x²+1)) dx = ∫₁² (2x)/(√u) · du/(2x) = ∫₁² u^(−1/2) du = [2u^(1/2)]₁² = 2√2 − 2

English: ∫₀¹ (2x)/(√(x²+1)) dx = ∫₁² (2x)/(√u) · du/(2x) = ∫₁² u^(−1/2) du = [2u^(1/2)]₁² = 2√2 − 2

2.4 三角换元 / Trigonometric Substitution

中文：当被积函数中包含 √(a² − x²)、√(a² + x²) 或 √(x² − a²) 这样的表达式时，三角换元是最有效的方法：

English: When the integrand contains expressions like √(a² − x²), √(a² + x²), or √(x² − a²), trigonometric substitution is the most effective approach:

• 对于 √(a² − x²)，令 x = a sin θ，利用 1 − sin²θ = cos²θ / For √(a² − x²), let x = a sin θ, using 1 − sin²θ = cos²θ
• 对于 √(a² + x²)，令 x = a tan θ，利用 1 + tan²θ = sec²θ / For √(a² + x²), let x = a tan θ, using 1 + tan²θ = sec²θ
• 对于 √(x² − a²)，令 x = a sec θ，利用 sec²θ − 1 = tan²θ / For √(x² − a²), let x = a sec θ, using sec²θ − 1 = tan²θ

三、两种方法的比较与选择 / 3. Comparing and Choosing Between the Two Methods

中文：选择使用哪种积分方法的关键在于识别被积函数的结构特征：

English: The key to choosing which integration method to use lies in recognizing the structural features of the integrand:

被积函数特征 / Integrand Feature	推荐方法 / Recommended Method
被积函数是两个不同类型函数的乘积（如 x·e^x, x·sin x）/ Product of two different function types	分部积分法 / Integration by Parts
被积函数中包含一个函数和其导数的乘积（如 f'(x)·[f(x)]^n）/ Contains a function multiplied by its derivative	换元积分法 / Integration by Substitution
被积函数中有明显的”内层函数”（复合函数）/ Clear “inner function” (composite function)	换元积分法 / Integration by Substitution
被积函数中包含 ln(x) 或反三角函数 / Contains ln(x) or inverse trig functions	分部积分法（u = ln(x) 等）/ Integration by Parts
被积函数中包含 √(a²±x²) 或 √(x²−a²) / Contains √(a²±x²) or √(x²−a²)	三角换元 / Trigonometric Substitution

四、考试常见陷阱与提示 / 4. Common Exam Pitfalls and Tips

• 忘记 +C / Don’t forget +C：不定积分必须加积分常数 C，这是 A-Level 考试中最常见的扣分点。/ Indefinite integrals must include the constant of integration C – this is the most common mark-losing point in A-Level exams.
• 定积分换元忘记改限 / Forgetting to change limits：使用换元法计算定积分时，必须将积分上下限也转换为新变量的值，否则答案会出错。/ When using substitution for definite integrals, you must convert the limits to the new variable’s values, or the answer will be incorrect.
• 分部积分中 u 的选择 / Choosing u in integration by parts：严格遵循 LIATE 规则。特别注意被积函数中如果只有 ln(x)，记得视为 ln(x)×1。/ Strictly follow the LIATE rule. Pay special attention to cases where the integrand is just ln(x) – remember to treat it as ln(x)×1.
• 循环型积分 / Cyclic integrals：遇到三角函数与指数函数的乘积时，需要分部积分两次，然后解方程。一定要小心代数运算中的正负号。/ When you encounter products of trig and exponential functions, you need to integrate by parts twice and then solve an equation. Be very careful with signs during algebraic manipulation.
• dx 的替换 / Replacing dx：换元时不要忘记将 dx 也替换为 du 的表达式。常见错误是只替换了被积函数中的 x，漏掉了 dx。/ When substituting, don’t forget to replace dx with an expression in du. A common mistake is only replacing x in the integrand while forgetting dx.

六、进阶综合应用 / 6. Advanced Combined Applications

6.1 分部积分与换元结合 / Combining Parts and Substitution

中文：在实际考试中，最具挑战性的题目往往需要将分部积分法和换元积分法结合使用。典型的情况是：先通过换元法简化被积函数的结构，然后对简化后的表达式使用分部积分法。

English: In actual exams, the most challenging questions often require combining Integration by Parts with Integration by Substitution. The typical scenario: first use substitution to simplify the integrand structure, then apply integration by parts to the simplified expression.

