Edexcel A-Level Maths Question 030: Integration by Substitution | Edexcel A-Level 数学第030题：换元积分法

📚 Edexcel A-Level Maths Question 030: Integration by Substitution | Edexcel A-Level 数学第030题：换元积分法

Integration by substitution is one of the most frequently examined techniques in the Edexcel A-Level Pure Mathematics specification. Question 030 from the pdfjoiner practice set focuses on the reverse chain rule, definite integrals with changed limits, and the correct handling of logarithmic and trigonometric forms. This article explains the core method, demonstrates worked examples step by step, highlights common mistakes, and gives exam-style practice so that you can secure full marks on this topic.

换元积分法（Integration by substitution）是 Edexcel A-Level 纯数学大纲中考查频率最高的技巧之一。pdfjoiner 练习集中的第030题重点考查逆向链式法则、改变积分限的定积分，以及对对数形式和三角形式的正确处理。本文讲解核心方法，逐步演示例题，强调常见错误，并提供考试风格的练习，帮助你在这个知识点上拿到满分。

1. Understanding the Reverse Chain Rule | 理解逆向链式法则

The reverse chain rule is the foundation of integration by substitution. If you recognise an integrand in the form f'(g(x)) g'(x), then its integral is simply f(g(x)) + C. In other words, when one factor is the derivative of the inner function, you can integrate the outer function directly.

逆向链式法则是换元积分法的基础。如果你能识别出被积函数具有 f'(g(x)) g'(x) 的形式，那么它的积分就是 f(g(x)) + C。换句话说，当一个因式是内层函数的导数时，你可以直接对外层函数积分。

∫ f'(g(x)) g'(x) dx = f(g(x)) + C

This is useful for simple cases such as ∫ 2x cos(x²) dx, where the derivative of x² is 2x. Recognising the pattern saves time and reduces algebraic errors.

这对于简单情况很有用，例如 ∫ 2x cos(x²) dx，其中 x² 的导数是 2x。识别出这个模式可以节省时间并减少代数错误。

2. The General Strategy for Substitution | 换元法的一般策略

When the integrand is more complicated, we introduce a new variable u = g(x). The differential du is then obtained from du/dx = g'(x), so du = g'(x) dx. The original integral is rewritten entirely in terms of u, integrated, and finally the original variable x is substituted back.

当被积函数更复杂时，我们引入一个新变量 u = g(x)。然后由 du/dx = g'(x) 得到微分 du = g'(x) dx。把原积分完全用 u 表示，进行积分，最后再把原变量 x 代回去。

Let u = g(x), then du = g'(x) dx

This method transforms a difficult integral into a standard one. It is essential to convert every part of the integrand, including dx, before integrating.

这种方法把一个困难的积分转化为标准积分。关键是在积分之前把被积函数的每一个部分（包括 dx）都转换完毕。

3. Choosing the Right Substitution u | 选择合适的代换变量 u

A good choice for u is usually the inner function of a composite function, or the part whose derivative already appears in the integrand. For example, in ∫ 3x² √(x³+1) dx, choose u = x³+1 because its derivative is 3x², which is present.

一个恰当的 u 通常选取复合函数的内层函数，或者选取其导数已经出现在被积函数中的部分。例如，在 ∫ 3x² √(x³+1) dx 中，选择 u = x³+1，因为它的导数是 3x²，正好出现在被积函数中。

If the derivative of u is not exactly present, it may still work if the remaining factor is a constant multiple. You can adjust by multiplying or dividing outside the integral, but never inside the differential carelessly.

如果 u 的导数没有完全出现，只要剩余因式是常数倍也可以使用。你可以在积分外乘以或除以常数进行调整，但绝不能随意在微分内部操作。

4. Worked Example 1: Basic Substitution | 例题1：基础换元

Let us integrate ∫ 2x e^(x²) dx. Choose u = x², then du/dx = 2x, so du = 2x dx. The integral becomes ∫ eᵘ du, which integrates to eᵘ + C. Substituting back gives e^(x²) + C.

我们来计算 ∫ 2x e^(x²) dx。选择 u = x²，则 du/dx = 2x，所以 du = 2x dx。积分变为 ∫ eᵘ du，积分结果为 eᵘ + C。代回原变量得到 e^(x²) + C。

∫ 2x e^(x²) dx = ∫ eᵘ du = eᵘ + C = e^(x²) + C

Always check your answer by differentiating. The derivative of e^(x²) is e^(x²) × 2x, which confirms the original integrand.

