📚 Reverse Chain Rule: Finding the Original Function from Composite Derivatives | 复合函数求原函数：链式法则的逆用
Integration is often described as the reverse of differentiation. When we differentiate a composite function, the chain rule produces a product of an outer derivative and an inner derivative. The reverse chain rule reverses this process: if you can recognise a derivative sitting inside an integrand, you can write down the original composite function directly.
积分常被描述为微分的逆运算。当我们对复合函数求导时，链式法则会得到一个“外层导数 × 内层导数”的乘积。复合函数求原函数（逆用链式法则）就是把这个过程倒过来：如果你能在被积函数中识别出某个导数结构，就可以直接写出原来的复合函数。
1. The Chain Rule in Differentiation | 微分中的链式法则
If y = g(f(x)), then the derivative is given by the chain rule:
若 y = g(f(x))，根据链式法则，其导数为：
dy/dx = g′(f(x)) × f′(x)
For example, if y = (x³ + 1)⁵, then f(x) = x³ + 1, g(f) = f⁵, and dy/dx = 5(x³ + 1)⁴ × 3x² = 15x²(x³ + 1)⁴.
例如，若 y = (x³ + 1)⁵，则 f(x) = x³ + 1，g(f) = f⁵，因此 dy/dx = 5(x³ + 1)⁴ × 3x² = 15x²(x³ + 1)⁴。
Notice that the derivative naturally splits into two factors: the derivative of the outer function evaluated at the inner function, and the derivative of the inner function itself.
注意，导数天然地分解为两个因子：外层函数在内层函数处的导数，以及内层函数本身的导数。
2. The Reverse Chain Rule Formula | 逆用链式法则的公式
The reverse chain rule states that if we can write an integrand in the form f′(x) × g′(f(x)), then its integral is simply g(f(x)) + C.
逆用链式法则指出：如果被积函数可以写成 f′(x) × g′(f(x)) 的形式，那么它的积分就是 g(f(x)) + C。
∫ f′(x) g′(f(x)) dx = g(f(x)) + C
In practice, this means you should look for an inner function f(x) whose derivative f′(x) also appears in the integrand, possibly up to a constant multiple.
在实践中，这意味着你需要寻找一个内层函数 f(x)，并且它的导数 f′(x) 也在被积函数中出现，可能只相差一个常数倍数。
When the derivative differs only by a constant, you can compensate by dividing by that constant after integrating.
当导函数只差一个常数倍时，可以在积分后除以该常数来补偿。
3. Recognising Patterns: f′(x) × [f(x)]ⁿ | 识别模式：f′(x) × [f(x)]ⁿ
The most common application is integrating expressions of the form f′(x) × [f(x)]ⁿ.
最常见的应用是积分形如 f′(x) × [f(x)]ⁿ 的表达式。
∫ f′(x) [f(x)]ⁿ dx = [f(x)]ⁿ⁺¹ / (n + 1) + C, n ≠ −1
For example, ∫ 3x²(x³ + 1)⁴ dx. Let f(x) = x³ + 1, so f′(x) = 3x². The integrand matches exactly, so the integral is (x³ + 1)⁵ / 5 + C.
例如，∫ 3x²(x³ + 1)⁴ dx。令 f(x) = x³ + 1，则 f′(x) = 3x²。被积函数完全匹配，所以积分结果为 (x³ + 1)⁵ / 5 + C。
If the derivative is not exact, adjust by a constant. Consider ∫ 2x(x² + 5)³ dx. Here f(x) = x² + 5 and f′(x) = 2x, so the integral is (x² + 5)⁴ / 4 + C.
如果导数不完全一致，就调整常数倍。考虑 ∫ 2x(x² + 5)³ dx。这里 f(x) = x² + 5，f′(x) = 2x，因此积分结果为 (x² + 5)⁴ / 4 + C。
4. Exponential Composites: ∫ f′(x) e^(f(x)) dx | 指数型复合函数：∫ f′(x) e^(f(x)) dx
Because the derivative of e^(f(x)) is f′(x) e^(f(x)), the reverse chain rule gives:
因为 e^(f(x)) 的导数是 f′(x) e^(f(x))，所以逆用链式法则可得：
∫ f′(x) e^(f(x)) dx = e^(f(x)) + C
For example, ∫ 4x³ e^(x⁴) dx. Since f(x) = x⁴ and f′(x) = 4x³, the integral is e^(x⁴) + C.
例如，∫ 4x³ e^(x⁴) dx。因为 f(x) = x⁴，f′(x) = 4x³，所以积分结果是 e^(x⁴) + C。
When the derivative is not an exact match, insert a constant factor. For instance, ∫ 3x² e^(x³) dx already matches exactly, giving e^(x³) + C.
当导数不完全匹配时，可引入常数因子。例如，∫ 3x² e^(x³) dx 已完全匹配，结果为 e^(x³) + C。
If the integrand is ∫ 6x² e^(x³) dx, the extra factor 2 can be pulled outside: 2∫ 3x² e^(x³) dx = 2e^(x³) + C.
