4 Case where the Antiderivative is Composite | 反导数为复合函数的情形

📚 4 Case where the Antiderivative is Composite | 反导数为复合函数的情形

In IB Mathematics, one of the most efficient integration techniques involves recognising when the integrand fits the pattern of a derivative of a composite function. This approach, often called the reverse chain rule, allows us to write down antiderivatives directly without performing a full substitution. In this article, we explore the case where the antiderivative itself is a composite function and examine the common types, worked examples, and pitfalls that appear in the syllabus.

在IB数学中，一种高效的积分技巧是识别被积函数何时符合复合函数导数的模式。这种方法通常称为反向链式法则，它使我们能够直接写出反导数，而无需进行完整的换元。本文探讨反导数本身为复合函数的情形，并研究大纲中常见的类型、演算示例及易错点。

---

1. The Chain Rule and Its Reverse | 链式法则及其逆过程

Differentiating a composite function f(g(x)) gives f'(g(x))·g' integrate such an expression, we recover the original composite function. In symbols, ∫ f'(g(x))·g' x = f(g(x)) + C. The challenge is to spot the inner function g(x) and the outer function f'(u) in the integrand, then integrate f' to obtain f.

对复合函数 f(g(x)) 求导，根据链式法则得到 f'(g(x))·g'(x)。如果我们对该表达式积分，就能恢复原来的复合函数。用符号表示即 ∫ f'(g(x))·g'(x) dx = f(g(x)) + C。难点在于从被积函数中识别内层函数 g(x) 和外层函数 f'(u)，再对 f' 积分得到 f。

d/dx [f(g(x))] = f'(g(x))·g'(x) ⇔ ∫ f'(g(x))·g'(x) dx = f(g(x)) + C

---

2. Recognising the Pattern f'(g(x))·g'(x) | 识别 f'(g(x))·g'(x) 形式

ivative of sin(x²) is cos(x²)·2x. Therefore ∫ 2x·cos(x²) dx = sin(x²) + C. Here g(x)=x², g'(x)=2x, and f'(u)=cos(u) so f(u)=sin(u).

要应用反向链式法则，需寻找一个乘积，其中一个因子显然是内层函数的导数，而另一个因子看起来像是外层函数在内层函数处的导数。一个典型示例：sin(x²) 的导数是 cos(x²)·2x。因此 ∫ 2x·cos(x²) dx = sin(x²) + C。这里 g(x)=x², g'(x)=2x, f'(u)=cos(u)，所以 f(u)=sin(u)。

---

3. Linear Inner Functions: ∫ f(ax+b) dx | 线性内层函数：∫ f(ax+b) dx

F is an antiderivative of f. This is because differentiating F(ax+b) yields f(ax+b)·a, so we must compensate by multiplying by 1/a.

当内层函数为线性时，比如 g(x)=ax+b，其导数为常数 a。反向链式法则就简化为一个非常实用的公式：∫ f(ax+b) dx = (1/a)·F(ax+b) + C，其中 F 是 f 的一个反导数。这是因为对 F(ax+b) 求导得到 f(ax+b)·a，因此需要乘以 1/a 来调整。

∫ f(ax+b) dx = (1/a)·F(ax+b) + C

---

4. Worked Example with Linear Inner Function | 线性内层函数的演算例子

Evaluate ∫ (2x+3)⁵ dx. Here f(u)=u⁵, inner function g(x)=2x+3, g' his can be verified by differentiating: d/dx[(1/12)(2x+3)⁶] = (1/12)·6(2x+3)⁵·2 = (2x+3)⁵.

计算 ∫ (2x+3)⁵ dx。这里 f(u)=u⁵，内层函数 g(x)=2x+3, g'(x)=2。u⁵ 的一个反导数是 u⁶/6。应用线性法则，得到 ∫ (2x+3)⁵ dx = (1/2)·((2x+3)⁶/6) + C = (1/12)(2x+3)⁶ + C。可通过求导验证：d/dx[(1/12)(2x+3)⁶] = (1/12)·6(2x+3)⁵·2 = (2x+3)⁵。

---

5. More General Composite Antiderivatives | 更一般的复合反导数

When the inner function is non-linear, we must check that the factor g'(x) appears explicitly. For instance, consider ∫ 2x·e^x² dx. Here g(x)=x² gives g'(x)=2x, which matches the factor, and the outer derivative is e^u whose antiderivative is e^u. Thus the integral is simply e^x² + C. The key is to identify the 'derivative pair' inside the product.

当内层函数为非线性时，必须检查因子 g'(x) 是否明显出现。例如，考虑 ∫ 2x·e^x² dx。这里 g(x)=x² 给出 g'(x)=2x，与因子匹配，而外层导数 e^u 的反导数就是 e^u。因此积分直接为 e^x² + C。关键是在乘积中识别出“导数对”。

---

6. Adjusting Constants: 'Almost' There | 调整常数：几乎匹配时

s check if you can force the exact derivative into the integrand by a constant adjustment.

