Integration of Functions of the Form f(ax+b) | 形如f(ax+b)型函数的积分法

📚 Integration of Functions of the Form f(ax+b) | 形如f(ax+b)型函数的积分法

In calculus, we often encounter integrals whose integrand is a composite of a known function f with a linear expression ax+b. Typical examples are ∫sin(3x+2) dx, ∫(2x−1)⁵ dx, and ∫e^{4x+7} dx. These integrals can be evaluated quickly if we apply a simple adjustment factor 1/a. This article explains the general rule, proves it by substitution, and provides worked examples suitable for A-Level mathematics.

在微积分中，我们经常会遇到被积函数是某个已知函数 f 与线性表达式 ax+b 复合而成的积分，例如 ∫sin(3x+2) dx、∫(2x−1)⁵ dx 和 ∫e^{4x+7} dx。只要应用一个简单的修正因子 1/a，这些积分就能快速求出。本文将解释这一通用法则，通过换元法给出证明，并提供适合 A-Level 数学的例题。

1. The General Rule | 一般法则

Suppose F is an antiderivative of f, so that F′ = f. Then for any constants a and b with a ≠ 0, we have:

设 F 是 f 的一个原函数，即 F′ = f。那么对于任意常数 a 和 b，且 a ≠ 0，有：

∫ f(ax+b) dx = (1/a) F(ax+b) + C

The factor 1/a appears because differentiating F(ax+b) with respect to x gives a·f(ax+b) by the chain rule. Therefore, to integrate f(ax+b), we first integrate f to obtain F, then replace the variable by ax+b, and finally divide by a.

出现因子 1/a 的原因是：根据链式法则，对 F(ax+b) 关于 x 求导得到 a·f(ax+b)。因此，要积分 f(ax+b)，我们先对 f 积分得到 F，然后将变量替换为 ax+b，最后除以 a。

2. Power Functions (ax+b)ⁿ | 幂函数 (ax+b)ⁿ 的积分

For n ≠ −1, the rule gives:

当 n ≠ −1 时，该法则给出：

∫ (ax+b)ⁿ dx = (ax+b)ⁿ⁺¹ / [a(n+1)] + C

Notice that the exponent increases by 1, and we divide by a(n+1). For example, ∫(2x+3)⁴ dx = (2x+3)⁵ / (2·5) + C = (2x+3)⁵ / 10 + C.

注意指数增加 1，并且要除以 a(n+1)。例如，∫(2x+3)⁴ dx = (2x+3)⁵ / (2·5) + C = (2x+3)⁵ / 10 + C。

If the power is negative, the same rule works as long as n ≠ −1. For instance, ∫1/(3x−1)² dx = ∫(3x−1)⁻² dx = (3x−1)⁻¹ / [3·(−1)] + C = −1/[3(3x−1)] + C.

如果指数为负，只要 n ≠ −1，同样的法则仍然适用。例如，∫1/(3x−1)² dx = ∫(3x−1)⁻² dx = (3x−1)⁻¹ / [3·(−1)] + C = −1/[3(3x−1)] + C。

3. Exponential Functions e^{ax+b} | 指数函数 e^{ax+b} 的积分

Since the derivative of e^{ax+b} is a·e^{ax+b}, reversing the chain rule gives:

因为 e^{ax+b} 的导数是 a·e^{ax+b}，逆用链式法则得到：

∫ e^{ax+b} dx = (1/a) e^{ax+b} + C

For example, ∫e^{5x−2} dx = (1/5)e^{5x−2} + C. This formula also covers functions like ∫2^{3x+1} dx if we rewrite 2^{3x+1} = e^{(3x+1)ln2}; the result is 2^{3x+1}/(3ln2) + C.

例如，∫e^{5x−2} dx = (1/5)e^{5x−2} + C。该公式也适用于 ∫2^{3x+1} dx 等函数，只需将 2^{3x+1} 改写为 e^{(3x+1)ln2}，结果为 2^{3x+1}/(3ln2) + C。

4. Trigonometric Functions | 三角函数

The linear adjustment applies directly to standard trigonometric integrals. The following table lists the most useful results:

线性修正因子可直接应用于标准三角积分。下表列出了最常用的结果：

Integral \| 积分	Result \| 结果
∫ 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
∫ csc²(ax+b) dx	−(1/a) cot(ax+b) + C
∫ sec(ax+b)tan(ax+b) dx	(1/a) sec(ax+b) + C
∫ csc(ax+b)cot(ax+b) dx	−(1/a) csc(ax+b) + C

As a check, differentiate (1/a)sin(ax+b) to obtain cos(ax+b), confirming the formula for ∫cos(ax+b) dx.

