A-Level Maths: Reverse Chain Rule for Integration | A-Level 数学：反链式法则求积分

📚 A-Level Maths: Reverse Chain Rule for Integration | A-Level 数学：反链式法则求积分

The reverse chain rule is one of the most powerful and frequently tested integration techniques in A-Level Mathematics. It allows us to integrate functions of the form f'(x) × g(f(x)) by reversing the chain rule from differentiation. This article breaks down the method, its applications, and the common pitfalls students encounter.

反链式法则是 A-Level 数学中最强大且最常考的积分技巧之一。它使我们能够通过逆转微分中的链式法则，来积分形如 f'(x) × g(f(x)) 的函数。本文将详细讲解该方法、其应用以及学生常犯的错误。

1. Understanding the Reverse Chain Rule | 理解反链式法则

The chain rule states that when differentiating a composite function, we multiply by the derivative of the inner function. Reversing this process, if we spot a function and its derivative appearing together in an integrand, we can integrate directly.

链式法则指出，对复合函数求微分时，我们需要乘以内层函数的导数。反过来，如果在被积函数中同时看到某个函数及其导数，我们就能直接积分。

For the general case where n ≠ -1:

对于一般情况 n ≠ -1：

∫ f'(x) × [f(x)]ⁿ dx = [f(x)]ⁿ⁺¹ / (n+1) + C

The key is recognising that f'(x) is present (possibly up to a constant factor) in the integrand.

关键是要识别出被积函数中存在 f'(x)（可能相差一个常数倍数）。

2. The Basic Pattern: Power Functions | 基本模式：幂函数

The simplest application involves integrating expressions like (ax + b)ⁿ. Since the derivative of (ax + b) is just the constant a, the reverse chain rule gives:

最简单的应用涉及积分 (ax + b)ⁿ 这类表达式。由于 (ax + b) 的导数就是常数 a，反链式法则给出：

∫ (ax + b)ⁿ dx = (ax + b)ⁿ⁺¹ / [a(n+1)] + C, n ≠ -1

Example 1: Find ∫ (2x + 3)⁵ dx.

例 1：求 ∫ (2x + 3)⁵ dx。

Here f'(x) = 2. We divide by this factor:

这里 f'(x) = 2。我们除以这个因子：

∫ (2x + 3)⁵ dx = (2x + 3)⁶ / (6 × 2) + C = (2x + 3)⁶ / 12 + C

Notice we multiply by 1/2 to compensate for the extra factor of 2 that would appear upon differentiation.

注意我们要乘以 1/2 来补偿微分时多出来的因子 2。

3. Trigonometric Functions | 三角函数

When integrating trigonometric composite functions, the trig function’s derivative appears as the multiplier. Common results include:

在积分三角复合函数时，三角函数的导数会作为乘数出现。常见结果包括：

∫ sin(ax + b) dx = -(1/a)cos(ax + b) + C
∫ cos(ax + b) dx = (1/a)sin(ax + b) + C
∫ sec²(ax + b) dx = (1/a)tan(ax + b) + C

Example 2: Evaluate ∫ 3x² · cos(x³) dx.

例 2：计算 ∫ 3x² · cos(x³) dx。

Notice that d/dx(x³) = 3x². The derivative of the inner function appears exactly. Therefore:

注意到 d/dx(x³) = 3x²。内层函数的导数恰好出现。因此：

∫ 3x² · cos(x³) dx = sin(x³) + C

Since the derivative of sin(x³) is cos(x³) × 3x² by the chain rule, the answer checks out perfectly.

因为根据链式法则，sin(x³) 的导数是 cos(x³) × 3x²，所以答案完全正确。

4. Exponential Functions | 指数函数

For exponential composites, the derivative of the exponent multiplies the expression. The general pattern is:

对于指数复合函数，指数的导数会乘以该表达式。一般模式为：

∫ f'(x) · e^[f(x)] dx = e^[f(x)] + C

Example 3: Integrate ∫ x · e^(x²) dx.

例 3：积分 ∫ x · e^(x²) dx。

Here f(x) = x² and f'(x) = 2x. We have x in the integrand, which is half of 2x, so:

这里 f(x) = x²，f'(x) = 2x。被积函数中有 x，是 2x 的一半，所以：

∫ x · e^(x²) dx = ½ e^(x²) + C

Why half? Because when we differentiate ½e^(x²), the chain rule produces ½ × 2x × e^(x²) = x · e^(x²).

