📚 Integration by Substitution: Complete A-Level Guide | 换元积分法：A-Level 完整指南
Integration by substitution is one of the most powerful techniques in Edexcel A-Level Pure Mathematics. It allows you to reverse the chain rule and integrate complicated expressions by changing the variable to make the integral much simpler.
换元积分法是 Edexcel A-Level 纯数学中最强大的技巧之一。它可以让你反过来使用链式法则，通过改变变量使被积函数更简单，从而对复杂表达式进行积分。
This guide covers the definition of the method, the reverse chain rule behind it, a reliable five-step routine, how to choose a substitution, three fully worked examples, definite integrals with changed limits, common examiner traps and a final revision checklist. Every idea is tied to the style of question that appears in Edexcel Paper 1 and Paper 2.
本指南涵盖换元积分法的定义、其背后的反向链式法则、可靠的五步流程、如何选择代换、三道完整例题、需要换限的定积分、考官常设的陷阱以及最后的复习清单。每个要点都紧扣 Edexcel Paper 1 与 Paper 2 的真题风格。
1. What is Integration by Substitution? | 什么是换元积分法？
In integration by substitution, you replace a difficult expression in x with a new variable u. The derivative du/dx is then used to rewrite dx in terms of du. This transforms the original integral into one in u that is easier to evaluate.
在换元积分法中，你把 x 中较难处理的部分替换成一个新变量 u。然后利用导数 du/dx 把 dx 改写成 du 的形式。这样原积分就转化为一个关于 u 的更容易计算的积分。
u, which comes from treating du/dx as a ratio of differentials. Every part of the original integrand - including dx itself - must be rewritten in terms of u before you integrate. Nothing in x may survive into the new integral.
核心关系是 dx = (dx/du) du，它来自把 du/dx 看作微分之比。原被积函数的每一部分，包括 dx 本身，都必须先改写为 u 的表达式才能开始积分。新的积分里不能再残留任何 x。
∫ f(x) dx = ∫ f(x(u)) × (dx/du) du
You have already used substitution when you integrated functions such as (ax + b)ⁿ by thinking of the inside function as u. At A-Level, the method is extended to more general expressions, including products, quotients, exponentials and trigonometric combinations.
你在积分 (ax + b)ⁿ 这类函数时已经用过代换思想，把内层函数看作 u。在 A-Level 阶段，这一方法会推广到更一般的表达式，包括乘积、商、指数函数和三角函数组合。
2. The Reverse Chain Rule | 反向链式法则
n rule read backwards. The chain rule states that if y = F(u) and u = g(x), then dy/dx = F'(u) × g'(x). Integrating both sides with respect to x undoes the differentiation.
换元并非随意的技巧，而是把链式法则倒过来读。链式法则指出，若 y = F(u) 且 u = g(x)，则 dy/dx = F'(u) × g'(x)。把两边对 x 积分，就抵消了求导这一步。
d/dx [F(g(x))] = F'(g(x)) × g'(x)
∫ F'(g(x)) × g'(x) dx = F(g(x)) + C
Setting u = g(x) so that du = g'(x) dx turns the left-hand side into ∫ F' h integrates immediately to F(u) + C. That is the whole mechanism. Recognising the pattern "function multiplied by the derivative of its inside" is the skill examiners are testing.
令 u = g(x)，于是 du = g'(x) dx，左边就变成 ∫ F'(u) du，可直接积出 F(u) + C。这就是全部机理。识别"函数乘以其内层函数的导数"这一模式，正是考官要考查的能力。
If the integrand contains the inside derivative only up to a constant multiple, the method still works: pull the constant outside the integral and adjust at the end.
如果被积函数里的内层导数只差一个常数倍，方法依然有效：把常数提到积分号外，最后再统一调整即可。
3. The Standard Five-Step Procedure | 标准五步流程
Most marks are lost through untidy working rather than genuinely hard mathematics. Following the same five steps every time keeps your solution clear and easy to mark.
