A-Level Maths: Integrating Composite Linear Functions | A-Level 数学：复合线性函数积分法

📚 A-Level Maths: Integrating Composite Linear Functions | A-Level 数学：复合线性函数积分法

This revision guide focuses on integrals of functions that contain a linear expression inside another function, such as (ax + b)ⁿ, e^(ax+b), sin(ax+b) or 1/(ax+b). These are called linear composite functions. Recognising the pattern is the key to integrating them quickly and accurately.

这篇复习指南聚焦于积分中被积函数内部包含线性表达式的情形，例如 (ax + b)ⁿ、e^(ax+b)、sin(ax+b) 或 1/(ax+b)。这类函数称为线性复合函数。识别出这种结构是快速准确积分的关键。

1. Recognising the Linear Composite Pattern | 识别复合线性函数结构

A composite linear function has the form f(ax+b), where a and b are constants, a ≠ 0, and f is the outer function. The inner function ax+b is linear, which is why the reverse chain rule applies.

复合线性函数具有 f(ax+b) 的形式，其中 a 和 b 为常数，且 a ≠ 0，f 为外层函数。内层函数 ax+b 是一次函数，因此可以使用逆链式法则。

Typical exam examples include powers, reciprocals, roots, exponentials and trigonometric expressions.

考试中典型的例子包括幂函数、倒数、根式、指数函数和三角表达式。

(2x+3)⁵, 1/(4x−1), √(7−3x), e^(2x+5), sin(3x−π/2)

2. Reverse Chain Rule: The Core Principle | 逆链式法则：核心原理

Recall that if y = f(ax+b), then dy/dx = a·f'(ax+b). Therefore, integrating f'(ax+b) with respect to x must give (1/a)·f(ax+b) + C.

回忆如果 y = f(ax+b)，则 dy/dx = a·f'(ax+b)。因此，对 f'(ax+b) 关于 x 积分必须得到 (1/a)·f(ax+b) + C。

∫ f'(ax+b) dx = (1/a) f(ax+b) + C

More generally, if F is an antiderivative of f, then:

更一般地，如果 F 是 f 的一个原函数，则：

∫ f(ax+b) dx = (1/a) F(ax+b) + C

3. Integrating (ax+b)ⁿ | 幂函数 (ax+b)ⁿ 的积分

For any rational power n ≠ −1, multiply the linear expression by the same expression inside the integration framework, then divide by the extra factor a(n+1).

对于任何有理幂 n ≠ −1，先按幂函数求出原函数，再除以额外因子 a(n+1) 即可。

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

Example: ∫ (2x+3)⁵ dx. Here a = 2 and n = 5, so the denominator is 2×6 = 12.

例：∫ (2x+3)⁵ dx。这里 a = 2，n = 5，所以分母为 2×6 = 12。

∫ (2x+3)⁵ dx = (2x+3)⁶/12 + C

Check by differentiating: d/dx[(2x+3)⁶/12] = (6/12)(2x+3)⁵·2 = (2x+3)⁵.

通过求导检验：d/dx[(2x+3)⁶/12] = (6/12)(2x+3)⁵·2 = (2x+3)⁵。

4. Reciprocal Functions: ∫ (ax+b)⁻¹ dx | 倒数函数：∫ (ax+b)⁻¹ dx

When n = −1, the power formula does not work. The correct result involves the natural logarithm.

当 n = −1 时，幂函数公式不适用。正确结果涉及自然对数。

∫ 1/(ax+b) dx = (1/a) ln|ax+b| + C

Example: ∫ 1/(3x+2) dx = (1/3) ln|3x+2| + C.

例：∫ 1/(3x+2) dx = (1/3) ln|3x+2| + C。

The absolute value is essential for logarithms because the argument must be positive.

绝对值符号是必要的，因为对数函数的真数必须为正。

If the denominator is linear and the numerator is a constant, use the logarithmic formula directly.
如果分母是线性函数且分子为常数，直接使用对数公式。

5. Roots and Negative Powers | 根式与负幂的积分

Roots must be rewritten as fractional powers before applying the power formula.

根式必须先改写为分数指数幂，再应用幂函数公式。

∫ √(ax+b) dx = ∫ (ax+b)^(1/2) dx = (2/(3a))(ax+b)^(3/2) + C

Worked example: ∫ √(4x−1) dx. Here a = 4 and n = 1/2.

例：∫ √(4x−1) dx。这里 a = 4，n = 1/2。

∫ (4x−1)^(1/2) dx = (1/(4×3/2))(4x−1)^(3/2) + C = (1/6)(4x−1)^(3/2) + C

Negative powers also work when n ≠ −1. For example:

负幂在 n ≠ −1 时同样适用。例如：

∫ 1/(x+1)² dx = ∫ (x+1)⁻² dx = (x+1)⁻¹/(−1) + C = −1/(x+1) + C

6. Exponential Functions with Linear Input | 含有线性输入端的指数函数

For an exponential function with a linear exponent, the integral is the same exponential divided by the coefficient of x.

对于线性指数的指数函数，其积分等于原指数函数除以 x 的系数。

∫ e^(ax+b) dx = (1/a) e^(ax+b) + C

Example: ∫ e^(3x+1) dx = (1/3) e^(3x+1) + C.