一定要通过求导来检查答案。e^(x²) 的导数是 e^(x²) × 2x，这与原始被积函数一致。

5. Worked Example 2: Definite Integral with Changed Limits | 例题2：改变积分限的定积分

For definite integrals, you must remember to change the limits whenever you substitute. Consider ∫₀¹ 3x² √(x³+1) dx. Let u = x³+1, so du = 3x² dx. When x = 0, u = 1; when x = 1, u = 2. The integral becomes ∫₁² √u du.

对于定积分，在换元时必须记住同时改变积分限。考虑 ∫₀¹ 3x² √(x³+1) dx。令 u = x³+1，则 du = 3x² dx。当 x = 0 时，u = 1；当 x = 1 时，u = 2。积分变为 ∫₁² √u du。

∫₁² √u du = ∫₁² u^½ du = [ (2/3) u^(3/2) ]₁² = (2/3)(2√2 – 1)

The final answer is (2/3)(2√2 – 1). If you forget to change the limits, you cannot evaluate the integral in terms of x without taking an extra step back.

最终答案是 (2/3)(2√2 – 1)。如果你忘记改变积分限，就不能直接代入 u 的值计算，必须再回到 x 变量，这会增加出错风险。

6. Integrating Fractions and Logarithmic Forms | 分式与对数形式的积分

When the integrand is a rational function whose numerator is the derivative of the denominator, substitution leads directly to a natural logarithm. For example, ∫ (2x+3)/(x²+3x) dx. Let u = x²+3x, then du = (2x+3) dx, so the integral is ∫ 1/u du = ln|u| + C.

当被积函数是有理函数，且分子是分母的导数时，换元会直接得到自然对数。例如，∫ (2x+3)/(x²+3x) dx。令 u = x²+3x，则 du = (2x+3) dx，所以积分是 ∫ 1/u du = ln|u| + C。

∫ (2x+3)/(x²+3x) dx = ln|x²+3x| + C

The absolute value is important because the logarithm is only defined for positive arguments, while u may take negative values on the domain of integration.

绝对值符号很重要，因为对数只对正数有定义，而 u 在积分区间内可能取负值。

7. Trigonometric Substitutions | 三角换元

Substitution also works well with powers of trigonometric functions. Consider ∫ sin³θ cosθ dθ. Let u = sinθ, then du = cosθ dθ. The integral simplifies to ∫ u³ du = u⁴/4 + C, so the answer is (1/4) sin⁴θ + C.

换元法同样适用于三角函数的幂。考虑 ∫ sin³θ cosθ dθ。令 u = sinθ，则 du = cosθ dθ。积分简化为 ∫ u³ du = u⁴/4 + C，所以答案是 (1/4) sin⁴θ + C。

∫ sin³θ cosθ dθ = (1/4) sin⁴θ + C

For integrals like ∫ tanθ dθ, rewrite tanθ as sinθ/cosθ and use u = cosθ. This gives -ln|cosθ| + C, which is equivalent to ln|secθ| + C.

对于 ∫ tanθ dθ 这样的积分，把 tanθ 写成 sinθ/cosθ，并用 u = cosθ。结果为 -ln|cosθ| + C，也就是 ln|secθ| + C。

8. Common Mistakes in Substitution | 换元中的常见错误

Students often lose marks by forgetting to change the limits in definite integrals. Another common error is leaving a mixture of x and u in the same integral. You must either convert everything to u or return fully to x.

学生经常因忘记改变定积分的积分限而失分。另一个常见错误是在同一个积分中混用 x 和 u。你必须把所有部分都转换为 u，或者完全回到 x 变量。

Forgetting to change limits in definite integrals | 忘记改变定积分中的积分限
Leaving both x and u inside the integral | 积分中同时残留 x 和 u
Incorrectly calculating du/dx | 错误计算 du/dx
Missing the constant factor when adjusting the integrand | 调整被积函数时漏掉常数因子

Check each step carefully. Writing down u, du/dx, and du explicitly before performing the integration reduces careless errors.

仔细检查每一步。在进行积分之前，明确写出 u、du/dx 和 du，可以减少粗心错误。

9. Exam Technique and Edexcel Marking Points | 考试技巧与 Edexcel 评分要点

In Edexcel A-Level papers, integration by substitution questions typically carry 5 to 7 marks. Marks are awarded for selecting a suitable substitution, finding du correctly, converting the integral, integrating the new expression, and substituting back or applying limits.

