A-Level Maths: The Complete Guide to Integration by Substitution | A-Level 数学：换元积分法全解析

📚 A-Level Maths: The Complete Guide to Integration by Substitution | A-Level 数学：换元积分法全解析

Integration by substitution is one of the most powerful techniques in A-Level calculus. It allows us to simplify complicated integrals by changing the variable, effectively reversing the chain rule. This guide provides a complete breakdown of the method, including when to use it, how to execute it step by step, and common pitfalls to avoid.

换元积分法是 A-Level 微积分中最强大的技巧之一。它通过改变变量来简化复杂的积分，本质上是链式法则的逆运算。本指南将全面解析这一方法，包括何时使用、如何逐步操作，以及需要避免的常见错误。

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

Integration by substitution is a method used to evaluate integrals by introducing a new variable, usually denoted as u or t . The goal is to transform a difficult integral into a simpler one that we can integrate directly. The method is based on the chain rule for differentiation.

换元积分法是通过引入一个新变量（通常记为 u 或 t ）来求解积分的方法。其目标是将一个困难的积分转化为可以直接求解的简单积分。该方法的理论基础是微积分中的链式法则。

If we have an integral of the form ∫ f(g(x)) · g'(x) dx, we can set u = g(x) , then du = g'(x) dx , and the integral becomes ∫ f(u) du.

如果我们有形式为 ∫ f(g(x)) · g'(x) dx 的积分，可以设 u = g(x) ，则 du = g'(x) dx ，原积分变为 ∫ f(u) du。

This is essentially the reverse of the chain rule: when you differentiate a composite function, you multiply by the derivative of the inner function; when you integrate, you divide by it (by substituting).

这本质上就是链式法则的逆运算：对复合函数求导时，需要乘以内层函数的导数；而积分时，通过换元来“抵消”这个导数。

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2. The Core Formula | 核心公式

Let u = g(x) be a differentiable function. Then for any continuous function f , we have:

设 u = g(x) 是一个可导函数。那么对于任意连续函数 f ，我们有：

∫ f(g(x)) g'(x) dx = ∫ f(u) du

In practice, we work with differentials: if u = g(x) , then dudx = g'(x) , which can be rearranged to du = g'(x) dx . This is not a rigorous algebraic manipulation but is accepted as a convenient notational device in A-Level Maths.

在实际操作中，我们使用微分形式：若 u = g(x) ，则 dudx = g'(x) ，可以改写为 du = g'(x) dx 。虽然这不是严格的代数运算，但在 A-Level 数学中被视为一种便捷的记号处理方式。

For definite integrals, we must also change the limits of integration. If the original limits are x = a and x = b , then the new limits are u(a) and u(b) .

对于定积分，我们还需要更换积分上下限。如果原来的限是 x = a 和 x = b ，那么新限为 u(a) 和 u(b) 。

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3. Step-by-Step Procedure | 分步操作流程

The method can be broken down into clear steps:

该方法可以分解为清晰的步骤：

• Step 1: Identify a suitable substitution u = g(x) that will simplify the integrand. Look for an inner function whose derivative is also present (up to a constant factor).
• Step 2: Compute dudx , then write dx = dug'(x) .
• Step 3: Substitute u and dx into the integral. The variable x should completely disappear.
• Step 4: Integrate with respect to u .
• Step 5: If the original variable was x , substitute back u = g(x) to express the answer in terms of x . If it is a definite integral, change the limits and then evaluate.

• 第一步：选择一个合适的代换 u = g(x) ，使被积函数简化。寻找内层函数，且其导数（差一个常数倍数）也出现在被积函数中。
• 第二步：计算 dudx ，然后写出 dx = dug'(x) 。
• 第三步：将 u 和 dx 代入积分，变量 x 应完全消失。
• 第四步：对 u 进行积分。
• 第五步：如果原变量是 x ，将 u = g(x) 代回，用 x 表达结果。如果是定积分，则更换上下限后直接计算。

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4. Example 1: Basic Linear Substitution | 例 1：基本线性代换

Evaluate ∫ (2x+1)⁵ dx.

求 ∫ (2x+1)⁵ dx。

Here the obvious substitution is u = 2x + 1 . Then dudx = 2 , so dx = du2 .

