📚 Integration Techniques: Substitution & By Parts | 积分技巧：换元法与分部积分法
Understanding integration beyond basic anti‑derivatives is essential for A‑Level Mathematics. This article covers two powerful methods: integration by substitution (including the reverse chain rule) and integration by parts. You will learn when and how to apply each technique, work through detailed examples, and discover tips to avoid common exam mistakes.
掌握超越基本反导的积分方法对 A‑Level 数学至关重要。本文介绍两种强大的技巧：换元积分法（含反向链式法则）和分部积分法。你将学会何时以及如何应用每种方法，通过详细示例演练，并发现避免常见考试错误的窍门。
1. Why Advanced Integration Techniques Matter | 为什么高级积分技巧很重要
At A‑Level you will meet functions that cannot be integrated simply by reversing the standard differentiation rules. Expressions such as ∫ 2x√(x²+1) dx or ∫ x·e× dx require systematic methods. Edexcel exam papers regularly test your ability to choose the correct technique and execute it accurately under timed conditions.
在 A‑Level 阶段你会遇到不能仅靠逆向标准微分规则来积分的函数。像 ∫ 2x√(x²+1) dx 或 ∫ x·e× dx 这样的表达式需要系统方法。Edexcel 试卷习惯性地考查你是否能在限时条件下选择合适的技巧并准确执行。
2. The Reverse Chain Rule as a Starting Point | 逆向链式法则作为起点
Before formal substitution, it is helpful to recognise integrals of the form ∫ f' r instance, ∫ 2x·cos(x²) dx = sin(x²) + C because d/dx(x²) = 2x.
在正式引入换元法之前，先识别形如 ∫ f'(x) · g(f(x)) dx 的积分会有帮助。如果你发现内层函数的导数作为因子出现，积分就会变得简单。例如 ∫ 2x·cos(x²) dx = sin(x²) + C，因为 d/dx(x²) = 2x。
This approach mirrors the chain rule for differentiation in reverse. Edexcel often awards marks for stating the integral by inspection, but showing a clear substitution is safer when multiple steps are involved.
这种方法逆向反映了微分中的链式法则。Edexcel 经常对肉眼直接写出积分给予分数，但当涉及多步时，写出明确的换元过程更加稳妥。
3. Formal Integration by Substitution | 正式换元积分法
The method of substitution sets u = g(x), so that du = g'(x) dx. The integral ∫ f(g(x)) · g' ter integrating with respect to u, we substitute back to express the result in terms of x.
换元法设定 u = g(x)，因此 du = g'(x) dx。积分 ∫ f(g(x)) · g'(x) dx 变为 ∫ f(u) du，通常更容易计算。对 u 积分后，再回代将结果用 x 表示。
Example: Evaluate ∫ 6x(3x² + 5)&sup4; dx.
The integral becomes ∫ u&sup4; du = (1/5)u&sup5; + C = (1/5)(3x² + 5)&sup5; + C.
例：求 ∫ 6x(3x² + 5)&sup4; dx。
设 u = 3x² + 5 → du/dx = 6x → du = 6x dx。积分变为 ∫ u&sup4; du = (1/5)u&sup5; + C = (1/5)(3x² + 5)&sup5; + C。
4. Choosing the Right Substitution | 选择正确的代换变量
Selecting u is the crucial first step. Look for an inner function whose derivative appears elsewhere in the integrand. Common choices are polynomials inside brackets, denominators of fractions, or arguments of trigonometric, exponential, or logarithmic functions.
选择 u 是关键的第一步。寻找内层函数，且其导数在积分表达式中别处出现。常见的选择有括号内的多项式、分数的分母、或三角函数、指数函数、对数函数的变量部分。
times;−² dx, let u = 3x−2. Practice helps you recognise patterns quickly.
如果被积函数含有 √(ax+b)，设 u = ax+b 经常有效。对于 ∫ e³×−² dx，设 u = 3x−2。练习有助于你快速识别模式。
5. Definite Integrals with Substitution | 定积分的换元法
When evaluating a definite integral ∫ab f(g(x)) g' new limits become u = g(a) and u = g(b). There is no need to return to x after integration – simply evaluate the u‑integral between the new limits.
计算定积分 ∫ab f(g(x)) g'(x) dx 时，你可以在换元的同时更改积分限。若 u = g(x)，新限变为 u = g(a) 和 u = g(b)。积分后无需再回到 x —— 直接在新的 u 限之间求值即可。
For example, ∫01 = x²+3, du = 2x dx. Limits: when x=0, u=3; when x=1, u=4. The integral becomes ∫34 2√u du = [ (4/3)u3/2 ]34 = ...
例如，∫01 4x√(x²+3) dx：设 u = x²+3，du = 2x dx。积分限：当 x=0 时 u=3；当 x=1 时 u=4。积分变为 ∫34 2√u du = [ (4/3)u3/2 ]34 = ...
6. Introduction to Integration by Parts | 分部积分法简介
be differentiated easily and the other integrated easily. The formula is derived from d/dx(u v) = u'v + u v'.
分部积分法是微分的乘法法则的逆运算。它适用于被积函数为两个函数相乘的情况，其中一个易于微分，另一个易于积分。公式源自 d/dx(u v) = u'v + u v'。
The standard formula is: ∫ u dv = u v – ∫ v du. In the A‑Level context we usually write ∫ u (dv/dx) dx = u v – ∫ v (du/dx) dx.