为什么要乘以二分之一？因为当我们对 ½e^(x²) 求导时，链式法则产生 ½ × 2x × e^(x²) = x · e^(x²)。

5. Fractional Forms: The Natural Logarithm | 分数形式：自然对数

When the integrand has the form f'(x)/f(x), the antiderivative is ln|f(x)| + C. This case corresponds to n = -1 in the general power rule:

当被积函数具有 f'(x)/f(x) 的形式时，其反导数为 ln|f(x)| + C。这一情况对应幂法则通式中的 n = -1：

∫ f'(x)/f(x) dx = ln|f(x)| + C

Example 4: Find ∫ 2x/(x² + 1) dx.

例 4：求 ∫ 2x/(x² + 1) dx。

Let f(x) = x² + 1. Then f'(x) = 2x, which appears in the numerator. Thus:

设 f(x) = x² + 1。则 f'(x) = 2x，恰好出现在分子中。因此：

∫ 2x/(x² + 1) dx = ln|x² + 1| + C

Since x² + 1 is always positive, the absolute value signs are technically unnecessary here, but it is good practice to include them.

由于 x² + 1 始终为正，绝对值符号在此处实际并非必需，但保留它们是良好的习惯。

6. Adjusting for Missing Constants | 调整缺失的常数

Often the derivative of the inner function does not appear exactly. In such cases, we may need to introduce a compensating factor. The general approach is:

内层函数的导数往往并非恰好出现。此时，我们可能需要引入一个补偿系数。一般方法是：

Identify the inner function f(x)
Compute f'(x) in your head
Compare with what appears in the integrand
Adjust by multiplying by the required constant

识别内层函数 f(x)
心算 f'(x)
与被积函数中出现的部分进行比较
乘以所需常数进行调整

Example 5: Evaluate ∫ sin(3x) · cos³(3x) dx.

例 5：计算 ∫ sin(3x) · cos³(3x) dx。

Let u = cos(3x). Then du/dx = -3sin(3x). The integrand contains sin(3x), which is (-1/3) of du/dx. Therefore:

设 u = cos(3x)。则 du/dx = -3sin(3x)。被积函数包含 sin(3x)，它是 du/dx 的 (-1/3)。因此：

∫ sin(3x) · cos³(3x) dx = -1/3 × cos⁴(3x)/4 + C = -cos⁴(3x)/12 + C

Check: d/dx[-cos⁴(3x)/12] = -1/12 × 4cos³(3x) × (-sin(3x)) × 3 = sin(3x) · cos³(3x). ✓

验证：d/dx[-cos⁴(3x)/12] = -1/12 × 4cos³(3x) × (-sin(3x)) × 3 = sin(3x) · cos³(3x)。✓

7. Substitution as a Formal Method | 换元法作为正式方法

The reverse chain rule is closely related to integration by substitution. When the pattern is not immediately obvious, a formal substitution can help:

反链式法则与换元积分法密切相关。当模式不是一目了然时，正式的换元可以帮助解题：

Example 6: Find ∫ x√(x² + 1) dx.

例 6：求 ∫ x√(x² + 1) dx。

Let u = x² + 1, then du = 2x dx, so x dx = ½ du:

令 u = x² + 1，则 du = 2x dx，所以 x dx = ½ du：

∫ x√(x² + 1) dx = ½ ∫ u^(½) du = ½ × (2/3)u^(3/2) + C = (1/3)(x² + 1)^(3/2) + C

Writing the substitution explicitly reduces the chance of errors and is often required for full marks in examinations.

明确写出换元过程能降低出错概率，在考试中也常是获得满分的必要条件。

8. Definite Integrals: Applying Limits | 定积分：应用上下限

For definite integrals using the reverse chain rule, we have two options: integrate first then substitute the limits, or change the limits when substituting. The second approach is often cleaner.

对于使用反链式法则的定积分，我们有两种选择：先积分再代上下限，或者在换元时同时变换上下限。第二种方法通常更简洁。

Example 7: Evaluate ∫₀¹ 6x²(2x³ + 1)⁴ dx.

例 7：计算 ∫₀¹ 6x²(2x³ + 1)⁴ dx。

Let u = 2x³ + 1. Then du = 6x² dx. The integrand contains exactly 6x², so:

令 u = 2x³ + 1。则 du = 6x² dx。被积函数恰好包含 6x²，所以：

When x = 0, u = 1; when x = 1, u = 3. Therefore:

当 x = 0 时，u = 1；当 x = 1 时，u = 3。因此：

∫₀¹ 6x²(2x³ + 1)⁴ dx = ∫₁³ u⁴ du = [u⁵/5]₁³ = 243/5 – 1/5 = 242/5

Changing the limits with the substitution avoids the need to substitute back, saving time and reducing arithmetic errors.