若被积函数是 ∫ 6x² e^(x³) dx，多余的因子 2 可以提到积分号外：2∫ 3x² e^(x³) dx = 2e^(x³) + C。
5. Logarithmic Composites: ∫ f′(x) / f(x) dx | 对数型复合函数：∫ f′(x) / f(x) dx
The derivative of ln|f(x)| is f′(x) / f(x). Therefore:
ln|f(x)| 的导数是 f′(x) / f(x)。因此：
∫ f′(x) / f(x) dx = ln|f(x)| + C
For example, ∫ 2x / (x² + 4) dx. Let f(x) = x² + 4, so f′(x) = 2x. The integral is ln|x² + 4| + C, and since x² + 4 is always positive, we may write ln(x² + 4) + C.
例如，∫ 2x / (x² + 4) dx。令 f(x) = x² + 4，则 f′(x) = 2x。积分结果为 ln|x² + 4| + C；由于 x² + 4 恒为正，可以写成 ln(x² + 4) + C。
This pattern also explains why ∫ tan(x) dx = ln|sec(x)| + C. Since tan(x) = sin(x)/cos(x), the numerator sin(x) is almost the negative derivative of cos(x).
这个模式也解释了为什么 ∫ tan(x) dx = ln|sec(x)| + C。因为 tan(x) = sin(x)/cos(x)，分子 sin(x) 几乎是 cos(x) 的导数的相反数。
∫ tan x dx = −ln|cos x| + C = ln|sec x| + C
6. Trigonometric Composites | 三角函数复合型
The derivative of sin(f(x)) is f′(x) cos(f(x)), and the derivative of cos(f(x)) is −f′(x) sin(f(x)).
sin(f(x)) 的导数是 f′(x) cos(f(x))，而 cos(f(x)) 的导数是 −f′(x) sin(f(x))。
Thus ∫ f′(x) cos(f(x)) dx = sin(f(x)) + C, and ∫ f′(x) sin(f(x)) dx = −cos(f(x)) + C.
因此 ∫ f′(x) cos(f(x)) dx = sin(f(x)) + C，且 ∫ f′(x) sin(f(x)) dx = −cos(f(x)) + C。
For example, ∫ 2x cos(x²) dx. Since f(x) = x² and f′(x) = 2x, the integral is sin(x²) + C.
例如，∫ 2x cos(x²) dx。因为 f(x) = x²，f′(x) = 2x，所以积分结果是 sin(x²) + C。
For ∫ x sin(x²) dx, only a factor 2 is missing: write x sin(x²) = ½ × 2x sin(x²), so the integral is −½ cos(x²) + C.
对于 ∫ x sin(x²) dx，只差一个因子 2：将 x sin(x²) 写成 ½ × 2x sin(x²)，因此积分结果为 −½ cos(x²) + C。
7. Linear Inner Functions: The Special Case | 一次内层函数的特殊情形
When the inner function is linear, say ax + b, the chain rule simply multiplies by a. Reversing this divides by a.
当内层函数为一次函数 ax + b 时，链式法则只是乘以 a。逆用时就除以 a。
∫ g′(ax + b) dx = g(ax + b) / a + C
For example, ∫ (3x + 1)⁵ dx. Here the inner derivative is 3, so the integral is (3x + 1)⁶ / (6 × 3) + C = (3x + 1)⁶ / 18 + C.
例如，∫ (3x + 1)⁵ dx。这里内层导数为 3，所以积分结果为 (3x + 1)⁶ / (6 × 3) + C = (3x + 1)⁶ / 18 + C。
Similarly, ∫ e^(5x) dx = e^(5x) / 5 + C, and ∫ sin(4x) dx = −cos(4x) / 4 + C.
类似地，∫ e^(5x) dx = e^(5x) / 5 + C，∫ sin(4x) dx = −cos(4x) / 4 + C。
This is often called integration of a function of a linear function, and it is the easiest case of the reverse chain rule.
这通常被称为“一次函数复合函数的积分”，是逆用链式法则中最简单的情形。
8. Integration by Substitution: A Formal Version | 换元积分法：更规范的形式
The reverse chain rule is closely related to integration by substitution. If we let u = f(x), then du = f′(x) dx, and the integral becomes simpler.
逆用链式法则与换元积分法密切相关。若令 u = f(x)，则 du = f′(x) dx，原积分会变得更简单。
For example, evaluate ∫ 4x³ √(x⁴ + 9) dx. Let u = x⁴ + 9, du = 4x³ dx, so the integral becomes ∫ √u du = (2/3) u^(3/2) + C = (2/3)(x⁴ + 9)^(3/2) + C.
例如，计算 ∫ 4x³ √(x⁴ + 9) dx。令 u = x⁴ + 9，则 du = 4x³ dx，原积分变为 ∫ √u du = (2/3) u^(3/2) + C = (2/3)(x⁴ + 9)^(3/2) + C。
Substitution gives a step-by-step procedure, but the reverse chain rule lets you write the answer directly when the pattern is visible.
换元法提供了逐步演算的过程，而当结构一目了然时，逆用链式法则可以直接写出答案。
In IB examinations, showing the substitution u = f(x) is often required for full marks, even if you can see the answer immediately.