多数失分源于书写不规范，而不是数学本身太难。每次都用同样的五步流程，能让解答清晰易判。
- Step 1 - Choose u. Pick the inside function, the awkward root, the denominator, the exponent or the trigonometric factor that pairs with its own derivative.
第 1 步 - 选择 u。 选取内层函数、难处理的根号、分母、指数部分，或与其导数配对出现的三角因子。 - Step 2 - Differentiate. rite du/dx explicitly, then rearrange to get dx on its own, for example dx = du / (du/dx).
第 2 步 - 求导。 明确写出 du/dx，再变形使 dx 单独在一边，例如 dx = du / (du/dx)。 - Step 3 - Substitute everything. Replace every x, including the one hidden inside dx, so the integral contains only u and du.
第 3 步 - 全面代换。 替换每一个 x，包括隐藏在 dx 里的那个，使积分中只剩 u 和 du。 - Step 4 - Integrate in u. This stage should be straightforward; if it is not, your choice of u was probably wrong.
第 4 步 - 对 u 积分。 这一步应当很简单；若不简单，多半是 u 选错了。 - Step 5 - Convert back and add C. Replace u with its expression in x, and never forget the constant of integration.
第 5 步 - 换回原变量并加 C。 把 u 换回 x 的表达式，并且千万不要漏掉积分常数。
For indefinite integrals, always finish in terms of x. For definite integrals, you may instead change the limits and never convert back at all.
对于不定积分，最后一定要回到 x 的表达式。对于定积分，你也可以改换上下限，从而完全不必换回 x。
4. Choosing the Right Substitution | 如何选择正确的代换
Edexcel often gives you the substitution in the question, for example "use the substitution u = x² + 1" does not, the following table covers almost every case that appears on the specification.
Edexcel 经常在题目中直接给出代换，例如"用代换 u = x² + 1"。若没有给出，下表几乎覆盖考纲中出现的所有情形。
| Pattern (English) | 模式（中文） | Substitution |
| Linear inside a power, (ax + b)ⁿ | 幂函数内是线性式，(ax + b)ⁿ | u = ax + b |
| Composite power, [g(x)]ⁿ | 复合幂函数，[g(x)]ⁿ | u = g(x) |
| Exponential e^(g(x)) times g'(x) | 指数 e^(g(x)) 乘以 g'(x) | u = g(x) |
| Fraction of the form g'(x) / g(x) | 形如 g'(x) / g(x) 的分式 | u = g(x), giving ln|g(x)| |
| ln x divided by x | ln x 除以 x | u = ln x |
| sinⁿx cos x or cosⁿx sin x | sinⁿx cos x 或 cosⁿx sin x | u = sin x or u = cos x |
| x√x + 1 type products | x√x + 1 类乘积 | u = x + 1, so x = u − 1 |
A useful rule of thumb: choose u so that its derivative already appears, or can easily be made to appear, elsewhere in the integrand. If different choices both work, prefer the one that removes the most awkward feature.
一个实用原则：让 u 的导数在被积函数中已经出现，或能轻松构造出来。若多种选择都可行，优先选择能消除最棘手部分的那个。
5. Worked Example 1: A Composite Power | 例题 1：复合幂函数
x² + 3 has derivative 2x, which is exactly the factor sitting outside, so this is a textbook reverse chain rule integral.
求 ∫ 2x(x² + 3)⁵ dx。括号内 x² + 3 的导数是 2x，恰好就是外面那个因子，因此这是标准的反向链式法则题型。
du = 2x dx. The entire front factor 2x dx is replaced by du, leaving a simple power of u.
令 u = x² + 3，则 du/dx = 2x，因此 du = 2x dx。前面的整个因子 2x dx 被 du 取代，只剩下 u 的简单幂函数。
∫ 2x(x² + 3)⁵ dx = ∫ u⁵ du = u⁶ / 6 + C
Substituting back gives the final answer (x² + 3)⁶ / 6 + C. Notice how little work was needed once the correct u was identified - the substitution did the hard part for you.
