📚 Integration by Parts (Edexcel Pure 4, p224) | 分部积分法（Edexcel 纯数4，第224页）
Integration by parts is one of the core reverse‑chain‑rule techniques required for Edexcel A‑Level Pure Mathematics 4. It transforms the integral of a product of two functions into a simpler form, often reducing the integral to a standard result or allowing it to be solved systematically. This article derives the formula from the product rule of differentiation, explains the LIATE strategy for choosing u and dv, and works through a wide range of examples — from basic polynomials times exponentials to repeated integration, logarithmic and inverse trigonometric integrals, definite integrals, and the special case where the original integral reappears. Exam‑style practice problems are included at the end to build fluency.
分部积分法是 Edexcel A‑Level 纯数 4 的核心积分技巧之一，源自乘积函数的求导法则。它将两个函数乘积的积分转化为更简单的形式，使积分能够归约为标准结果或系统求解。本文从乘积法则推导分部积分公式，讲解选择 u 和 dv 的 LIATE 策略，并通过丰富例题展示多项式和指数函数、重复分部积分、对数与反三角积分、定积分以及积分回代自身等典型情境。文末配有考试风格练习题以巩固熟练度。
1. The Formula and Derivation | 公式与推导
th respect to x gives ∫ u dv/dx dx = uv − ∫ v du/dx dx. In compact differential notation this is written as ∫ u dv = uv − ∫ v du. The formula allows us to trade one integral for another that is hopefully easier to evaluate.
乘积法则指出，对于两个可导函数 u(x) 与 v(x)，有 d/dx (u v) = u dv/dx + v du/dx。移项并对 x 积分可得 ∫ u dv/dx dx = uv − ∫ v du/dx dx。用简洁的微分记号常写作 ∫ u dv = uv − ∫ v du。该公式使我们能够将一个积分替换为另一个更容易计算的积分。
∫ u dv = uv − ∫ v du
2. Choosing u and dv: The LIATE Rule | 选择 u 和 dv：LIATE 规则
Success with integration by parts depends heavily on the choice of u. A well‑established heuristic is the LIATE rule, which prioritises functions that simplify when differentiated. The acronym stands for Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential. Choose u as the function type that appears first in this list; everything else becomes part of dv.
分部积分成功的关键在于 u 的选择。一个成熟的启发式方法是 LIATE 规则，优先选择求导后能够简化的函数类型。该缩略词代表 对数 (Logarithmic)、反三角 (Inverse trigonometric)、代数 (Algebraic)、三角 (Trigonometric)、指数 (Exponential)。将列表中首先出现的函数类型选为 u，其余部分放入 dv。
| Priority | Function type | Examples |
| 1 | Logarithmic | ln x, logₐ x |
| 2 | Inverse trigonometric | arcsin x, arctan x |
| 3 | Algebraic | x, x², polynomials |
| 4 | Trigonometric | sin x, cos x, tan x |
| 5 | Exponential | eˣ, aˣ |
In Edexcel exams, many products involve ln x (put u = ln x), xⁿ eˣ = xⁿ), or x sin x (put u = x). Following LIATE almost always produces a simpler integral.
在 Edexcel 考试中，许多乘积涉及 ln x（令 u = ln x）、xⁿ eˣ（令 u = xⁿ）或 x sin x（令 u = x）。遵循 LIATE 几乎总能生成更简单的积分。
3. Basic Integration by Parts Examples | 基本分部积分例题
Example 1: Evaluate ∫ x eˣ dx. Let u = x (Algebraic, priority 3) and dv = eˣ dx (Exponential, priority 5). Then du = dx and v = eˣ. Substituting: ∫ x eˣ dx = x eˣ − ∫ eˣ dx = x eˣ − eˣ + C.
例1：求 ∫ x eˣ dx。令 u = x（代数，优先级3），dv = eˣ dx（指数，优先级5）。则 du = dx, v = eˣ。代入得：∫ x eˣ dx = x eˣ − ∫ eˣ dx = x eˣ − eˣ + C。
∫ x eˣ dx = x eˣ − eˣ + C
Example 2: Then du = dx, v = sin x. Hence ∫ x cos x dx = x sin x − ∫ sin x dx = x sin x + cos x + C.