例：∫ e^(3x+1) dx = (1/3) e^(3x+1) + C。

If the base is not e, use the general exponential rule.

如果底数不是 e，使用一般指数函数求积规则。

∫ p^(ax+b) dx = p^(ax+b) / [a ln p] + C

Example: ∫ 2^(x−1) dx = 2^(x−1)/(ln 2) + C.

例：∫ 2^(x−1) dx = 2^(x−1)/(ln 2) + C。

7. Trigonometric Linear Composites | 三角线性复合函数的积分

The standard trigonometric integrals adapt in the same way: each result is divided by a.

标准三角积分公式以相同方式调整：每个结果都除以 a。

∫ 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: ∫ cos(2x−π/3) dx = (1/2) sin(2x−π/3) + C.

例：∫ cos(2x−π/3) dx = (1/2) sin(2x−π/3) + C。

Always check the sign for sine. The derivative of cos(ax+b) is −a sin(ax+b), so the integral of sin(ax+b) must be negative.

注意正弦积分的正负号。cos(ax+b) 的导数为 −a sin(ax+b)，所以 sin(ax+b) 的积分必然带负号。

8. Definite Integrals: Changing Limits or Back-Substitution | 定积分：换限或回代

With a definite integral, you can either integrate directly in x and substitute the original limits, or use the substitution u = ax+b and change the limits. Both methods are accepted by exam boards.

对于定积分，你可以直接对 x 积分并代入原上下限，也可以使用代换 u = ax+b 并更换上下限。两种方法在考试中均可接受。

Example: Evaluate ∫₀¹ (2x+1)³ dx.

例：计算 ∫₀¹ (2x+1)³ dx。

Method 1 – direct formula:

方法一：直接使用公式：

∫₀¹ (2x+1)³ dx = [(2x+1)⁴/8]₀¹ = (3⁴ − 1⁴)/8 = (81 − 1)/8 = 10

Method 2 – substitution: let u = 2x+1, so du = 2 dx. When x = 0, u = 1; when x = 1, u = 3.

方法二：代换法：令 u = 2x+1，则 du = 2 dx。当 x = 0，u = 1；当 x = 1，u = 3。

∫₀¹ (2x+1)³ dx = (1/2)∫₁³ u³ du = [u⁴/8]₁³ = 10

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

Forgetting to divide by the coefficient a.
忘记除以系数 a。
Using the power formula for n = −1 and producing (ax+b)⁰/0.
对 n = −1 错误使用幂函数公式，得到 (ax+b)⁰/0。
Omitting the absolute value in ln|ax+b|.
省略 ln|ax+b| 中的绝对值符号。
Dropping the arbitrary constant C for indefinite integrals.
在不定积分中遗漏任意常数 C。
Sign errors when integrating sine or sec² functions.
对正弦或 sec² 函数积分时出现符号错误。

Forgetting the factor 1/a is the single most common mark-loss error in this topic.

忘记因子 1/a 是本主题中最常见的失分错误。

10. Constant Factors and Simple Simplifications | 常数因子与简单化简

If a constant multiplies the composite function, bring it outside the integral and apply the standard formula.

如果复合函数前乘以一个常数，可将常数移到积分号外，再应用标准公式。

Example: ∫ 5(2x+3)⁴ dx = 5 × (2x+3)⁵/(2×5) + C = (2x+3)⁵/2 + C.

例：∫ 5(2x+3)⁴ dx = 5 × (2x+3)⁵/(2×5) + C = (2x+3)⁵/2 + C。

However, do not apply the reverse chain rule to products such as x(x+1)², because x is not a constant factor. In such cases, expand or use integration by parts.

但是，对于 x(x+1)² 这类乘积，不能直接使用逆链式法则，因为 x 不是常数因子。此时需展开或用分部积分法。

11. Exam-Style Questions and Solutions | 真题风格练习

Question 1: Find ∫ 1/(5x−2) dx.

练习 1：求 ∫ 1/(5x−2) dx。

Solution: (1/5) ln|5x−2| + C

Question 2: Evaluate ∫₀^(π/2) cos(2x) dx.

练习 2：计算 ∫₀^(π/2) cos(2x) dx。

Solution: [(1/2) sin(2x)]₀^(π/2) = (1/2)(sin π − sin 0) = 0

Question 3: Find ∫ √(3x+1) dx.

练习 3：求 ∫ √(3x+1) dx。

Solution: (2/9)(3x+1)^(3/2) + C

12. Summary Table and Final Advice | 总结表与最终建议

The following standard results should be memorised for the exam.

以下标准结论应在考前熟记。

Integral	Result
∫ (ax+b)ⁿ dx, n ≠ −1	(ax+b)^(n+1) / [a(n+1)] + C
∫ 1/(ax+b) dx	(1/a) ln\|ax+b\| + C
∫ e^(ax+b) dx	(1/a) e^(ax+b) + C
∫ 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

Always identify a and b first, then apply the correct standard formula. For every indefinite integral, add + C. For definite integrals, remember to use the correct limits.

做题时先确定 a 和 b，再应用正确的标准公式。对不定积分务必加上 + C；对定积分则要正确使用上下限。

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A-Level Maths: Integrating Composite Linear Functions | A-Level 数学：复合线性函数积分法