📚 Reverse Chain Rule | 反向链式法则
The reverse chain rule is a fundamental integration technique that allows us to integrate composite functions by recognising the derivative of the inner function. It is a direct consequence of reversing the chain rule for differentiation and is essential for tackling a wide range of integrals in the Edexcel A-Level Mathematics syllabus.
反向链式法则是一种通过识别内层函数的导数来积分复合函数的基本技巧。它是微分链式法则的逆向应用，对于应对 Edexcel A-Level 数学大纲中各类积分题目至关重要。
1. Integration as Reverse Differentiation | 积分作为微分的逆运算
Integration is the process of finding a function given its derivative. Since the chain rule tells us how to differentiate composite functions, reversing this process enables us to integrate functions that are the result of a chain rule differentiation.
积分是给定导数后寻找原函数的过程。由于链式法则告诉我们如何对复合函数求导，逆向使用这一过程就能使我们积分那些由链式法则求导得到的函数。
For every differentiation rule, there is a corresponding integration rule. The reverse chain rule is the integration counterpart to the chain rule for differentiation, just as the power rule for integration corresponds to the power rule for differentiation.
每一种微分法则都有对应的积分法则。反向链式法则就是微分链式法则的积分对应物，正如幂函数积分法则对应于幂函数微分法则。
2. The Chain Rule for Differentiation | 微分链式法则回顾
Recall the chain rule: if y = f(g(x)), then dy/dx = f'(g(x)) · g' ction evaluated at the inner function, multiplied by the derivative of the inner function.
回顾链式法则：若 y = f(g(x))，则 dy/dx = f'(g(x)) · g'(x)。简言之，复合函数的导数等于外层函数在内层函数处的导数，乘以内层函数的导数。
(3x + 2)&sup5;, then let u = 3x+2, so y = u&sup5;. Then dy/dx = 5u⁴ × 3 = 15(3x+2)⁴.
例如，若 y = (3x + 2)&sup5;，令 u = 3x+2，则 y = u&sup5;。于是 dy/dx = 5u⁴ × 3 = 15(3x+2)⁴。
Notice how the derivative of the inner function (3) appears as a factor in the final result. This is the key observation that drives the reverse chain rule.
注意内层函数（3x+2）的导数 3 作为一个因子出现在最终结果中。这是驱动反向链式法则的关键观察。
3. The Concept of Reverse Chain Rule | 反向链式法则的概念
The reverse chain rule is used when an integral contains a function and its derivative in a specific pattern. In essence, if we recognise that the integrand is of the form f'(g(x)) · g'(x), we can immediately write the answer as f(g(x)) + C.
当一个被积函数中包含某个函数及其导数并呈现特定模式时，就可以使用反向链式法则。本质上，如果我们识别出被积函数形如 f'(g(x)) · g'(x)，就可以直接写出答案 f(g(x)) + C。
This means that instead of performing a full substitution, we mentally reverse the differentiation process. It is a shortcut that saves time and reduces errors once mastered.
这意味着我们不需要进行完整的代换，而是心算出微分的逆过程。一旦掌握，它就是一个省时且减少错误的捷径。
4. Recognising the Pattern: f'(g(x)) g'(x) | 识别模式：f'(g(x)) g'(x)
The standard reverse chain rule can be written as:
标准反向链式法则可写作：
∫ f'(g(x)) · g'(x) dx = f(g(x)) + C
To apply this, identify an 'inner' function g(x) and check whether its derivative g'(x) appears as a factor in the integrand. The remaining part must be exactly f'(g(x)) for some function f.
应用此法则时，要确定一个“内层”函数 g(x)，并检查其导数 g'(x) 是否作为因子出现在被积函数中。其余部分必须恰为某个函数 f 的导数 f'(g(x))。
When the inner function is linear, that is g(x) = ax + b, then g' nt. This makes the pattern especially easy to spot and is the most common case at A-Level.
当内层函数为线性函数，即 g(x) = ax + b 时，g'(x) = a 为常数。这使得该模式特别容易识别，这也是 A-Level 中最常见的情形。
5. Integrating (ax + b)ⁿ | 积分形如 (ax + b)ⁿ 的幂函数
For integrals of the form ∫ (ax + b)ⁿ dx, where n ≠ -1, we can use the reverse chain rule by thinking about differentiating (ax + b)ⁿ⊃⁺¹.
对于形如 ∫ (ax + b)ⁿ dx（n ≠ -1）的积分，我们可以通过思考 (ax + b)ⁿ⊃⁺¹ 的微分来使用反向链式法则。
Differentiating (ax + b)ⁿ⊃⁺¹ gives (n+1)(ax + b)ⁿ · a. Therefore, to get (ax + b)ⁿ alone, we divide by a(n+1). The formula is:
对 (ax + b)ⁿ⊃⁺¹ 求导得到 (n+1)(ax + b)ⁿ · a。因此，为得到单独的 (ax + b)ⁿ，我们需要除以 a(n+1)。公式如下：
∫ (ax + b)ⁿ dx = (1/a) × (ax + b)ⁿ⊃⁺¹ / (n+1) + C, n ≠ -1
3, n = 4. Apply the formula: (1/2) × (2x+3)&sup5; / 5 + C = (1/10)(2x+3)&sup5; + C.