换元时同时变换上下限，可以避免代回原变量，既节省时间又减少计算错误。

9. Table of Standard Reverse Chain Rule Results | 反链式法则标准结果表

The following table summarises the most important patterns students should memorise:

下表总结了学生应当牢记的最重要的模式：

Integrand 被积函数	Integral 积分结果
f'(x)[f(x)]ⁿ (n ≠ -1)	[f(x)]ⁿ⁺¹/(n+1) + C
f'(x)/f(x)	ln\|f(x)\| + C
f'(x) · e^[f(x)]	e^[f(x)] + C
f'(x) · sin(f(x))	-cos(f(x)) + C
f'(x) · cos(f(x))	sin(f(x)) + C
f'(x) · sec²(f(x))	tan(f(x)) + C

These six patterns cover the vast majority of reverse chain rule questions in A-Level examinations.

这六种模式覆盖了 A-Level 考试中绝大多数反链式法则题目。

10. Common Mistakes and Examiner Tips | 常见错误与考官建议

Students frequently lose marks on reverse chain rule questions for predictable reasons. Here are the most common errors and how to avoid them:

学生常因可预见的原因在反链式法则题目中失分。以下是最常见的错误及避免方法：

Forgetting the constant multiple: Always check whether the derivative of your chosen inner function appears exactly; if not, adjust with the appropriate factor.
Mistaking the sign: Remember that d/dx[cos x] = -sin x, so integrating sin requires a negative sign.
Ignoring the +C: Indefinite integrals always require the constant of integration.
Confusing power rules: When n = -1, use the logarithm rule, not the power rule.

忘记常数倍数：始终检查所选内层函数的导数是否恰好出现；如果没有，用相应系数调整。
搞错正负号：记住 d/dx[cos x] = -sin x，因此积分 sin 需要负号。
忽略 +C：不定积分永远需要积分常数。
混淆幂法则：当 n = -1 时，使用对数法则而非幂法则。

Examiners look for clear working. Even if you can see the answer by inspection, showing the identification of f(x) and f'(x) earns method marks and protects you from careless errors.

考官看重清晰的解题过程。即使你能直接看出答案，写出识别 f(x) 和 f'(x) 的过程也能获得方法分，并防止粗心错误。

11. Practice Questions | 练习题目

Test your understanding with these exam-style questions:

用这些考试风格的题目检验你的理解：

Question 1: Find ∫ (3x² + 2)(x³ + 2x)⁴ dx.

题目 1：求 ∫ (3x² + 2)(x³ + 2x)⁴ dx。

Question 2: Evaluate ∫₀^π/2 sin³x · cos x dx.

题目 2：计算 ∫₀^π/2 sin³x · cos x dx。

Question 3: Find ∫ e^(3x)/(e^(3x) + 5) dx.

题目 3：求 ∫ e^(3x)/(e^(3x) + 5) dx。

Question 4: Find ∫ 4x · sec²(x²) dx.

题目 4：求 ∫ 4x · sec²(x²) dx。

Solutions: (1) (x³ + 2x)⁵/5 + C; (2) sin⁴(π/2)/4 – 0 = 1/4; (3) (1/3)ln|e^(3x) + 5| + C; (4) 2tan(x²) + C.

答案：(1) (x³ + 2x)⁵/5 + C；(2) sin⁴(π/2)/4 – 0 = 1/4；(3) (1/3)ln|e^(3x) + 5| + C；(4) 2tan(x²) + C。

12. Summary | 总结

The reverse chain rule is a cornerstone of A-Level integration. By recognising patterns where a function and its derivative appear together, you can integrate complex composite functions quickly and accurately. Master this technique, and you will find that many seemingly difficult exam questions become routine.

反链式法则是 A-Level 积分学的基石。通过识别函数与其导数同时出现的模式，你可以快速准确地积分复杂的复合函数。掌握这一技巧后，你会发现许多看似困难的考试题目变得驾轻就熟。

Remember the golden rule: identify the inner function f(x), check whether f'(x) appears (possibly up to a constant factor), and apply the corresponding standard result. With sufficient practice, this process becomes intuitive, and you will be able to handle a wide range of integration problems with confidence.

记住黄金法则：识别内层函数 f(x)，检查 f'(x) 是否出现（可能相差一个常数因子），然后应用相应的标准结果。通过充分练习，这个过程会变得自然而直觉化，你将能够自信地处理各种积分问题。

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

A-Level Maths: Reverse Chain Rule for Integration | A-Level 数学：反链式法则求积分