Integration by Substitution for Edexcel A-Level Pure Maths | Edexcel A-Level 纯数学：换元积分法完全指南

📚 Integration by Substitution for Edexcel A-Level Pure Maths | Edexcel A-Level 纯数学：换元积分法完全指南

Integration by substitution is one of the most powerful tools in the Edexcel A-Level Pure Mathematics specification. It appears in both the Pure 1 and Pure 2 units and is essential for tackling a wide range of integrals that cannot be solved by direct reverse differentiation.

换元积分法是 Edexcel A-Level 纯数学大纲中最强大的工具之一。它出现在 Pure 1 和 Pure 2 单元中，对于解决大量无法通过直接逆微分求出的积分至关重要。

1. What Is Integration by Substitution? | 什么是换元积分法？

Integration by substitution is a technique used to rewrite a complicated integral in terms of a new variable, usually u. The goal is to transform the original integral into a simpler standard form that can be integrated directly.

换元积分法是一种将复杂积分用新变量（通常是 u）重新表示的方法。其目标是将原积分转化为可以直接积分的更简单标准形式。

When you let u = g(x), you also need to convert the differential dx into du by using the derivative du/dx = g'(x). This gives the key relation:

当你令 u = g(x) 时，还需要利用导数 du/dx = g'(x) 把微分 dx 转换为 du。从而得到关键关系：

du = g'(x) dx

This relation is the heart of the method. Without correctly replacing dx with du, the integral will not be valid.

这个关系是该方法的核心。如果没有正确地将 dx 替换为 du，积分将不成立。

2. The Reverse Chain Rule Foundation | 反向链式法则的基础

Integration by substitution is closely linked to the reverse chain rule. If you recognise an integrand of the form f'(g(x))g'(x), then the integral is simply f(g(x)) + C.

换元积分法与反向链式法则密切相关。如果你识别出被积函数形如 f'(g(x))g'(x)，那么它的积分就是 f(g(x)) + C。

For example, consider ∫ 2x(x² + 1)⁵ dx. Here the inner function is x² + 1, and its derivative 2x appears in front of the bracket. This pattern tells us that substitution will work perfectly.

例如，考虑 ∫ 2x(x² + 1)⁵ dx。这里内层函数是 x² + 1，其导数 2x 出现在括号前面。这一模式告诉我们换元法会非常有效。

By letting u = x² + 1, we get du = 2x dx, so the integral becomes ∫ u⁵ du = u⁶/6 + C. Substituting back gives (x² + 1)⁶/6 + C.

令 u = x² + 1，则 du = 2x dx，因此积分变为 ∫ u⁵ du = u⁶/6 + C。回代后得到 (x² + 1)⁶/6 + C。

3. Choosing the Substitution u | 选择换元变量 u

Choosing the right substitution is a skill that improves with practice. In Edexcel exams, the substitution is often given, but you still need to know how to apply it correctly.

选择合适的换元变量是一项需要练习的技能。在 Edexcel 考试中，换元变量通常会给出，但你仍需知道如何正确运用。

If the substitution is not given, look for an inner function whose derivative also appears in the integrand. Common choices include u = x² + a, u = ax + b, or u = cos x when sin x is present.

如果题目没有给出换元变量，可以寻找一个内层函数，并且其导数也出现在被积函数中。常见的选择包括 u = x² + a、u = ax + b，或当存在 sin x 时令 u = cos x。

Integrand feature	Good substitution
f(ax + b) with derivative a	u = ax + b
g(x² + a) with x outside	u = x² + a
g(sin x) with cos x outside	u = sin x
g(cos x) with sin x outside	u = cos x

Once you have chosen u, always write down du/dx and then express du in terms of dx before substituting.

选定 u 之后，一定要先写出 du/dx，再将 du 用 dx 表示出来，然后再进行替换。

4. Worked Example: Indefinite Integral | 例题：不定积分

Let us work through a complete indefinite integral using substitution. Find ∫ 3x²√(x³ + 1) dx.

让我们通过一个完整的不定积分例题来演示换元法。求 ∫ 3x²√(x³ + 1) dx。

Step 1: Let u = x³ + 1. Then du/dx = 3x², so du = 3x² dx. This matches the factor 3x² dx in the original integral exactly.

步骤 1：令 u = x³ + 1。则 du/dx = 3x²，因此 du = 3x² dx。这与原积分中的 3x² dx 完全匹配。

∫ 3x²√(x³ + 1) dx = ∫ √u du = ∫ u^(1/2) du

Step 2: Integrate with respect to u. ∫ u^(1/2) du = (2/3)u^(3/2) + C.

步骤 2：对 u 进行积分。∫ u^(1/2) du = (2/3)u^(3/2) + C。

Step 3: Substitute back u = x³ + 1 to get (2/3)(x³ + 1)^(3/2) + C. This is the final answer.

步骤 3：回代 u = x³ + 1，得到 (2/3)(x³ + 1)^(3/2) + C。这就是最终答案。

5. Worked Example: Definite Integral | 例题：定积分

For definite integrals, you must either change the limits to the new u-values or return to the original variable before evaluating. Changing limits is usually cleaner.

对于定积分，你必须将积分限转换为新的 u 值，或者在求值前回到原变量。通常转换积分限会更简洁。

Find ∫ from 0 to 1 of 2x√(x² + 1) dx. Let u = x² + 1, so du = 2x dx. When x = 0, u = 1; when x = 1, u = 2.

