📚 Integration by Substitution for Edexcel A-Level Maths | Edexcel A-Level数学换元积分法精讲
In Edexcel A-Level Mathematics, integration by substitution is one of the most powerful techniques for evaluating integrals that cannot be tackled by standard formulae alone. This method reverses the chain rule and is examined regularly in Pure Mathematics, especially when the integrand contains a composite function multiplied by some derivative of the inner function.
在Edexcel A-Level数学中，换元积分法是处理无法仅靠标准公式求解的积分的最强大技巧之一。该方法反向运用链式法则，在纯数学考试中经常出现，尤其是当被积函数包含复合函数乘以内层函数的某阶导数时。
This guide covers the core idea, how to choose a substitution, the standard four-step routine, worked examples, definite integrals with changed limits, trigonometric and exponential cases, and the mistakes that cost marks most often.
本文涵盖核心思想、如何选择替换变量、标准四步流程、典型例题、定积分上下限的变换、三角与指数型换元，以及最容易失分的常见错误。
1. The Core Idea | 核心思想
Integration by substitution is essentially the chain rule in reverse. If you know that the derivative of f(g(x)) is f'(g(x))g'(x), then integrating f'(g(x))g'(x) with respect to x recovers f(g(x)) plus a constant. Substitution lets you rename the inner function g(x) as u so that the integral becomes much simpler.
换元积分本质上是链式法则的逆运算。如果你知道 f(g(x)) 的导数为 f'(g(x))g'(x)，那么对 f'(g(x))g'(x) 关于 x 积分就会还原为 f(g(x)) 加常数。换元可以让你把内层函数 g(x) 重新命名为 u，使积分形式变得简单得多。
Formally, if u = g(x) is differentiable, then du/dx = g'(x), so du = g' ery x and every dx in the original integral must be replaced so that only u and du remain.
严格地说，若 u = g(x) 可导，则 du/dx = g'(x)，因此 du = g'(x) dx。原积分中的每一个 x 和每一个 dx 都必须被替换，最终只留下 u 与 du。
∫ f'(g(x)) g'(x) dx = ∫ f'(u) du = f(u) + C
2. Spotting When Substitution Is Needed | 识别何时需要换元
el paper, you will often see a clue that substitution is needed: the integrand contains an 'inner function' 5 is 2x, and we already have x in the integrand, so a substitution u = x² + 5 makes sense.
在Edexcel试卷中，你通常会看到需要换元积分的线索：被积函数包含一个"内层函数"，且其导数出现在旁边，只是导数可能相差一个常数因子。例如，在 ∫ x(x² + 5)⁶ dx 中，x² + 5 的导数是 2x，而被积函数中已经有 x，因此设 u = x² + 5 是合理的。
Typical signals include a bracket raised to a power, a square root of a linear or quadratic expression, a denominator of the form ax + b or x² + k, and expressions inside sin, cos, tan, e or ln.
典型信号包括：一个括号的幂次、线性或二次表达式的平方根、形如 ax + b 或 x² + k 的分母，以及 sin、cos、tan、e 或 ln 内部的表达式。
If the derivative of the inner function is nowhere to be seen, substitution is usually not the intended method; integration by parts or a trigonometric identity is more likely.
如果内层函数的导数完全看不到，那么通常并不是要用换元法，而更可能是分部积分或三角恒等变形。
3. Choosing the Substitution | 选择替换变量
inator, or inside a trigonometric or exponential function. You should also write du/dx = g'(x) and then express dx in terms of du.
通常选择 u = g(x) 作为复合函数的内层部分，例如括号内、根号内、分母中或三角函数与指数函数内部的表达式。你还需要写出 du/dx = g'(x)，然后将 dx 用 du 表示。
When the integrand contains x dx rather than 6x dx, you can multiply or divide by a constant to match the given integral. This step is essential before you replace every x and dx in the original problem.
当积分式子中含有 x dx 而不是 6x dx 时，你可以通过乘除一个常数来与被给的积分匹配。在替换原题中每一个 x 与 dx 之前，这一步是必不可少的。
n du/dx = 3, so dx = du/3, and the constant 1/3 is simply taken outside the integral sign.
例如，若 u = 3x + 1，则 du/dx = 3，因此 dx = du/3，常数 1/3 直接提到积分号外面即可。
4. The Four-Step Routine | 标准四步流程
ntiate to get du/dx. Step 2: rearrange to obtain dx (or x dx) in terms of du. Step 3: substitute everything so the integral is entirely in terms of u. Step 4: integrate with respect to u, then back-substitute x if the integral is indefinite.
