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A-Level数学积分技巧:掌握∫(ax+b)ⁿdx与指数函数积分 | Integration of (ax+b)

2026-05-10 04:29 · 本文部分内容由AI辅助创作

Integration of functions of the form (ax+b)ⁿ 是 A-Level 数学纯数模块中的核心技能。看似简单——”把幂次加 1,除以新幂次和 x 系数”——但考试中频繁以变形形式出现,是许多考生的失分重灾区。本文系统梳理 (ax+b) 类型积分的五大变体,配套练习题解析,助你彻底攻克这一考点。

Integrating functions of the form (ax+b)ⁿ is a cornerstone skill in A-Level Pure Mathematics. The rule seems simple — “add 1 to the power, divide by the new power and the coefficient of x” — but exam questions routinely disguise it, making this a common pitfall. This article systematically covers five key variants of (ax+b) integration with worked examples to help you master the topic.

📐 核心公式 | The Core Formula

对于 n ≠ −1 的情况:
∫(ax+b)ⁿ dx = (ax+b)ⁿ⁺¹ / [a(n+1)] + C
这就是”反向链式法则”(reverse chain rule)的直接应用。关键点:不仅要除以新幂次 (n+1),还要除以内部函数 ax+b 的导数 a。忘记除 a 是最常见的错误。

For n ≠ −1:
∫(ax+b)ⁿ dx = (ax+b)ⁿ⁺¹ / [a(n+1)] + C
This is a direct application of the reverse chain rule. The critical point: not only must you divide by the new power (n+1), you must also divide by a, the derivative of the inner function ax+b. Forgetting to divide by a is the single most common mistake.

🔢 知识点一:基本幂函数积分 | Basic Power Integration

例 1:∫(2x + 5)³ dx
解:n = 3, a = 2 → ∫(2x+5)³ dx = (2x+5)⁴ / (2 × 4) = (2x+5)⁴ / 8 + C

Example 1: ∫(2x + 5)³ dx
Solution: n = 3, a = 2 → ∫(2x+5)³ dx = (2x+5)⁴ / (2 × 4) = (2x+5)⁴ / 8 + C

例 2:∫(8 − 5x)⁴ dx
注意:这里 a = −5(不是 5!)→ ∫(8−5x)⁴ dx = (8−5x)⁵ / (−5 × 5) = −(8−5x)⁵ / 25 + C

Example 2: ∫(8 − 5x)⁴ dx
Watch out: here a = −5 (not 5!) → ∫(8−5x)⁴ dx = (8−5x)⁵ / (−5 × 5) = −(8−5x)⁵ / 25 + C

📏 知识点二:分母形式的积分 | Integrating Fractions with Linear Denominators

这是 n = −1 的特殊情况——公式不再适用!当被积函数为 1/(ax+b) 时:
∫ 1/(ax+b) dx = (1/a) · ln|ax+b| + C

This is the n = −1 special case — the power formula breaks! For integrands of the form 1/(ax+b):
∫ 1/(ax+b) dx = (1/a) · ln|ax+b| + C

例 3:∫ 1/(4x−3) dx = (1/4) · ln|4x−3| + C
例 4:∫ 3/(2x+1) dx = (3/2) · ln|2x+1| + C (常数因子提出后再积分)

Example 3: ∫ 1/(4x−3) dx = (1/4) · ln|4x−3| + C
Example 4: ∫ 3/(2x+1) dx = (3/2) · ln|2x+1| + C (factor out the constant, then integrate)

⚡ 知识点三:指数函数积分 | Integrating Exponential Functions with Linear Exponents

指数函数 eax+b 的积分也遵循反向链式法则:
∫ eax+b dx = (1/a) · eax+b + C

Integrating eax+b also follows the reverse chain rule:
∫ eax+b dx = (1/a) · eax+b + C

例 5:∫ e2x−3 dx = (1/2) e2x−3 + C
例 6:∫ 5e7−3t dt = 5 × (−1/3) e7−3t = −(5/3) e7−3t + C

Example 5: ∫ e2x−3 dx = (1/2) e2x−3 + C
Example 6: ∫ 5e7−3t dt = 5 × (−1/3) e7−3t = −(5/3) e7−3t + C

📐 知识点四:定积分应用 — 求曲线下方面积 | Definite Integrals — Area Under a Curve

定积分的核心步骤:先求不定积分 → 代入上下限 → 相减。关键陷阱:当 a 为负数且幂次为偶数时,符号处理需格外小心。

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例 7:计算 ∫₀¹ (3x+1)² dx
解:F(x) = (3x+1)³ / (3×3) = (3x+1)³ / 9
F(1) = 64/9, F(0) = 1/9 → 结果 = 63/9 = 7

Example 7: Evaluate ∫₀¹ (3x+1)² dx
Solution: F(x) = (3x+1)³ / 9, F(1) = 64/9, F(0) = 1/9 → Result = 7

例 8:求 y = 12/(2x+1)³ 在 x=0 到 x=1 之间与坐标轴围成的面积
y = 12(2x+1)⁻³ → ∫ 12(2x+1)⁻³ dx = 12 × (2x+1)⁻² / (−2×2) = −3(2x+1)⁻² + C
面积 = [−3/(2x+1)²]₀¹ = −3/9 − (−3/1) = −1/3 + 3 = 8/3

Example 8: Area bounded by y = 12/(2x+1)³, x=0, x=1, and axes.
y = 12(2x+1)⁻³ → ∫ 12(2x+1)⁻³ dx = 12 × (2x+1)⁻² / (−2×2) = −3(2x+1)⁻² + C
Area = [−3/(2x+1)²]₀¹ = −3/9 − (−3/1) = 8/3

🧪 知识点五:已知导数求原函数 | Finding f(x) from f'(x)

当题目给出 f'(x) 和曲线上的一个点时,先积分得到含常数 C 的 f(x),再代入已知点求 C。这是考试中最容易拿分也最容易丢分的题型——积分正确但忘记解 C,至少扣 2 分。

When given f'(x) and a point on the curve: first integrate to get f(x) with unknown constant C, then substitute the point to find C. This is simultaneously the easiest-to-score and easiest-to-lose-marks question type — correct integration followed by forgetting to solve for C costs at least 2 marks.

例 9:f'(x) = 8(2x−3)³, 曲线过点 (2, 6), 求 f(x)
f(x) = ∫ 8(2x−3)³ dx = 8 × (2x−3)⁴ / (2×4) = (2x−3)⁴ + C
代入 (2, 6):6 = (4−3)⁴ + C → C = 5 → f(x) = (2x−3)⁴ + 5

Example 9: f'(x) = 8(2x−3)³, curve passes through (2, 6). Find f(x).
f(x) = ∫ 8(2x−3)³ dx = (2x−3)⁴ + C, then 6 = 1⁴ + C → C = 5 → f(x) = (2x−3)⁴ + 5

💡 学习建议 | Study Tips


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