求 ∫ 从 0 到 1 的 2x√(x² + 1) dx。令 u = x² + 1，则 du = 2x dx。当 x = 0 时，u = 1；当 x = 1 时，u = 2。

∫ from 0 to 1 of 2x√(x² + 1) dx = ∫ from 1 to 2 of √u du

Now integrate: ∫ from 1 to 2 of u^(1/2) du = [ (2/3)u^(3/2) ] from 1 to 2.

现在进行积分：∫ 从 1 到 2 的 u^(1/2) du = [ (2/3)u^(3/2) ] 从 1 到 2。

Evaluating gives (2/3)(2^(3/2) – 1^(3/2)) = (2/3)(2√2 – 1).

求值得到 (2/3)(2^(3/2) – 1^(3/2)) = (2/3)(2√2 – 1)。

6. Substitution with Trigonometric Functions | 三角函数的换元

Trigonometric integrals often require substitution when the derivative of one trig function appears with another. A common example is ∫ sin x cos²x dx.

当某个三角函数的导数与另一个三角函数同时出现时，三角积分通常需要换元。一个常见例子是 ∫ sin x cos²x dx。

Let u = cos x. Then du/dx = -sin x, so du = -sin x dx. Rearranging gives -du = sin x dx.

令 u = cos x。则 du/dx = -sin x，因此 du = -sin x dx。移项得到 -du = sin x dx。

∫ sin x cos²x dx = -∫ u² du = -u³/3 + C

Substituting back u = cos x gives -cos³x/3 + C. Always check the sign carefully when the derivative is negative.

回代 u = cos x 得到 -cos³x/3 + C。当导数为负时，务必仔细检查符号。

7. Common Pitfalls and How to Avoid Them | 常见错误与避免方法

Many students lose marks on substitution questions because of small but critical errors. The most frequent mistake is forgetting to change the limits in a definite integral.

许多学生在换元题上失分，是因为一些细小但关键的错误。最常见的错误是在定积分中忘记改变积分限。

Another common error is writing du but still leaving dx in the integrand. Always replace the entire original differential with the new expression involving du.

另一个常见错误是写了 du 但被积函数中仍留有 dx。一定要用包含 du 的新表达式完整替换原来的微分。

Avoid these mistakes by following a clear routine: write down u, compute du/dx, express du = … dx, change the limits if necessary, and then rewrite the whole integral before integrating.

通过遵循清晰的步骤可以避免这些错误：写下 u，计算 du/dx，表示出 du = … dx，必要时改变积分限，然后在积分前重新写出整个积分。

8. Substitution in Differential Equations | 微分方程中的换元

Substitution also appears when solving separable first-order differential equations. For example, consider dy/dx = x/y.

换元法也出现在求解可分离变量的一阶微分方程中。例如，考虑 dy/dx = x/y。

Separate the variables to get y dy = x dx. Then integrate both sides: ∫ y dy = ∫ x dx.

分离变量得到 y dy = x dx。然后两边积分：∫ y dy = ∫ x dx。

y²/2 = x²/2 + C

This leads to the general solution y² = x² + 2C, where 2C is just an arbitrary constant. This method is closely linked to the idea of substitution when integrating each side.

这得到通解 y² = x² + 2C，其中 2C 只是任意常数。这种方法与两边积分时的换元思想密切相关。

9. Exam-Style Question Walkthrough | 真题演练

Let us attempt a typical Edexcel-style question. Find ∫ x/(x² + 1) dx.

让我们尝试一道典型的 Edexcel 风格题目。求 ∫ x/(x² + 1) dx。

Let u = x² + 1. Then du/dx = 2x, so du = 2x dx. Rearranging gives x dx = (1/2)du.

令 u = x² + 1。则 du/dx = 2x，因此 du = 2x dx。移项得到 x dx = (1/2)du。

∫ x/(x² + 1) dx = (1/2)∫ 1/u du = (1/2)ln|u| + C

Substituting back gives (1/2)ln|x² + 1| + C. Since x² + 1 is always positive, the absolute value bars can be omitted.

回代得到 (1/2)ln|x² + 1| + C。由于 x² + 1 恒为正，绝对值符号可以省略。

In an exam, you should present your substitution clearly, show the du conversion, and write the final answer in terms of the original variable.

在考试中，你应该清晰地展示换元过程，写出 du 的转换，并以原变量写出最终答案。

10. Quick Revision Summary | 快速复习总结

Integration by substitution is a systematic method for simplifying integrals. Always choose u as the inner function when possible, then convert du and dx correctly.

换元积分法是一种简化积分的系统方法。尽可能选择内层函数作为 u，然后正确转换 du 和 dx。

Write u = g(x) and compute du/dx.
Express du = g'(x) dx and replace all x-parts completely.
For definite integrals, change the limits to u-values.
Integrate with respect to u, then substitute back for indefinite integrals.
Check for negative signs and missing constants.

写下 u = g(x) 并计算 du/dx。表示出 du = g'(x) dx 并完整替换所有含 x 的部分。对于定积分，将积分限转换为 u 值。对 u 积分，然后对不定积分回代。检查负号和缺失的常数。

With consistent practice, substitution becomes a reliable way to gain marks on the Edexcel Pure Mathematics papers.

通过持续练习，换元法会成为你在 Edexcel 纯数学试卷中可靠拿分的方法。

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Integration by Substitution for Edexcel A-Level Pure Maths | Edexcel A-Level 纯数学：换元积分法完全指南