在 IB 考试中，即使你能立即看出答案，通常也需要写出换元 u = f(x) 的步骤才能获得满分。
9. Definite Integrals and Changing Limits | 定积分与换限
For definite integrals, you can either find the antiderivative and evaluate at the original limits, or change the limits when using substitution.
对于定积分，你可以先求出原函数再代入原来的上下限，也可以在换元时同时更换积分限。
Evaluate ∫₀¹ 2x e^(x²) dx. Let u = x². When x = 0, u = 0; when x = 1, u = 1. The integral becomes ∫₀¹ e^u du = e − 1.
计算 ∫₀¹ 2x e^(x²) dx。令 u = x²。当 x = 0 时，u = 0；当 x = 1 时，u = 1。原积分变为 ∫₀¹ e^u du = e − 1。
Using the reverse chain rule directly: the antiderivative is e^(x²), so the definite integral is e^(1²) − e^(0²) = e − 1. Both methods agree.
直接使用逆用链式法则：原函数为 e^(x²)，因此定积分为 e^(1²) − e^(0²) = e − 1。两种方法结果一致。
When changing limits, never mix old and new variables inside the same integral.
换限时，绝不能在同一个积分中混用新旧变量。
10. Common Mistakes and How to Avoid Them | 常见错误与避坑指南
Mistake 1: Forgetting the inner derivative. ∫ (x² + 1)³ dx is not (x² + 1)⁴ / 4 + C, because the derivative of x² + 1 is 2x, which is absent.
错误一：忘记内层导数。∫ (x² + 1)³ dx 的结果不是 (x² + 1)⁴ / 4 + C，因为 x² + 1 的导数是 2x，而它并不出现在被积函数中。
Mistake 2: Forgetting to divide by the constant multiple when the inner derivative differs by a factor.
错误二：当内层导数相差一个倍数时，忘记除以该倍数。
Mistake 3: Using the reverse chain rule with an inner function that is not actually present inside the outer function.
错误三：外层函数内部并没有真正包含所选的内层函数，却仍使用逆用链式法则。
Mistake 4: Forgetting the absolute value in ∫ f′(x)/f(x) dx = ln|f(x)| + C.
错误四：在 ∫ f′(x)/f(x) dx = ln|f(x)| + C 中忘记加绝对值。
Mistake 5: Forgetting the constant of integration C for indefinite integrals.
错误五：求不定积分时忘记常数 C。
Always test your answer by differentiating it mentally before moving on.
继续做题前，先在心里对答案求导验证一下。
11. Worked Examples | 典型例题精讲
Example 1: Find ∫ 6x(x² + 3)⁵ dx.
例 1：求 ∫ 6x(x² + 3)⁵ dx。
Let f(x) = x² + 3, f′(x) = 2x. The integrand has 6x = 3 × 2x, so the integral is 3 × (x² + 3)⁶ / 6 + C = (x² + 3)⁶ / 2 + C.
令 f(x) = x² + 3，f′(x) = 2x。被积函数中 6x = 3 × 2x，所以积分结果为 3 × (x² + 3)⁶ / 6 + C = (x² + 3)⁶ / 2 + C。
Example 2: Find ∫ cos(x) / (1 + sin(x)) dx.
例 2：求 ∫ cos(x) / (1 + sin(x)) dx。
Let f(x) = 1 + sin(x), f′(x) = cos(x). The integral is ln|1 + sin(x)| + C.
令 f(x) = 1 + sin(x)，f′(x) = cos(x)。积分结果为 ln|1 + sin(x)| + C。
Example 3: Evaluate ∫₀^π sin(x) cos²(x) dx.
例 3：计算 ∫₀^π sin(x) cos²(x) dx。
Let u = cos(x), du = −sin(x) dx. When x = 0, u = 1; when x = π, u = −1. The integral becomes ∫₁⁻¹ −u² du = ∫₋₁¹ u² du = 2/3.
令 u = cos(x)，则 du = −sin(x) dx。当 x = 0 时，u = 1；当 x = π 时，u = −1。原积分变为 ∫₁⁻¹ −u² du = ∫₋₁¹ u² du = 2/3。
12. Summary | 总结
The reverse chain rule is one of the most powerful integration strategies. Always look for an inner function f(x) whose derivative f′(x) appears in the integrand.
逆用链式法则是最强大的积分策略之一。做题时始终寻找内层函数 f(x)，并检查它的导数 f′(x) 是否出现在被积函数中。
If the derivative appears exactly, integrate directly as g(f(x)) + C. If it appears up to a constant multiple, compensate by dividing by that multiple. If it does not appear at all, the reverse chain rule cannot be applied.
如果导数恰好出现，直接积分得 g(f(x)) + C；如果相差一个常数倍数，就除以该倍数；如果完全没有出现，则不能使用逆用链式法则。
Mastering this technique through pattern recognition will save you time in exams and deepen your understanding of the relationship between differentiation and integration.
通过模式识别掌握这一技巧，不仅能帮你在考试中节省时间，还能加深你对微分与积分之间关系的理解。
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