中文：例5 – 先换元后分部：计算 ∫ x³ e^(x²) dx

English: Example 5 – Substitution then Parts: Evaluate ∫ x³ e^(x²) dx

中文：第一步：观察到 x³ = x · x²，而 e^(x²) 是 x² 的复合函数。令 u = x²，则 du/dx = 2x，即 x dx = du/2。重写积分：∫ x³ e^(x²) dx = ∫ x² · x · e^(x²) dx = ∫ u · e^u · (du/2) = (1/2)∫ u e^u du

English: Step 1: Observe that x³ = x · x², and e^(x²) is a composite function of x². Let u = x², then du/dx = 2x, so x dx = du/2. Rewrite the integral: ∫ x³ e^(x²) dx = ∫ x² · x · e^(x²) dx = ∫ u · e^u · (du/2) = (1/2)∫ u e^u du

中文：第二步：对 (1/2)∫ u e^u du 使用分部积分法。令 U = u, dV = e^u du，则 dU = du, V = e^u：(1/2)∫ u e^u du = (1/2)(u e^u – ∫ e^u du) = (1/2)(u e^u – e^u) + C

English: Step 2: Apply integration by parts to (1/2)∫ u e^u du. Let U = u, dV = e^u du, then dU = du, V = e^u: (1/2)∫ u e^u du = (1/2)(u e^u – ∫ e^u du) = (1/2)(u e^u – e^u) + C

中文：第三步：回代 u = x²，得到最终答案：(1/2)(x² e^(x²) – e^(x²)) + C = (1/2)e^(x²)(x² – 1) + C

English: Step 3: Substitute back u = x² to get the final answer: (1/2)(x² e^(x²) – e^(x²)) + C = (1/2)e^(x²)(x² – 1) + C

6.2 分部积分与有理函数 / Integration by Parts with Rational Functions

中文：例6 – 分部积分结合有理函数：计算 ∫ ln(x+1) / x² dx

English: Example 6 – Parts with Rational Functions: Evaluate ∫ ln(x+1) / x² dx

中文：令 u = ln(x+1), dv = (1/x²)dx = x^(-2) dx。则 du = (1/(x+1))dx, v = -1/x。代入分部积分公式：∫ ln(x+1)/x² dx = -ln(x+1)/x – ∫ (-1/x)(1/(x+1))dx = -ln(x+1)/x + ∫ 1/(x(x+1)) dx

English: Let u = ln(x+1), dv = (1/x²)dx = x^(-2) dx. Then du = (1/(x+1))dx, v = -1/x. Substituting: ∫ ln(x+1)/x² dx = -ln(x+1)/x – ∫ (-1/x)(1/(x+1))dx = -ln(x+1)/x + ∫ 1/(x(x+1)) dx

中文：现在使用部分分式分解 1/(x(x+1)) = 1/x – 1/(x+1)。因此 ∫ 1/(x(x+1))dx = ∫ (1/x – 1/(x+1))dx = ln|x| – ln|x+1| + C。最终答案：-ln(x+1)/x + ln|x| – ln|x+1| + C

English: Now use partial fraction decomposition: 1/(x(x+1)) = 1/x – 1/(x+1). So ∫ 1/(x(x+1))dx = ∫ (1/x – 1/(x+1))dx = ln|x| – ln|x+1| + C. Final answer: -ln(x+1)/x + ln|x| – ln|x+1| + C

七、Edexcel历年真题分析 / 7. Edexcel Past Paper Analysis

中文：研究 Edexcel A-Level 纯数学历年真题可以发现，积分题目呈现出以下几种高频考查模式：

English: Analysing Edexcel A-Level Pure Mathematics past papers reveals several high-frequency question patterns for integration:

中文：模式一 – 分部积分 + 解微分方程：考试题常要求学生先用分部积分求出某个积分的表达式，然后将结果代入微分方程求解。这是 Pure Mathematics Paper 2 中的常见大题，通常占 8-12 分。

English: Pattern 1 – Integration by Parts + Solving Differential Equations: Exam questions often require students to first evaluate an integral using integration by parts, then substitute the result into a differential equation. This is a common long question in Pure Mathematics Paper 2, typically worth 8-12 marks.

中文：模式二 – 换元 + 定积分求面积/体积：涉及旋转体体积或曲线下面积的计算时，通常需要先通过换元法求出不定积分，然后将积分结果代入定积分限。这类题目考查学生的综合能力，分值通常在 10-15 分。

English: Pattern 2 – Substitution + Definite Integrals for Area/Volume: When calculating volumes of revolution or areas under curves, you typically need to first find the indefinite integral via substitution, then evaluate with the definite limits. These questions test comprehensive skills and are usually worth 10-15 marks.

中文：模式三 – 参数积分（Parametric Integration）：当曲线以参数方程 x = f(t), y = g(t) 给出时，面积公式变为 ∫ y dx = ∫ g(t) f'(t) dt。这类题目同时考查了参数微分和积分的结合能力。

English: Pattern 3 – Parametric Integration: When a curve is given in parametric form x = f(t), y = g(t), the area formula becomes ∫ y dx = ∫ g(t) f'(t) dt. These questions test the combined ability to handle parametric differentiation and integration.

