📚 Integrating f(αx + b) | 积分 f(αx + b)
In A-Level Mathematics, integrating expressions of the form f(αx + b) is a core skill built on the reverse chain rule. Once you master the simple adjustment of dividing by α, you can integrate powers, exponentials, reciprocals and trigonometric functions quickly and accurately.
在 A-Level 数学中，积分形如 f(αx + b) 的表达式是建立在逆链式法则基础上的核心技能。一旦掌握除以 α 这一简单调整，你就能快速、准确地积分幂函数、指数函数、倒数函数和三角函数。
1. What Does f(αx + b) Mean? | 什么是 f(αx + b)？
f(αx + b) is a composite function: the input x is first mapped to αx + b, and then the outer function f is applied. The constant α controls the horizontal stretch or compression, while b shifts the graph horizontally. In integration, α must not be zero, otherwise the expression would not depend on x in the intended way.
f(αx + b) 是一个复合函数：输入 x 先被变换为 αx + b，然后再对外层函数 f 求值。常数 α 控制水平方向的伸缩，b 控制水平平移。在积分中，α 不能为零，否则表达式就不会按预期方式依赖于 x。
For example, if f(u) = u⁴, then f(2x + 3) = (2x + 3)⁴. If f(u) = eᵘ, then f(3x + 1) = e³ˣ⁺¹. The challenge is to reverse the chain rule that would appear if we differentiated these expressions.
例如，如果 f(u) = u⁴，那么 f(2x + 3) = (2x + 3)⁴。如果 f(u) = eᵘ，那么 f(3x + 1) = e³ˣ⁺¹。积分的关键在于逆转我们对这些表达式求导时所用的链式法则。
2. The Reverse Chain Rule | 逆链式法则
By the chain rule, differentiating f(αx + b) gives α f ‘(αx + b). Therefore, if the integrand is exactly of the form α f ‘(αx + b), integration immediately returns f(αx + b). This is the reverse chain rule.
根据链式法则，对 f(αx + b) 求导得到 α f ‘(αx + b)。因此，如果被积函数恰好具有 α f ‘(αx + b) 的形式，积分就直接得到 f(αx + b)。这就是逆链式法则。
∫ f ‘(αx + b) dx = (1/α) f(αx + b) + C
The factor 1/α appears because derivative multiplication by α must be undone by division by α when integrating.
公式中的因子 1/α 出现的原因是：求导时乘以 α 的效应，必须在积分时通过除以 α 来抵消。
3. General Formula for Indefinite Integration | 不定积分的一般公式
For a general outer function f, we can use the substitution u = αx + b. Then du/dx = α, so dx = du/α. The integral becomes:
对于一般的外层函数 f，我们可以使用换元 u = αx + b。则 du/dx = α，因此 dx = du/α。积分变为：
∫ f(αx + b) dx = (1/α) ∫ f(u) du
After integrating f(u) with respect to u, replace u by αx + b. This shows why many standard integrals for f(αx + b) simply take the usual integral of f and divide by α.
对 f(u) 关于 u 积分后，再将 u 替换为 αx + b。这表明为什么许多形如 f(αx + b) 的标准积分，只需对 f 作常规积分后除以 α 即可。
This formula works for any function f that is continuous on the relevant interval, provided α ≠ 0.
这个公式适用于在相关区间上连续的任意函数 f，只要 α ≠ 0。
4. Using u-Substitution | 使用 u 换元法
Although the general formula is convenient, exam questions often want you to show substitution working. Consider ∫ e³ˣ⁺¹ dx. Let u = 3x + 1, so du/dx = 3 and dx = du/3.
尽管一般公式很方便，但考试题目往往要求写出换元过程。以 ∫ e³ˣ⁺¹ dx 为例。令 u = 3x + 1，则 du/dx = 3，所以 dx = du/3。
∫ e³ˣ⁺¹ dx = (1/3) ∫ eᵘ du = (1/3) eᵘ + C = (1/3) e³ˣ⁺¹ + C
The method is always the same: write u = αx + b, isolate dx, integrate with respect to u, and back-substitute. This systematic approach reduces errors in more complicated integrals.