有时内层导数差一个常数因子。例如，在 ∫ x·cos(x²) dx 中我们希望有 2x 而不是 x。我们仍然可以通过乘以并除以 2 来使用反向链式法则：∫ x·cos(x²) dx = (1/2) ∫ 2x·cos(x²) dx = (1/2)·sin(x²) + C。始终检查是否可以通过常数调整将准确导数强行放入被积函数。

---

广告 / Sponsored

7. Trigonometric Functions with Composite Argument | 复合参数的三角函数

chain rule applies beautifully to trigonometric integrands. We have standard results: ∫ g'(x)·cos(g(x)) dx = sin(g(x)) + C, ∫ g'(x)·sin(g(x)) dx = -cos(g(x)) + C, and ∫ g'(g(x)) dx = tan(g(x)) + C. For example, ∫ 3x²·sin(x³) dx = -cos(x³) + C because g(x)=x³, g'(x)=3x², and ∫ sin(u) du = -cos(u).

反向链式法则非常适用于三角被积函数。我们有标准结果：∫ g'(x)·cos(g(x)) dx = sin(g(x)) + C，∫ g'(x)·sin(g(x)) dx = -cos(g(x)) + C，以及 ∫ g'(x)·sec²(g(x)) dx = tan(g(x)) + C。例如，∫ 3x²·sin(x³) dx = -cos(x³) + C，因为 g(x)=x³, g'(x)=3x²，而 ∫ sin(u) du = -cos(u)。

---

8. Exponential and Logarithmic Composites | 指数与对数复合函数

amined patterns involve exponential and logarithmic functions. The general results are ∫ g'(x)·e^g(x) dx = e^g(x) + C and ∫ g'(x)/g(x) dx = ln|g(x)| + C. For instance, ∫ (2x)/(x²+1) dx matches the logarithmic form with g(x)=x²+1 and g' ive is ln(x²+1) + C. Always check the numerator against the derivative of the denominator.

两个最常考的模型涉及指数与对数函数。一般结果为 ∫ g'(x)·e^g(x) dx = e^g(x) + C 以及 ∫ g'(x)/g(x) dx = ln|g(x)| + C。例如，∫ (2x)/(x²+1) dx 符合对数形式，其中 g(x)=x²+1, g'(x)=2x，因此反导数为 ln(x²+1) + C。始终对照分子与分母的导数进行检查。

---

9. Rational Powers and the Reverse Chain Rule | 有理幂与反向链式法则

For powers of a function, we have ∫ g'ⁿ dx = [g(x)]ⁿ⁺¹/(n+1) + C, provided n ≠ -1. This extends the linear inner case. Take ∫ x·√x²+1 dx: we note g(x)=x²+1, g'(x)=2x, so x dx is (1/2) g'(x) dx. Thus the integral equals (1/2) ∫ u^½ du where u=x²+1, giving (1/2)·(2/3)u^3/2 = (1/3)(x²+1)^3/2 + C.

对于函数的幂次，有 ∫ g'(x)·[g(x)]ⁿ dx = [g(x)]ⁿ⁺¹/(n+1) + C，其中 n ≠ -1。这扩展了线性内层的情形。以 ∫ x·√x²+1 dx 为例：注意到 g(x)=x²+1, g'(x)=2x，所以 x dx 是 (1/2) g'(x) dx。因此积分等于 (1/2) ∫ u^½ du，其中 u=x²+1，得到 (1/2)·(2/3)u^3/2 = (1/3)(x²+1)^3/2 + C。

---

10. Common Mistakes and Pitfalls | 常见错误与陷阱

(1/2x)·sin(x²) + C is wrong because 2x is not constant and cannot be factored out. Another mistake is forgetting the constant adjustment. Always verify by differentiating your proposed antiderivative: it should return the original integrand.

一个常见错误是将线性公式应用于非线性内层函数而缺少导数因子。例如，将 ∫ cos(x²) dx 写成 (1/2x)·sin(x²) + C 是错误的，因为 2x 不是常数，不能提出。另一个错误是忘记常数调整。务必通过对你提出的反导数求导来验证：它应该回到原被积函数。

---

11. Summary of Key Patterns | 关键模式总结

Mastering the case where the antiderivative is composite hinges on recognising a handful of standard forms and adjusting constants when needed. The table below summarises the core integrals that every IB student should be able to spot rapidly.

掌握反导数为复合函数的情形，关键在于识别少数几种标准形式，并在需要时调整常数。下表总结了每位IB学生都应迅速识别的核心积分。

Integrand Form	Antiderivative	Conditions
f(ax+b)	(1/a)·F(ax+b) + C	F' = f
g'(x)·f'(g(x))	f(g(x)) + C	f' integrable
g'(x)·[g(x)]ⁿ	[g(x)]ⁿ⁺¹/(n+1) + C	n ≠ -1
g'(x)·eg(x)	eg(x) + C
g'(x)/g(x)	ln|g(x)| + C	g(x) ≠ 0
g'(x)·cos(g(x))	sin(g(x)) + C

The reverse chain rule transforms what looks like a complicated integral into a straightforward recognition exercise. With practice, you can integrate these cases in a single step and save valuable time in examinations.

反向链式法则将看似复杂的积分转化为直接的识别练习。通过练习，你可以一步完成这些情形的积分，并在考试中节省宝贵时间。

---

Published by TutorHao | IB Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）