中文：模式四 – 分部积分法的多次使用（Reduction Formula）：某些积分如 ∫ x^n e^x dx 或 ∫ sin^n x dx 需要反复使用分部积分，最终得到一个递推公式（Reduction Formula）。这是 A-Level 进阶内容，在 Further Mathematics 中更常见。

English: Pattern 4 – Repeated Parts (Reduction Formula): Certain integrals like ∫ x^n e^x dx or ∫ sin^n x dx require repeated application of integration by parts, ultimately yielding a reduction formula. This is advanced A-Level content, more commonly seen in Further Mathematics.

八、练习建议与学习路径 / 8. Practice Recommendations and Study Pathway

中文：为了真正掌握这两种积分方法，建议按照以下学习路径逐步推进：

English: To truly master these two integration methods, we recommend following this progressive study pathway:

中文：第一阶段 – 基本模式识别（1-2周）：每天练习 5 道基础积分题，重点训练”看到题目就能判断应该用哪种方法”的能力。不需要追求速度，关键是建立准确的方法选择直觉。推荐使用 Edexcel 教科书每个章节末尾的 Mixed Exercise。

English: Stage 1 – Basic Pattern Recognition (1-2 weeks): Practise 5 basic integration problems daily, focusing on the ability to “see a problem and immediately judge which method to use.” Speed is not the priority – building accurate method-selection intuition is key. Use the Mixed Exercises at the end of each chapter in the Edexcel textbook.

中文：第二阶段 – 综合应用（2-4周）：选择包含多种积分方法的混合练习题，训练在复杂问题中灵活切换方法的能力。特别注意”先换元后分部”和”先分部后换元”两种组合方式的区别。

English: Stage 2 – Combined Application (2-4 weeks): Choose mixed practice sets that require multiple integration methods, training the ability to flexibly switch between methods in complex problems. Pay special attention to the difference between “substitution then parts” and “parts then substitution.”

中文：第三阶段 – 真题模拟（4-6周）：限时完成 Edexcel 历年真题中的积分题目。从 2018-2023 年的 Pure Mathematics Papers 1 和 2 中选择所有积分相关题目，每题控制在 5-10 分钟内完成。完成后对照评分标准（Mark Scheme），学习如何写出满分解答过程。

English: Stage 3 – Past Paper Simulation (4-6 weeks): Complete integration questions from Edexcel past papers under timed conditions. Select all integration-related questions from the 2018-2023 Pure Mathematics Papers 1 and 2, aiming to finish each within 5-10 minutes. After completing, check against the mark scheme to learn how to write full-mark solutions.

中文：第四阶段 – 错题整理与反思（持续进行）：建立一个错题本（Error Log），记录每道做错的积分题，标注错误类型（方法选择错误、代数运算错误、符号错误、忘记+C等）。定期回顾，确保同类错误不再重复出现。

English: Stage 4 – Error Logging and Reflection (ongoing): Create an error log to record every incorrect integration problem, noting the error type (wrong method choice, algebraic error, sign error, forgot +C, etc.). Review regularly to ensure similar errors don’t recur.

中文：Integration（积分）是 A-Level 纯数学中最能区分学生水平的知识板块。掌握了分部积分法和换元积分法，你就具备了应对 Edexcel 考试中 90% 以上积分题目的能力。保持稳定的练习节奏，关注细节（尤其是 +C 和绝对值符号），你一定可以在考试中取得理想的成绩。

English: Integration is the topic area in A-Level Pure Mathematics that most effectively distinguishes student ability. By mastering Integration by Parts and Integration by Substitution, you will be equipped to handle over 90% of integration questions in the Edexcel exam. Maintain a steady practice rhythm, pay attention to details (especially +C and absolute value signs), and you will certainly achieve the results you want in the exam.

五、总结 / 5. Summary

中文：分部积分法和换元积分法是 A-Level 数学中两种互补的积分工具。分部积分法擅长处理不同类型函数的乘积，而换元积分法擅长处理复合函数。掌握这两种方法的关键在于大量练习 – 先识别被积函数的结构特征，再选择合适的积分策略。建议每周至少练习 5-10 道积分题，逐步建立对积分方法的”直觉”。Edexcel 的历年真题（Past Papers）中有大量高质量的积分题目，是备考的最佳材料。

English: Integration by Parts and Integration by Substitution are two complementary integration tools in A-Level Mathematics. Integration by Parts excels at handling products of different function types, while Integration by Substitution excels at handling composite functions. The key to mastering both methods lies in extensive practice – first identify the structural features of the integrand, then select the appropriate integration strategy. We recommend practising at least 5-10 integration problems per week to gradually develop “intuition” for integration methods. Edexcel past papers contain a wealth of high-quality integration questions and are the best resource for exam preparation.

中文：如果你在练习中遇到困难，可以先从识别”这是否是一个乘积？”和”被积函数中是否有明显的复合结构？”这两个问题入手。90% 以上的 A-Level 积分题目都可以通过这两个问题的答案找到正确的解法。

English: If you encounter difficulties while practising, start by asking two questions: “Is this a product?” and “Is there an obvious composite structure in the integrand?” Over 90% of A-Level integration problems can be solved correctly by answering these two questions.