作为验证，对 (1/a)sin(ax+b) 求导得到 cos(ax+b)，从而确认 ∫cos(ax+b) dx 的公式正确。

5. Reciprocal Functions 1/(ax+b) | 倒数函数 1/(ax+b) 的积分

The case n = −1 is special because the power rule fails. The correct result uses the natural logarithm:

n = −1 的情形很特殊，因为幂法则失效。正确结果需要使用自然对数：

∫ 1/(ax+b) dx = (1/a) ln|ax+b| + C

For example, ∫1/(2x+3) dx = (1/2)ln|2x+3| + C. The absolute value is important because the logarithm is only defined for positive numbers, while ax+b may be negative. Some exam boards ignore absolute values when ax+b is known to be positive; always follow the notation required by your syllabus.

例如，∫1/(2x+3) dx = (1/2)ln|2x+3| + C。绝对值很重要，因为对数仅在正数时有定义，而 ax+b 可能为负。某些考试局在 ax+b 已知为正时忽略绝对值；请务必遵循你所在教学大纲的要求。

6. Substitution Proof: Let u = ax+b | 换元法证明：令 u = ax+b

The general rule can be proved rigorously by substitution. Let u = ax+b. Then du/dx = a, so dx = du/a. Therefore:

一般法则可以通过换元法严格证明。令 u = ax+b，则 du/dx = a，所以 dx = du/a。因此：

∫ f(ax+b) dx = ∫ f(u) · (1/a) du = (1/a)∫ f(u) du = (1/a)F(u) + C = (1/a)F(ax+b) + C

This proof shows exactly where the factor 1/a comes from: it is the reciprocal of the derivative of u = ax+b. The substitution u = ax+b is especially useful when the integrand is more complicated than a simple standard form.

这个证明清楚地展示了因子 1/a 的来源：它是 u = ax+b 的导数的倒数。当被积函数比简单的标准形式更复杂时，换元 u = ax+b 尤其有用。

7. Definite Integrals: Changing the Limits | 定积分：更换积分限

For a definite integral, we can either integrate first and then substitute the original limits, or change the limits to u-values. If x runs from p to q, then u = ax+b runs from ap+b to aq+b. Hence:

对于定积分，我们可以先积分再用原来的上下限代入，也可以将积分限换成 u 的值。若 x 从 p 到 q，则 u = ax+b 从 ap+b 到 aq+b。因此：

∫ₚᵠ f(ax+b) dx = (1/a)∫_{ap+b}^{aq+b} f(u) du

Example: Evaluate ∫₀¹ (2x+1)³ dx. Let u = 2x+1. When x = 0, u = 1; when x = 1, u = 3. Then the integral becomes (1/2)∫₁³ u³ du = (1/2)[u⁴/4]₁³ = (1/8)(81 − 1) = 10.

例：计算 ∫₀¹ (2x+1)³ dx。令 u = 2x+1。当 x = 0 时，u = 1；当 x = 1 时，u = 3。于是积分变为 (1/2)∫₁³ u³ du = (1/2)[u⁴/4]₁³ = (1/8)(81 − 1) = 10。

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

Forgetting to divide by a. The most frequent error is writing ∫cos(3x) dx = sin(3x) + C. The correct answer is (1/3)sin(3x) + C.

忘记除以 a。最常见的错误是把 ∫cos(3x) dx 写成 sin(3x) + C。正确答案是 (1/3)sin(3x) + C。
Using the power rule when n = −1. ∫(ax+b)⁻¹ dx is not (ax+b)⁰/0; it must be treated as a logarithm.

在 n = −1 时使用幂法则。∫(ax+b)⁻¹ dx 不是 (ax+b)⁰/0；必须按对数处理。
Dropping the absolute value in logarithmic results, which can lead to domain errors.