方法总是相同：令 u = αx + b，解出 dx，对 u 积分，然后回代。这种系统化方法能减少较复杂积分中的错误。
5. Integrating Powers of (αx + b) | 积分 (αx + b) 的幂
For n ≠ −1, the power rule extends directly:
当 n ≠ −1 时，幂函数法则可以直接推广：
∫ (αx + b)ⁿ dx = (αx + b)ⁿ⁺¹ / [α(n + 1)] + C
This follows from differentiating (αx + b)ⁿ⁺¹, which gives α(n + 1)(αx + b)ⁿ. Dividing by α(n + 1) reverses the derivative.
这是因为对 (αx + b)ⁿ⁺¹ 求导得到 α(n + 1)(αx + b)ⁿ。除以 α(n + 1) 即可逆转求导结果。
Example: ∫ (2x + 3)⁵ dx = (2x + 3)⁶ / (2 × 6) + C = (2x + 3)⁶ / 12 + C.
例如：∫ (2x + 3)⁵ dx = (2x + 3)⁶ / (2 × 6) + C = (2x + 3)⁶ / 12 + C。
Do not forget the extra division by α. A quick derivative check confirms: d/dx [(2x+3)⁶/12] = 6(2x+3)⁵ × 2 / 12 = (2x+3)⁵.
不要忘记额外除以 α。快速求导检验：d/dx [(2x+3)⁶/12] = 6(2x+3)⁵ × 2 / 12 = (2x+3)⁵。
6. Integrating Exponentials | 积分指数函数
The exponential function eᵘ is its own derivative, so integrating e^(αx + b) is straightforward:
指数函数 eᵘ 的导数等于它本身，因此积分 e^(αx + b) 非常简单：
∫ e^(αx + b) dx = (1/α) e^(αx + b) + C
Example: ∫ e^(3x + 1) dx = (1/3) e^(3x + 1) + C.
例如：∫ e^(3x + 1) dx = (1/3) e^(3x + 1) + C。
For exponential functions with other bases, use ∫ a^(αx + b) dx = a^(αx + b) / (α ln a) + C, but Edexcel questions most often focus on base e.
对于其他底的指数函数，可使用 ∫ a^(αx + b) dx = a^(αx + b) / (α ln a) + C，但 Edexcel 试题最常考查以 e 为底的指数函数。
7. Integrating Reciprocals | 积分倒数函数
The reciprocal function has a special case because integrating u⁻¹ gives ln|u|. Therefore:
倒数函数是特殊情况，因为对 u⁻¹ 积分得到 ln|u|。因此：
∫ 1/(αx + b) dx = (1/α) ln|αx + b| + C
The absolute value is important because the logarithm is only defined for positive arguments, while αx + b can be negative. Writing ln|αx + b| covers both cases.
绝对值非常重要，因为对数只对正数自变量有定义，而 αx + b 可能为负。写成 ln|αx + b| 可以覆盖两种情形。
Example: ∫ 1/(2x + 5) dx = (1/2) ln|2x + 5| + C, provided 2x + 5 ≠ 0.
例如：∫ 1/(2x + 5) dx = (1/2) ln|2x + 5| + C，前提是 2x + 5 ≠ 0。
If the numerator is a constant k, take it outside: ∫ k/(αx + b) dx = (k/α) ln|αx + b| + C.
如果分子是常数 k，可以提到积分号外：∫ k/(αx + b) dx = (k/α) ln|αx + b| + C。
8. Integrating Trigonometric Functions | 积分三角函数
Trigonometric integrals with linear arguments follow the same divide-by-α rule:
具有线性角变量的三角积分遵循同样的除以 α 规则：
∫ sin(αx + b) dx = −(1/α) cos(αx + b) + C
∫ cos(αx + b) dx = (1/α) sin(αx + b) + C
∫ sec²(αx + b) dx = (1/α) tan(αx + b) + C
Watch the sign on sine: integrating sin gives negative cosine, and the negative is kept outside the 1/α factor.
注意正弦积分的符号：积分 sin 得到负的 cos，负号保留在 1/α 因子之外。
Example: ∫ cos(4x − π/2) dx = (1/4) sin(4x − π/2) + C.