在 Edexcel A-Level 试卷中，换元积分题通常占 5 到 7 分。评分标准包括选择合适的代换、正确求出 du、转换积分、对新表达式积分，以及代回变量或代入积分限。

Even if you make a later slip, showing a clear substitution and du earns method marks. For definite integrals, always write the changed limits explicitly to show the examiner that you have understood the process.

即使后面的计算出现错误，清晰写出代换和 du 也能得到方法分。对于定积分，一定要明确写出改变后的积分限，向阅卷老师展示你已经理解了这个过程。

10. Harder Exam-Style Question 030 | 考试风格第030题强化练习

Try this challenge which reflects the style of Question 030: Evaluate ∫ (1 + 2x)⁵ × 2 dx from x = 0 to x = 1. Let u = 1 + 2x, then du = 2 dx. When x = 0, u = 1; when x = 1, u = 3. The integral becomes ∫₁³ u⁵ du = [u⁶/6]₁³ = (3⁶ – 1⁶)/6.

尝试这道体现第030题风格的挑战题：计算 ∫₀¹ (1 + 2x)⁵ × 2 dx。令 u = 1 + 2x，则 du = 2 dx。当 x = 0 时，u = 1；当 x = 1 时，u = 3。积分变为 ∫₁³ u⁵ du = [u⁶/6]₁³ = (3⁶ – 1⁶)/6。

∫₁³ u⁵ du = (3⁶ – 1⁶) / 6 = (729 – 1) / 6 = 728 / 6 = 364 / 3

Therefore the exact value is 364/3. To verify, you could expand (1+2x)⁵ and integrate term by term, but substitution is far more efficient and is the expected method at A-Level.

因此精确值是 364/3。为了验证，你可以展开 (1+2x)⁵ 再逐项积分，但换元法要高效得多，也是 A-Level 阶段的要求方法。

11. Tips for Choosing u in Exam Questions | 考试中选择 u 的技巧

When the integrand contains a composite function, look for an inner function whose derivative is also present. In expressions with powers, roots, exponentials, or denominators, the inner function is usually the best candidate for u.

当被积函数包含复合函数时，寻找一个其导数也同时出现的内层函数。在含有幂、根号、指数或分母的表达式中，内层函数通常是最佳的 u 候选。

If the derivative is missing a constant factor only, you can still use the substitution. For example, ∫ x cos(x²) dx can be handled with u = x², du = 2x dx, so x dx = (1/2) du. The integral becomes (1/2) ∫ cos u du.

如果导数只是缺少一个常数因式，你仍然可以使用换元。例如，∫ x cos(x²) dx 可以用 u = x²，du = 2x dx，所以 x dx = (1/2) du。积分变为 (1/2) ∫ cos u du。

∫ x cos(x²) dx = (1/2) sin(x²) + C

Practise identifying u quickly by scanning the integrand for a function and its derivative. This skill is essential for success in the Edexcel Pure Mathematics units.

通过扫描被积函数中的函数及其导数来快速识别 u，需要大量练习。这个技能对于在 Edexcel 纯数学单元中取得成功至关重要。

12. Summary and Key Takeaways | 总结与关键要点

Integration by substitution is a powerful tool that reverses the chain rule. The key steps are to choose u, compute du, rewrite the integral fully in terms of u, integrate, and then substitute back or apply new limits.

换元积分法是逆向链式法则的强大工具。关键步骤是选择 u，计算 du，把积分完全用 u 表示，积分，然后代回变量或应用新的积分限。

Choose u as the inner function whose derivative is present | 选择导数出现的那个内层函数作为 u
Change limits in definite integrals immediately | 在定积分中立即改变积分限
Do not mix x and u in the same integrand | 不要在同一被积函数中混用 x 和 u
Verify your answer by differentiation | 通过求导验证答案

Mastering this technique will help you answer a wide range of Edexcel A-Level questions confidently and accurately, including the challenging Question 030 style problems from practice papers.

掌握这一技巧将帮助你自信而准确地解答 Edexcel A-Level 的各种题目，包括练习卷中具有挑战性的第030题风格问题。

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Edexcel A-Level Maths Question 030: Integration by Substitution | Edexcel A-Level 数学第030题：换元积分法