标准公式为：∫ u dv = u v – ∫ v du。在 A‑Level 语境中，我们通常写作 ∫ u (dv/dx) dx = u v – ∫ v (du/dx) dx。
7. Step‑by‑Step Procedure for Integration by Parts | 分部积分法分步流程
1) Identify the two parts of the product: choose u (to differentiate) and dv/dx (to integrate).
2) Compute du/dx and v = ∫ (dv/dx) dx.
3) Apply the formula ∫ u (dv/dx) dx = u v – ∫ v (du/dx) dx.
4) Evaluate the new integral, which should be simpler.
1) 识别乘积中的两部分：选择 u（进行微分）和 dv/dx（进行积分）。
2) 计算 du/dx 以及 v = ∫ (dv/dx) dx。
3) 代入公式 ∫ u (dv/dx) dx = u v – ∫ v (du/dx) dx。
4) 计算新的积分，它应当更简单。
If the new integral is still complex, you may need a second application of integration by parts or a different choice of u and dv/dx.
若新积分仍较复杂，你可能需要再次使用分部积分法，或者调整 u 和 dv/dx 的选择。
8. Classic Example: Polynomial Times Exponential | 经典示例：多项式乘以指数函数
en differentiated) and dv/dx = e× (easy to integrate). Then du/dx = 1, v = e×.
计算 ∫ x e× dx。选择 u = x（微分后会简化），且 dv/dx = e×（容易积分）。则 du/dx = 1，v = e×。
Using the formula: ∫ x e× dx = x e× – ∫ e× · 1 dx = x e× – e× + C = e×(x – 1) + C.
应用公式：∫ x e× dx = x e× – ∫ e× · 1 dx = x e× – e× + C = e×(x – 1) + C。
Note how the polynomial degree reduces: we turned an integral with x into one with only the constant 1.
注意多项式的次数降低了：我们将含有 x 的积分变为仅含常数 1 的积分。
9. Integrating Natural Logarithm | 积分自然对数
re is an apparent product: ln x times 1. Set u = ln x, dv/dx = 1. Then du/dx = 1/x, v = x.
对于 ∫ ln x dx，存在一个隐式乘积：ln x 乘以 1。设 u = ln x，dv/dx = 1。则 du/dx = 1/x，v = x。
So ∫ ln x dx = x ln x – ∫ x · (1/x) dx = x ln x – ∫ 1 dx = x ln x – x + C.
所以 ∫ ln x dx = x ln x – ∫ x · (1/x) dx = x ln x – ∫ 1 dx = x ln x – x + C。
This trick (multiplying by 1) appears frequently in Edexcel questions where the integrand seems not to be a product at first glance.
这个技巧（乘以 1）经常出现在 Edexcel 考题中，被积函数乍看之下不像乘积。
10. Repeated Integration by Parts | 多次分部积分
= e³×. The resulting integral ∫ v du will involve x e³×. Apply parts again with u = x.
形如 ∫ x² e³× dx 的积分需要应用该方法两次，因为多项式次数为 2。首先设 u = x²;，dv/dx = e³×。得到的新积分 ∫ v du 会包含 x e³×。再次使用分部积分并设 u = x。
Complete working: ∫ x² e³× dx = (1/3)x² e³× – (2/3)∫ x e³× dx. Then ∫ x e³× dx = (1/3)x e³× – (1/9) e³×. Combine terms to get final answer.
完整过程：∫ x² e³× dx = (1/3)x² e³× – (2/3)∫ x e³× dx。然后 ∫ x e³× dx = (1/3)x e³× – (1/9) e³×。合并各项得到最终答案。
The LIATE rule (Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential) helps prioritise u: higher in the list means better to choose as u.
LIATE 规则（对数、反三角、代数、三角、指数）有助于为 u 排序：在列表中越靠前的函数越应选作 u。
11. Substitution vs Integration by Parts: How to Decide | 换元法与分部积分：如何抉择
Use substitution when the integrand contains a composite function and the derivative of the inner function appears. Use parts when you see a product of two distinct types of functions (e.g., x and sin x, or ln x and x³). Some integrals require both techniques in sequence.
当被积函数含有复合函数且内层函数的导数出现时，使用换元法。当看到两个不同类型函数的乘积（例如 x 和 sin x，或 ln x 和 x³）时，使用分部积分。某些积分需依次结合两种技巧。
If unsure, try to rewrite the integral in different forms. Edexcel mark schemes reward a clear statement of the chosen method and a correct first step even if the execution is incomplete.
如果不确定，尝试将积分改写成不同形式。即便未能完全解出，Edexcel 评分方案也会对你明确写出所选方法和正确的第一步给予分数。
12. Common Pitfalls and Exam Tips | 常见陷阱与考试贴士
ly: if u = 3x−5, then dx = du/3, not du. For definite integrals, change limits or re‑substitute before applying original limits – sign before the new integral; many candidates lose easy marks by writing u v + ∫ v du.
始终正确处理微分：若 u = 3x−5，则 dx = du/3，而不是 du。对于定积分，要么更改积分限，要么回代后再用原积分限 —— 绝不要在同一行中混用两种方法。在分部积分中，记住新积分前的负号；许多考生因误写 u v + ∫ v du 而痛失容易得分。
Practise questions from Edexcel past papers: look for integrals like ∫ sin&supmin;¹ x dx, ∫ x² ln x dx, and ∫ e²× sin 3x dx where parts must be repeated cyclically. Time yourself to build speed and confidence.
练习 Edexcel 历年真题：寻找诸如 ∫ sin&supmin;¹ x dx、∫ x² ln x dx 和 ∫ e²× sin 3x dx 等积分，其中分部积分需循环重复。计时练习以提高速度与信心。
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