例如：∫ cos(4x − π/2) dx = (1/4) sin(4x − π/2) + C。
You may also meet ∫ cosec²(αx + b) dx = −(1/α) cot(αx + b) + C and ∫ sec(αx + b)tan(αx + b) dx = (1/α) sec(αx + b) + C.
你可能还会遇到 ∫ cosec²(αx + b) dx = −(1/α) cot(αx + b) + C 以及 ∫ sec(αx + b)tan(αx + b) dx = (1/α) sec(αx + b) + C。
9. Definite Integrals and Changed Limits | 定积分与换限
For a definite integral ∫ₐᵇ f(αx + b) dx, you can first find the indefinite integral and then apply the original limits. Alternatively, use substitution and change the limits: when x = a, u = αa + b; when x = b, u = αb + b.
对于定积分 ∫ₐᵇ f(αx + b) dx，你可以先求出不定积分，然后代入原上下限。也可以使用换元法并替换上下限：当 x = a 时，u = αa + b；当 x = b 时，u = αb + b。
Example: evaluate ∫₀¹ e^(2x + 1) dx. Using the indefinite integral gives [½ e^(2x + 1)] from 0 to 1 = ½ e³ − ½ e = ½(e³ − e).
例如：计算 ∫₀¹ e^(2x + 1) dx。使用不定积分得到 [½ e^(2x + 1)] 从 0 到 1 = ½ e³ − ½ e = ½(e³ − e)。
Using substitution u = 2x + 1 changes the limits from x=0 to u=1 and x=1 to u=3, so the integral becomes (1/2)∫₁³ eᵘ du = ½(e³ − e). Both methods give the same result.
使用换元 u = 2x + 1，下限从 x=0 变为 u=1，上限从 x=1 变为 u=3，因此积分变为 (1/2)∫₁³ eᵘ du = ½(e³ − e)。两种方法结果相同。
10. Common Mistakes and Checking Answers | 常见错误与检验答案
The most common error is forgetting to divide by α. For example, writing ∫ e²ˣ dx = e²ˣ + C is wrong; the correct answer is (1/2)e²ˣ + C.
最常见的错误是忘记除以 α。例如，把 ∫ e²ˣ dx 写成 e²ˣ + C 是错误的；正确答案是 (1/2)e²ˣ + C。
Another frequent mistake is missing the new power denominator when integrating powers: ∫ (3x+2)⁴ dx is not (3x+2)⁵/5 + C; it must be (3x+2)⁵/(3×5) + C.
另一个常见错误是积分幂函数时漏掉新幂次的分母：∫ (3x+2)⁴ dx 不是 (3x+2)⁵/5 + C；正确答案必须是 (3x+2)⁵/(3×5) + C。
Always check your result by differentiation. If you differentiate your answer, you should recover the original integrand exactly. This is the fastest way to catch sign and factor errors.
始终通过求导来检验你的结果。如果对答案求导后能完全恢复原被积函数，就说明积分正确。这是发现符号和系数错误的最快方法。
For definite integrals, if you change variables, do not mix original x limits with u substitution. Change the limits or clearly return to x before substituting limits.
对于定积分，如果进行了换元，不要把原来的 x 上下限直接用于 u。要么替换上下限，要么在代回 x 后再代入原来的上下限。
11. Exam-Style Worked Examples | 考试风格例题
Example 1: Find ∫ (4x − 1)⁷ dx.
例题 1：求 ∫ (4x − 1)⁷ dx。
∫ (4x − 1)⁷ dx = (4x − 1)⁸ / (4 × 8) + C = (4x − 1)⁸ / 32 + C
Example 2: Find ∫ cos(3x + π/4) dx.
例题 2：求 ∫ cos(3x + π/4) dx。
∫ cos(3x + π/4) dx = (1/3) sin(3x + π/4) + C
Example 3: Evaluate ∫₀^(π/6) sin(2x) dx.
例题 3：计算 ∫₀^(π/6) sin(2x) dx。
∫ sin(2x) dx = −(1/2) cos(2x) + C
[−(1/2) cos(2x)] from 0 to π/6 = −(1/2) cos(π/3) + (1/2) cos(0) = −1/4 + 1/2 = 1/4
Example 4: Find ∫ 4/(5x + 2) dx.
例题 4：求 ∫ 4/(5x + 2) dx。
∫ 4/(5x + 2) dx = (
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