示例：∫ (2x + 3)⁴ dx。这里 a = 2, b = 3, n = 4。套用公式得到：(1/2) × (2x+3)&sup5; / 5 + C = (1/10)(2x+3)&sup5; + C。
of (1/10)(2x+3)&sup5; is (1/10)×5(2x+3)⁴×2 = (2x+3)⁴, confirming correctness.
始终通过微分来检验答案。(1/10)(2x+3)&sup5; 的导数为 (1/10)×5(2x+3)⁴×2 = (2x+3)⁴，确认无误。
6. Integrating eax + b | 积分指数函数 eax + b
The exponential function eˣ is its own derivative. When the exponent is linear, the reverse chain rule gives a simple adjustment factor.
指数函数 eˣ 的导数就是它本身。当指数是线性函数时，反向链式法则给出一个简单的调整因子。
Since d/dx [eax+b] = a eax+b, it follows that ∫ eax+b dx = (1/a) eax+b + C.
由于 d/dx [eax+b] = a eax+b，因此 ∫ eax+b dx = (1/a) eax+b + C。
Example: ∫ e5x-2 dx = (1/5) e5x-2 + C. The factor 1/5 compensates for the derivative of the inner function 5x-2.
示例：∫ e5x-2 dx = (1/5) e5x-2 + C。因子 1/5 补偿了内层函数 5x-2 的导数。
This extends to any base: ∫ akx+c dx = (1/(k ln a)) akx+c + C, but at A-Level the focus is predominantly on base e.
这可以拓展到任意底数：∫ akx+c dx = (1/(k ln a)) akx+c + C，但在 A-Level 中重点主要放在底数 e 上。
7. Integrating Trigonometric Functions of (ax + b) | 积分三角函数 (ax + b)
unctions. We know that d/dx [sin(ax+b)] = a cos(ax+b) and d/dx [cos(ax+b)] = -a sin(ax+b).
类似的逻辑适用于正弦和余弦函数。我们知道 d/dx [sin(ax+b)] = a cos(ax+b) 以及 d/dx [cos(ax+b)] = -a sin(ax+b)。
Thus, by reversing these, we obtain the standard integrals:
因此，通过逆向推导，我们得到标准积分公式：
∫ cos(ax+b) dx = (1/a) sin(ax+b) + C
∫ sin(ax+b) dx = -(1/a) cos(ax+b) + C
;/3) dx = (1/4) sin(4x + π/3) + C. The inner derivative is 4, so we place 1/4 in front.
示例：∫ cos(4x + π/3) dx = (1/4) sin(4x + π/3) + C。内层函数的导数为 4，因此我们在前面乘以 1/4。
b): ∫ sec²(ax+b) dx = (1/a) tan(ax+b) + C, as the derivative of tan is sec².
同样的模式也适用于 sec²(ax+b)：∫ sec²(ax+b) dx = (1/a) tan(ax+b) + C，因为 tan 的导数就是 sec²。
8. Adjusting for Constants Not Exactly Matching | 调整不完全匹配的常数
Sometimes the integrand contains the inner derivative multiplied by a constant that does not exactly match what we need. In such cases, we can still use the reverse chain rule by introducing a compensating factor.
有时被积函数包含内层导数乘以一个不完全匹配的常数。这种情况下，我们仍然可以通过引入一个补偿因子来使用反向链式法则。
Consider ∫ x² (x³ + 1)&sup5; dx. Let g(x) = x³ + 1, then g' up3;+1)&sup5; dx = (1/3) × (1/6)(x³+1)&sup6; + C = (1/18)(x³+1)&sup6; + C.
考虑 ∫ x² (x³ + 1)&sup5; dx。令 g(x) = x³ + 1，则 g'(x) = 3x²;。我们有 x² dx，但我们需要 3x²。因此我们将积分写作 sup3;+1)&sup5; dx = (1/3) × (1/6)(x³+1)&sup6; + C = (1/18)(x³+1)&sup6; + C。
The general rule: if the integrand is k · g'(x) · f(g(x)), we can pull the constant k outside, then apply the reverse chain rule, and finally multiply the result by k.
一般规则：若被积函数为 k · g'(x) · f(g(x))，我们可以将常数 k 提出积分号，再应用反向链式法则，最后将结果乘以 k。
Never forget to adjust for the constant; omitting the compensating factor is one of the most common errors.