第一步：写出 u = g(x) 并求导得到 du/dx。第二步：变形求出 dx（或 x dx）关于 du 的表达式。第三步：全部代入，使积分完全以 u 表示。第四步：对 u 积分，若为不定积分则再回代为 x。
A clean presentation matters in Edexcel exams: show du/dx explicitly, show dx, and never mix x and u in the same integrand. Mixing the two variables is a standard reason for losing accuracy marks.
在Edexcel考试中，书写规范非常重要：明确写出 du/dx、明确写出 dx，并且绝不要在同一次积分中混用 x 与 u。两个变量混用是丢失准确分的常见原因。
5. Handling Constant Factors | 处理常数因子
rite x dx = du/2, which introduces a factor of ½ that carries through to the final answer.
最常见的错误来源是漏掉常数。如果 du = 2x dx，但被积函数中只有 x dx，你就必须写 x dx = du/2，这会引入一个 ½ 因子并一直带到最终答案。
k ∫ f(u) du for any constant k. This keeps the working tidy and reduces arithmetic slips.
常数可以随意提到积分号外，因为对任意常数 k 都有 ∫ k·f(u) du = k ∫ f(u) du。这样可以让过程更整洁，减少算错的风险。
Always check at the end: differentiate your answer and see whether you recover the original integrand. A quick reverse check catches nearly every constant error.
最后一定要检验：把答案求导，看看能否还原出原来的被积函数。快速反向验证几乎能抓住所有常数错误。
6. Worked Example: ∫ x(x² + 5)⁶ dx | 例题精讲一
2x, so du = 2x dx and x dx = du/2. Substituting gives the integral of a simple power of u.
设 u = x² + 5，则 du/dx = 2x，所以 du = 2x dx，即 x dx = du/2。代入后得到一个关于 u 的简单幂函数积分。
∫ x(x² + 5)⁶ dx = ½ ∫ u⁶ du = u⁷ / 14 + C
final answer in terms of x. Notice that the constant ½ appears because the derivative 2x was only partly present in the integrand.
把 u = x² + 5 回代，即得以 x 表示的最终答案。注意常数 ½ 的出现，是因为导数 2x 在被积函数中只出现了一部分。
∫ x(x² + 5)⁶ dx = (x² + 5)⁷ / 14 + C
(x² + 5)⁷ / 14 gives 7(x² + 5)⁶ · 2x / 14 = x(x² + 5)⁶, which confirms the result exactly.
对 (x² + 5)⁷ / 14 求导得 7(x² + 5)⁶ · 2x / 14 = x(x² + 5)⁶，完全验证了结果。
7. Worked Example: Linear Substitutions | 例题精讲二：线性替换
= 3 and dx = du/3. The integral becomes one third of the integral of u⁵ with respect to u.
对于 ∫ (3x + 1)⁵ dx，取 u = 3x + 1，则 du/dx = 3，dx = du/3。原积分变成 u⁵ 关于 u 的积分再乘以三分之一。
∫ (3x + 1)⁵ dx = (1/3) · u⁶ / 6 + C = (3x + 1)⁶ / 18 + C
Another standard case is ∫ x√x² + 4 dx. With u = x² + 4, du = 2x dx, so the integral equals ½ ∫ u1/2 du = ⅓ (x² + 4)3/2 + C.
另一个标准情形是 ∫ x√x² + 4 dx。令 u = x² + 4，则 du = 2x dx，故积分等于 ½ ∫ u1/2 du = ⅓ (x² + 4)3/2 + C。
² + x + 3) dx, where the numerator is exactly the derivative of the denominator, giving ln|x² + x + 3| + C.
类似的类型是 ∫ (2x + 1)/(x² + x + 3) dx，其分子恰好是分母的导数，结果为 ln|x² + x + 3| + C。
8. Definite Integrals and Changing Limits | 定积分与上下限变换
y in u, or integrate in u, back-substitute to x, and then apply the original x limits. The first route is usually faster and safer.
对于定积分有两条可行路径：要么用 u = g(x) 把上下限换算成 u 的取值并全程用 u 计算；要么先用 u 积分，再回代为 x，最后代原来的 x 上下限。第一条路径通常更快也更稳妥。
If x runs from a to b, then u runs from g(a) to g(b). Never keep the x limits while the integrand is written in u; that single slip invalidates the whole answer.
