Integration Techniques: Substitution, By Parts and Applications | Edexcel A-Level 数学积分技巧：换元法、分部积分与应用

📚 Integration Techniques: Substitution, By Parts and Applications | Edexcel A-Level 数学积分技巧：换元法、分部积分与应用

Integration is one of the largest and most technique-driven topics in Edexcel A-Level Mathematics. Whether you are preparing for Pure 1, Pure 2 or the full A-Level papers, you must be able to select the correct integration method quickly: direct reverse differentiation, substitution, integration by parts, or algebraic preparation such as partial fractions. This article walks through the core techniques with exam-style examples and common pitfalls.

积分是 Edexcel A-Level 数学中内容最多、也最看重方法的主题之一。无论你备考 Pure 1、Pure 2 还是完整的 A-Level 试卷，都必须能够快速选择正确的积分方法：直接逆用微分、换元法、分部积分法，或者先通过部分分式等进行代数整理。本文将结合考试常见题型和易错点，系统梳理核心技巧。

1. The Integration Toolkit: Standard Results | 积分工具箱：标准结果

Before learning advanced methods, you must memorise the fundamental antiderivatives. These are the reverse of the differentiation rules you already know.

在学习高级方法之前，你必须熟记基本原函数。它们是你已掌握的微分法则的逆运算。

∫ xⁿ dx = xⁿ⁺¹/(n+1) + C, n ≠ -1

∫ eˣ dx = eˣ + C   |   ∫ 1/x dx = ln|x| + C

∫ cos x dx = sin x + C   |   ∫ sin x dx = -cos x + C

∫ sec²x dx = tan x + C   |   ∫ tan x dx = ln|sec x| + C

For example, since d/dx(sin x) = cos x, it follows that ∫ cos x dx = sin x + C. The constant C is essential for indefinite integrals because differentiation removes any constant term.

例如，因为 d/dx(sin x) = cos x，所以 ∫ cos x dx = sin x + C。不定积分中的常数 C 必不可少，因为微分会消去任何常数项。

2. Integration by Substitution: The Core Idea | 换元积分法：核心思想

Substitution is used when an integrand contains a composite function multiplied by the derivative of the inner function. The idea is to let u equal the inner function so that the integral simplifies to a standard form.

当被积函数包含一个复合函数乘以内层函数的导数时，常使用换元积分法。其思想是令 u 等于内层函数，从而使积分化简为标准形式。

Step 1: Choose u = g(x) whose derivative g′(x) appears as a factor.
Step 2: Replace g′(x) dx by du.
Step 3: Integrate with respect to u.
Step 4: Substitute back to express the answer in terms of x.

步骤 1：选择 u = g(x)，使其导数 g′(x) 作为因式出现。
步骤 2：将 g′(x) dx 替换为 du。
步骤 3：对 u 积分。
步骤 4：回代，用 x 表示结果。

Find ∫ 2x(x²+1)³ dx.

求 ∫ 2x(x²+1)³ dx。

Let u = x²+1, then du/dx = 2x, so du = 2x dx.

∫ 2x(x²+1)³ dx = ∫ u³ du = u⁴/4 + C = (x²+1)⁴/4 + C

令 u = x²+1，则 du/dx = 2x，所以 du = 2x dx。于是 ∫ 2x(x²+1)³ dx = ∫ u³ du = u⁴/4 + C = (x²+1)⁴/4 + C。

3. Substitution in Definite Integrals: Changing Limits | 定积分换元：替换上下限

When evaluating a definite integral by substitution, you must either change the limits to the

Published by TutorHao | A-Level Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）