切勿忘记调整常数；遗漏补偿因子是学生最常犯的错误之一。
9. Using the Reverse Chain Rule with More Complex Functions | 对更复杂函数运用反向链式法则
reverse chain rule can also handle rational and logarithmic forms. For instance, ∫ (g'(x))/(g(x)) dx = ln|g(x)| + C, which is a direct application of the reverse chain rule with f(u) = ln|u|.
反向链式法则也可以处理有理式和对数形式。例如，∫ (g'(x))/(g(x)) dx = ln|g(x)| + C，这是反向链式法则的直接应用，其中 f(u) = ln|u|。
Another common pattern is ∫ g'(x) e^(g(x)) dx = e^(g(x)) + C. This works because the derivative of e^(g(x)) is g'(x) e^(g(x)).
另一个常见模式是 ∫ g'(x) e^(g(x)) dx = e^(g(x)) + C。这是因为 e^(g(x)) 的导数恰好为 g'(x) e^(g(x))。
When the inner function is not linear, inspection becomes vital. Look for a function and its derivative. Example: ∫ 2x cos(x²) dx. Notice that the derivative of x² is 2x, which is present. Thus the integral is sin(x²) + C.
当内层函数不是线性时，观察就显得至关重要。要寻找一个函数及其导数。示例：∫ 2x cos(x²) dx。注意到 x² 的导数是 2x，而 2x 正好出现。因此该积分等于 sin(x²) + C。
10. Substitution vs Reverse Chain Rule | 代换法与反向链式法则的比较
e is essentially substitution done mentally. In substitution, we set u = g(x), then du = g' u. The reverse chain rule skips the explicit substitution and goes straight to the result.
反向链式法则本质上是心算完成的代换法。在代换法中，我们设 u = g(x)，则 du = g'(x) dx，并将积分重新用 u 表达。反向链式法则跳过了显式代换步骤，直接得到结果。
For linear inner functions, the reverse chain rule is much faster and is the expected method in Edexcel exams. However, when the inner derivative is not perfectly present, some students find substitution clearer.
对于线性内层函数，反向链式法则要快得多，且是 Edexcel 考试中期望使用的方法。然而，当内层导数不完全匹配时，一些学生可能会觉得代换法更清晰。
It is beneficial to practice both methods so that you can choose the most efficient one for any given integral. In an exam, the reverse chain rule saves valuable time.
对两种方法都进行练习是有益的，这样你就可以对任意给定积分选择最高效的方法。在考试中，反向链式法则能节省宝贵的时间。
11. Common Mistakes and How to Avoid Them | 常见错误及如何避免
One frequent mistake is forgetting to divide by the coefficient of x when integrating linear functions. Always ask: 'If I differentiate my answer, do I get back the original integrand?'
一个常见错误是积分线性函数时忘记除以 x 的系数。始终要问自己：“如果我将我的答案微分，能否得到原来的被积函数？”
Another error is misidentifying the inner function. Make sure the function you pick as g(x) is indeed having its derivative present as a factor. Double-check the derivative in your head.
另一个错误是错误识别内层函数。要确保你选作 g(x) 的函数，其导数确实作为因子出现。心算导数时要反复检查。
Students also sometimes mishandle constants, either forgetting to bring them outside or adjusting incorrectly. Write the integral with the constant factor separated before applying the rule.
学生有时也会处理常数不当，要么忘记将其提出，要么调整错误。在应用法则前，先将带有常数因子的积分分离写出来。
Finally, avoid applying the reverse chain rule when the inner function is not differentiable or when the pattern simply isn't there. In those cases, use substitution or other techniques.
最后，当内层函数不可微或者模式根本不存在时，要避免使用反向链式法则。那种情况下，应使用代换法或其他技巧。
12. Exam-Style Questions and Tips | 考试题型与应试技巧
Edexcel frequently asks students to integrate functions like (3x - 2)⁷, e0.5x, sin(2x + 1), or more involved integrands like x√(x²+5). Always look for the inner derivative first.
Edexcel 经常要求学生积分诸如 (3x - 2)⁷、e0.5x、sin(2x + 1) 之类的函数，或更复杂的被积函数如 x√(x²+5)。务必首先寻找内层函数的导数。
For definite integrals, apply the reverse chain rule to find the antiderivative, then substitute the limits as usual. The adjustment constant is not needed for definite integrals; just evaluate at bounds.
对于定积分，先运用反向链式法则求出原函数，然后照常代入上下限。定积分不需要使用调整常数，直接计算上下限即可。
When practicing, cover the full range: polynomials, exponentials, trigonometric, and rational functions. Try to increase speed by doing mental checks rather than writing full substitution.
练习时要覆盖全部类型：多项式、指数、三角函数和有理式。试着通过心算检查来提速，而不是写出完整的代换过程。
Finally, remember that the reverse chain rule is a must-know integration technique. Combined with the product rule and integration by parts, it forms the backbone of A-Level integration.
最后，请记住反向链式法则是必须掌握的积分技巧。它与乘积法则、分部积分结合在一起，构成了 A-Level 积分知识体系的主干。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图