若 x 从 a 变到 b，则 u 从 g(a) 变到 g(b)。当被积函数已经写成 u 的形式时，绝不要继续沿用 x 的上下限；这一处疏忽会让整个答案作废。
+ 1 gives limits u = 1 and u = 5, so the value is [u⁴/4] from 1 to 5 = (625 − 1)/4 = 156.
例如，∫₀² 2x(x² + 1)³ dx 令 u = x² + 1，上下限变为 u = 1 与 u = 5，故结果为 [u⁴/4] 由 1 到 5 = (625 − 1)/4 = 156。
9. Trigonometric Substitutions | 三角换元
ubstitution. For ∫ sin³x cos x dx, take u = sin x, so du = cos x dx, giving sin⁴x / 4 + C.
许多三角积分可以用简单换元解决。对于 ∫ sin³x cos x dx，取 u = sin x，则 du = cos x dx，结果为 sin⁴x / 4 + C。
The classic result ∫ tan x dx = ln|sec x| + C comes from u = cos x, since du = −sin x dx and the integral becomes −∫ du/u = −ln|cos x| + C.
经典结果 ∫ tan x dx = ln|sec x| + C 来自 u = cos x，因为 du = −sin x dx，积分变为 −∫ du/u = −ln|cos x| + C。
For radicals such as √a² − x² root into a cos θ, and dx becomes a cos θ dθ. Remember to convert back using a right-angled triangle or inverse trigonometric functions.
对于 √a² − x² 这类根式，换元 x = a sin θ 可把根号化为 a cos θ，同时 dx 变为 a cos θ dθ。记得借助直角三角形或反三角函数回代。
10. Exponential and Logarithmic Integrals | 指数与对数型换元
Integrals of the form ∫ g'(x) e^(g(x)) dx collapse immediately with u = g(x). For instance, ∫ 2x ex² dx = ex² + C.
形如 ∫ g'(x) e^(g(x)) dx 的积分，令 u = g(x) 后立即化简。例如 ∫ 2x ex² dx = ex² + C。
Similarly, ∫ 6x/(3x² + 1) dx uses u = 3x² + 1 with du = 6x dx, giving ln|3x² + 1| + C. The pattern is always a numerator equal to the derivative of the denominator.
同样地，∫ 6x/(3x² + 1) dx 令 u = 3x² + 1，du = 6x dx，得到 ln|3x² + 1| + C。规律始终是分子等于分母的导数。
When the numerator is a constant multiple of that derivative, adjust by a constant only, never by a variable; variable factors must be absorbed into the substitution itself.
当分子是该导数的一个常数倍时，只能调整常数，绝不能用变量去调整；含变量的因子必须被换元本身吸收。
11. Common Exam Mistakes | 常见失分点
Forgetting to replace dx, leaving a mixture of x and u in the integrand.
忘记替换 dx，导致积分中同时出现 x 与 u。
Losing the constant factor, such as writing x dx = du instead of du/2.
漏掉常数因子，例如把 x dx 写成 du 而不是 du/2。
Omitting + C for indefinite integrals, which costs a mark every time.
不定积分忘记加 + C，每次都会丢分。
Keeping the original x limits in a definite integral already written in u.
在已写成 u 的定积分中仍沿用原来的 x 上下限。
Back-substituting incorrectly, or forgetting to substitute back at all.
回代错误，或者完全忘记把 u 换回 x。
12. Exam Strategy and Mark Scheme Insight | 考试策略与评分要点
Edexcel mark schemes typically award a method mark for a sensible substitution and correct differentiation, and accuracy marks for the correct integral and the correct final form. Writing the substitution and du/dx clearly earns that first method mark even if later arithmetic slips.
Edexcel评分标准通常对合理的换元与正确的求导给出方法分，对正确的积分结果与最终形式给出准确分。即使后面算错，清晰写出换元式与 du/dx 也能拿到第一个方法分。
If a question says 'use the substitution u = ...' miss the specific method mark. Show the change of limits explicitly in definite integrals.
如果题目明确要求"使用换元 u = ..."，你就必须使用该换元；其他正确方法也可能得分，但可能拿不到指定的方法分。在定积分中要明确写出上下限的变化。
Finally, always finish with a simplified answer, include units or exact values where required, and use a quick differentiation check when time allows. Substitution is a routine skill, so consistent practice with past-paper questions is the fastest route to full marks.
最后，答案一定要化简，必要时给出精确值，时间允许时用求导进行快速验证。换元是一项程序性技能，坚持刷真题是通向满分的捷径。
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