在对数结果中忽略绝对值，这可能导致定义域错误。
Forgetting to change the limits when using substitution in a definite integral.

在定积分中使用换元法时忘记更换积分限。
Sign errors when a is negative. For example, ∫1/(1−x) dx = −ln|1−x| + C, not +ln|1−x| + C.

当 a 为负时出现符号错误。例如，∫1/(1−x) dx = −ln|1−x| + C，而不是 +ln|1−x| + C。

9. Worked Example: Polynomial Combination | 综合例题：多项式组合

Evaluate ∫(3x+2)⁴ + 5/(3x+2) dx.

计算 ∫(3x+2)⁴ + 5/(3x+2) dx。

We integrate term by term. For the first term, n = 4 and a = 3, so ∫(3x+2)⁴ dx = (3x+2)⁵/(3·5) = (3x+2)⁵/15. For the second term, ∫5/(3x+2) dx = 5·(1/3)ln|3x+2| = (5/3)ln|3x+2|. Therefore the full integral is:

我们逐项积分。对于第一项，n = 4，a = 3，所以 ∫(3x+2)⁴ dx = (3x+2)⁵/(3·5) = (3x+2)⁵/15。对于第二项，∫5/(3x+2) dx = 5·(1/3)ln|3x+2| = (5/3)ln|3x+2|。因此完整积分为：

(3x+2)⁵/15 + (5/3)ln|3x+2| + C

This example shows how the linear rule combines smoothly with algebraic simplification.

这个例子说明线性法则如何与代数化简顺利结合。

10. Worked Example: Trigonometric with Definite Limits | 综合例题：三角函数的定积分

Evaluate ∫₀^{π/6} cos(3x − π/4) dx.

计算 ∫₀^{π/6} cos(3x − π/4) dx。

Using the formula for ∫cos(ax+b) dx with a = 3, an antiderivative is (1/3)sin(3x − π/4). Now substitute the limits:

利用 ∫cos(ax+b) dx 的公式，a = 3，一个原函数是 (1/3)sin(3x − π/4)。现在代入上下限：

(1/3)[sin(π/2 − π/4) − sin(0 − π/4)] = (1/3)[sin(π/4) − sin(−π/4)] = (1/3)(√2/2 + √2/2) = √2/3

Notice that sin(−π/4) = −√2/2, so the two terms add rather than cancel.

注意 sin(−π/4) = −√2/2，因此两项是相加而不是相消。

11. Practice Problems | 练习

Try the following questions before checking the answers below.

请先尝试完成以下题目，再核对下方答案。

∫(4x−1)³ dx

∫(4x−1)³ dx
∫e^{2x+5} dx

∫e^{2x+5} dx
∫sin(5x) dx

∫sin(5x) dx
∫₀¹ 1/(2x+1)² dx

∫₀¹ 1/(2x+1)² dx
∫sec²(3x−1) dx

∫sec²(3x−1) dx

Answers: 1) (4x−1)⁴/16 + C. 2) (1/2)e^{2x+5} + C. 3) −(1/5)cos(5x) + C. 4) [−1/(2(2x+1))]₀¹ = −1/6 + 1/2 = 1/3. 5) (1/3)tan(3x−1) + C.

答案：1) (4x−1)⁴/16 + C。2) (1/2)e^{2x+5} + C。3) −(1/5)cos(5x) + C。4) [−1/(2(2x+1))]₀¹ = −1/6 + 1/2 = 1/3。5) (1/3)tan(3x−1) + C。

12. Summary | 总结

To integrate a function of the form f(ax+b), integrate f normally to get F, replace the variable by ax+b, and divide by a. This simple rule handles powers, exponentials, trigonometric functions, and reciprocals. It can always be justified by the substitution u = ax+b, and for definite integrals the limits can be changed accordingly. Mastery of this technique saves time and reduces errors in exams.

要积分形如 f(ax+b) 的函数，先正常对 f 积分得到 F，将变量替换为 ax+b，再除以 a。这条简单法则适用于幂函数、指数函数、三角函数和倒数函数。它总可以用换元 u = ax+b 来证明，对于定积分可以相应更换积分限。熟练掌握这一技巧能节省时间，并在考试中减少错误。

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Integration of Functions of the Form f(ax+b) | 形如f(ax+b)型函数的积分法