代回原变量，最终答案为 (x² + 3)⁶ / 6 + C。一旦选对 u，计算量其实很小——难题都由代换本身解决了。
[(x² + 3)⁶ / 6] = 6(x² + 3)⁵ × 2x / 6 = 2x(x² + 3)⁵, which matches the original integrand.
你随时可以用求导验证：d/dx [(x² + 3)⁶ / 6] = 6(x² + 3)⁵ × 2x / 6 = 2x(x² + 3)⁵，与原被积函数完全一致。
6. Worked Example 2: Trigonometric Substitution | 例题 2：三角代换
nd is a power of sin x multiplied by cos x, which is the derivative of sin x, so the pattern is identical to Example 1 with trigonometry instead of algebra.
求 ∫ sin³x cos x dx。被积函数是 sin x 的幂乘以 cos x，而 cos x 正是 sin x 的导数，因此题型与例题 1 完全相同，只是把代数换成了三角。
cos x and du = cos x dx. Then sin³x cos x dx becomes u³ du, which integrates in one line.
令 u = sin x，则 du/dx = cos x，du = cos x dx。于是 sin³x cos x dx 变为 u³ du，一行即可积出。
∫ sin³x cos x dx = ∫ u³ du = u⁴ / 4 + C = sin⁴x / 4 + C
dx you would set u = cos x, giving du = −sin x dx and hence a factor of −1 from the sign.
对称情形同理：对于 ∫ cos⁵x sin x dx，应设 u = cos x，得到 du = −sin x dx，因此会多出一个 −1 的符号因子。
∫ u⁵ du = −u⁶ / 6 + C. Sign slips here are one of the most frequent sources of lost marks.
务必小心这个负号。由 du = −sin x dx 得 sin x dx = −du，因此积分变成 −∫ u⁵ du = −u⁶ / 6 + C。这里的符号失误是最常见的失分点之一。
7. Worked Example 3: Exponential and Logarithmic Cases | 例题 3：指数与对数情形
Exponential integrands follow the same logic. For ∫ 2x ex² dx, the exponent x² has derivative 2x, which again matches the front factor.
指数型被积函数遵循同样的逻辑。对于 ∫ 2x ex² dx，指数上的 x² 求导得 2x，又与前面的因子吻合。
Let u = x², so du = 2x dx. Then the integral becomes ∫ eu du = eu + C, and converting back gives ex² + C.
令 u = x²，则 du = 2x dx。积分变为 ∫ eu du = eu + C，换回原变量得 ex² + C。
sts u = ln x, with du = (1/x) dx. The integral becomes ∫ u du = u² / 2 + C = ½(ln x)² + C.
对于 ∫ (ln x) / x dx，出现 ln x 及其导数 1/x，立刻提示取 u = ln x，du = (1/x) dx。积分变为 ∫ u du = u² / 2 + C = ½(ln x)² + C。
A closely related standard result, which you should recognise instantly, is ∫ g'(x) / g(x) dx = ln|g(x)| + C. For example ∫ (2x + 1) / (x² + x + 3) dx = ln|x² + x + 3| + C, because the numerator is exactly the derivative of the denominator.
一个密切相关、应当一眼认出的标准结果是 ∫ g'(x) / g(x) dx = ln|g(x)| + C。例如 ∫ (2x + 1) / (x² + x + 3) dx = ln|x² + x + 3| + C，因为分子恰是分母的导数。
8. Definite Integrals and Changing Limits | 定积分与换限
With a definite integral you have two legitimate routes. Either convert back to x and then apply the original x-limits, or change the limits into u-values before integrating and never convert back. The second route is usually faster and less error-prone.
处理定积分有两条合法路径：要么换回 x 再用原来的 x 上下限，要么在积分前就把上下限改为 u 的值，从而完全不必换回。第二条通常更快、更不易出错。
³ dx with u = x² + 1, so du = 2x dx and x dx = ½ du. When x = 0, u = 1; when x = 1, u = 2.
用 u = x² + 1 计算 ∫₀¹ x(x² + 1)³ dx，则 du = 2x dx，x dx = ½ du。当 x = 0 时 u = 1；当 x = 1 时 u
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