这里显然设 u = 2x + 1 。则 dudx = 2 ，所以 dx = du2 。

Substituting gives:

代入得：

∫ (2x+1)⁵ dx = ∫ u⁵ · 12 du = 12 · ^u6⁄₆ + C = ^u6⁄₁₂ + C

Finally, replace u with 2x+1 :

最后，将 u 代回为 2x+1 ：

∫ (2x+1)⁵ dx = ^(2x+1)6⁄₁₂ + C

This is a classic example of a linear substitution, where the derivative of u is a constant.

这是一个典型的线性代换，其中 u 的导数是常数。

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5. Example 2: Recognising the Derivative Pattern | 例 2：识别导数模式

Evaluate ∫ 2x √x² + 1 dx.

求 ∫ 2x √x² + 1 dx。

Notice that the derivative of x² + 1 is 2x , which appears as a factor. Let u = x² + 1 . Then dudx = 2x , so du = 2x dx .

注意 x² + 1 的导数是 2x ，而它正好是积分中的因子。设 u = x² + 1 ，则 dudx = 2x ，所以 du = 2x dx 。

The integral becomes:

积分变为：

∫ √u du = ∫ u^1/2 du = 23 u^3/2 + C

Substituting back:

代回原变量：

∫ 2x √x² + 1 dx = 23 (x²+1)^3/2 + C

This pattern is extremely common: look for a function and its derivative inside the integrand.

这种模式非常常见：在被积函数中寻找一个函数及其导数。

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6. Definite Integrals: Changing the Limits | 定积分：更换上下限

When evaluating a definite integral using substitution, you do not need to substitute back; instead, change the limits of integration.

在使用换元法求定积分时，不需要代回原变量，而是更换积分的上下限。

Example: Evaluate ∫₀¹ 2x √x² + 1 dx.

例：求 ∫₀¹ 2x √x² + 1 dx。

Let u = x² + 1 . Then du = 2x dx . When x = 0 , u = 1 ; when x = 1 , u = 2 . The integral becomes:

设 u = x² + 1 ，则 du = 2x dx 。当 x = 0 时， u = 1 ；当 x = 1 时， u = 2 。积分变为：

∫₁² √u du = [ 23 u^3/2 ]₁² = 23 (2^3/2 - 1^3/2) = 23 (2√2 - 1)

It is a common mistake to forget to change the limits. Always compute the new limits before integrating a definite integral.

忘记更换上下限是常见的错误。在计算定积分时，一定要先算出新的上下限。

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7. Substitutions Involving Trigonometric Functions | 涉及三角函数的代换

Trigonometric identities can be used as substitutions, especially when dealing with expressions like √a² - x² or √a² + x².

三角函数恒等式可用于代换，尤其是处理形如 √a² - x² 或 √a² + x² 的表达式时。

Example: Evaluate ∫ ¹⁄_{√1 - x²} dx.

例：求 ∫ ¹⁄_{√1 - x²} dx。

Let x = sin u , so dx = cos u du . Then √1 - x² = √1 - sin² u = √cos² u = cos u (for u in the range where cos u ≥ 0).

设 x = sin u ，则 dx = cos u du 。于是 √1 - x² = √1 - sin² u = √cos² u = cos u（在 cos u ≥ 0 的范围内）。

The integral becomes:

积分变为：

∫ cos ucos u du = ∫ 1 du = u + C = arcsin x + C

This is how the standard result ∫ ¹⁄_{√1 - x²} dx = arcsin x + C is derived.

这就是标准结果 ∫ ¹⁄_{√1 - x²} dx = arcsin x + C 的推导过程。

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8. Handling Constants: Adjustment Factor | 处理常数：调整因子

Sometimes the derivative of u differs from the remaining factor by a constant. You can compensate by inserting a constant factor outside the integral.

有时 u 的导数与剩余因子相差一个常数倍数。你可以在积分号外插入一个常数因子来补偿。

Example: Evaluate ∫ x √x² + 3 dx.

例：求 ∫ x √x² + 3 dx。

Let u = x² + 3 . Then dudx = 2x , so x dx = 12 du .

设 u = x² + 3 ，则 dudx = 2x ，所以 x dx = 12 du 。

The integral becomes:

积分变为：

∫ x √x² + 3 dx = ∫ 12 √u du = 12 · 23 u^3/2 + C = 13 (x²+3)^3/2 + C

Notice that we wrote x dx = 12 du instead of dx = du2x . This is often more efficient because it directly replaces the whole x dx group.

注意我们写了 x dx = 12 du 而不是 dx = du2x 。这通常更高效，因为它直接替换了整个 x dx 组。

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9. Which Substitution to Choose? | 如何选择合适的代换？

Choosing the right substitution is a skill. The general rules are:

选择合适的代换是一项技能。一般规则如下：

• Linear inner function: For (ax+b)ⁿ, use u = ax+b .
• Function and its derivative: If you see f(g(x)) · g'(x) , use u = g(x) .
• Radicals with √a² - x² : Use x = a sin u .
• Radicals with √a² + x² : Use x = a tan u .
• Rational functions with 1x : Use u = ln x when appropriate.

• 线性内层函数：对于 (ax+b)ⁿ，使用 u = ax+b 。
• 函数及其导数：如果看到 f(g(x)) · g'(x) ，使用 u = g(x) 。
• 含有 √a² - x² 的根式：使用 x = a sin u 。
• 含有 √a² + x² 的根式：使用 x = a tan u 。
• 含有 1x 的分式：适当情况下使用 u = ln x 。

Practice is essential. With time, you will recognise patterns more quickly.

练习至关重要。随着时间推移，你会更快地识别出模式。

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10. Common Mistakes and How to Avoid Them | 常见错误及避免方法

Here are the most frequent errors students make with integration by substitution:

以下是学生在换元积分法中最常犯的错误：

• Forgetting to change the limits in definite integrals. Always evaluate the new limits before integrating.
• Not substituting back in indefinite integrals. Always return to the original variable.
• Incorrectly computing dx . Be careful with the algebra when rearranging dudx .
• Choosing an unhelpful substitution. If the substitution makes the integral more complicated, try a different one.
• Missing the constant factor. When du = k dx , don't forget to insert 1k outside the integral.

• 忘记更换上下限：在定积分中，积分之前一定要计算新的上下限。
• 不定积分忘记代回原变量：最终答案必须用原变量表示。
• 错误计算 dx 。在重新排列 dudx 时，注意代数运算。
• 选择了无用的代换。如果代换使积分更复杂，尝试其他代换。
• 遗漏常数因子。当 du = k dx 时，不要忘记在积分号外插入 1k 。

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11. Advanced Example: Using Substitution Twice | 进阶示例：两次换元

Some integrals require more than one substitution or combine with other techniques.

有些积分需要多次换元，或与其他技巧结合使用。

Example: Evaluate ∫ x √x+1 dx.

例：求 ∫ x √x+1 dx。

Let u = x+1 . Then x = u - 1 and dx = du . Substituting:

设 u = x+1 ，则 x = u - 1 ， dx = du 。代入：

∫ (u-1) √u du = ∫ (u^3/2 - u^1/2) du = 25 u^5/2 - 23 u^3/2 + C

Finally, replace u with x+1 :

最后将 u 代回为 x+1 ：

25 (x+1)^5/2 - 23 (x+1)^3/2 + C

This demonstrates that sometimes you need to express x explicitly in terms of u , not just replace dx .

这说明有时你需要用 u 显式表达 x ，而不仅仅替换 dx 。

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12. Summary and Exam Tips | 总结与考试技巧

Integration by substitution is a core technique in A-Level Mathematics. To master it:

换元积分法是 A-Level 数学的核心技巧。要掌握它：

• Always check whether the integrand contains a function and its derivative.
• Write down the substitution clearly and compute dx or the appropriate differential group.
• For definite integrals, change the limits immediately.
• Simplify the integral if possible before integrating.
• Never forget the constant of integration for indefinite integrals.

• 始终检查被积函数是否包含函数及其导数。
• 清晰地写出代换，并计算 dx 或相应的微分组。
• 对于定积分，立即更换上下限。
• 积分前尽量化简被积函数。
• 不定积分不要忘记积分常数。

In the exam, show all steps. A clear substitution can earn method marks even if you make a small arithmetic error. Practice past paper questions to build confidence and speed.

在考试中，要展示完整步骤。即使有小的计算错误，清晰的代换也能获得方法分。练习历年真题以增强信心和速度。

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