例2：求 ∫ x cos x dx。令 u = x, dv = cos x dx，则 du = dx, v = sin x。因此 ∫ x cos x dx = x sin x − ∫ sin x dx = x sin x + cos x + C。
∫ x cos x dx = x sin x + cos x + C
4. Repeated Integration by Parts | 重复分部积分
nction, we apply integration by parts twice, reducing the power of x each time. For ∫ x² eˣ dx, first set u = x², dv = eˣ dx → ∫ x² eˣ dx = x² eˣ − ∫ 2x eˣ dx. Then apply integration by parts to the new integral ∫ 2x eˣ dx using u = 2x, dv = eˣ dx, which yields 2x eˣ − 2eˣ. Combining, we get x² eˣ − 2x eˣ + 2eˣ + C.
当被积函数含有 x² 或更高次幂乘以指数或三角函数时，需要两次应用分部积分，每次降低 x 的幂次。例如 ∫ x² eˣ dx，先设 u = x², dv = eˣ dx → ∫ x² eˣ dx = x² eˣ − ∫ 2x eˣ dx。然后对新积分 ∫ 2x eˣ dx 用 u = 2x, dv = eˣ dx 进行分部积分，得到 2x eˣ − 2eˣ。合并得 x² eˣ − 2x eˣ + 2eˣ + C。
∫ x² eˣ dx = eˣ (x² − 2x + 2) + C
Example 3: sin x dx = 2x sin x + 2 cos x. Result: ∫ x² sin x dx = −x² cos x + 2x sin x + 2 cos x + C.
例3：求 ∫ x² sin x dx。第一步 u = x², dv = sin x dx → ∫ x² sin x dx = −x² cos x + ∫ 2x cos x dx。对 ∫ 2x cos x dx 令 u = 2x, dv = cos x dx → 2x sin x − ∫ 2 sin x dx = 2x sin x + 2 cos x。结果为 ∫ x² sin x dx = −x² cos x + 2x sin x + 2 cos x + C。
5. Integration by Parts Leading Back to the Original Integral | 回归原积分的分部积分
For products of exponentials and trigonometric functions, like ∫ eˣ n x dx, two applications of integration by parts return the original integral on the right‑hand side, allowing us to solve for it algebraically. Let I = ∫ eˣ sin x dx. First, u = sin x, dv = eˣ dx ⇒ I = eˣ sin x − ∫ eˣ cos x dx. Now apply integration by parts to ∫ eˣ cos x dx with u = cos x, dv = eˣ dx → eˣ cos x + ∫ eˣ sin x dx = eˣ cos x + I. Substituting back: I = eˣ sin x − (eˣ cos x + I) ⇒ I = eˣ sin x − eˣ cos x − I ⇒ 2I = eˣ (sin x − cos x). Hence I = ½ eˣ (sin x − cos x) + C.
对指数与三角函数的乘积，如 ∫ eˣ sin x dx，两次分部积分后等号右侧会出现原积分，从而可代数求解。设 I = ∫ eˣ sin x dx。首先 u = sin x, dv = eˣ dx ⇒ I = eˣ sin x − ∫ eˣ cos x dx。再对 ∫ eˣ cos x dx 使用分部积分，令 u = cos x, dv = eˣ dx → eˣ cos x + ∫ eˣ sin x dx = eˣ cos x + I。代回得 I = eˣ sin x − (eˣ cos x + I) ⇒ 2I = eˣ (sin x − cos x)，故 I = ½ eˣ (sin x − cos x) + C。
∫ eˣ sin x dx = ½ eˣ (sin x − cos x) + C
6. Definite Integration by Parts | 定积分的分部积分法
nite integral. Example: Evaluate ∫₁² ln x dx. This is a classic case where ln x is best chosen as u, even though dv = dx seems pure. u = ln x, dv = dx ⇒ = x. Then ∫₁² ln x dx = [x ln x]₁² − ∫₁² 1 dx = (2 ln 2 − 1 ln 1) − (2 − 1) = 2 ln 2 − 1.
对于定积分，公式为 ∫ₐᵇ u dv = [uv]ₐᵇ − ∫ₐᵇ v du。先计算 uv 在上下限的值，再求解剩下的定积分。例如：求 ∫₁² ln x dx。这是典型例子，最好选 ln x 为 u，尽管 dv = dx 看似简单。u = ln x, dv = dx ⇒ du = 1/x dx, v = x。则 ∫₁² ln x dx = [x ln x]₁² − ∫₁² 1 dx = (2 ln 2 − 1 ln 1) − (2 − 1) = 2 ln 2 − 1。
∫₁² ln x dx = 2 ln 2 − 1
Another common definite integral uses repeated integration by parts, e.g. ∫₀¹ x² eˣ dx. Applying the indefinite result obtained earlier: [eˣ (x² − 2x + 2)]₀¹ = e¹(1 −2 +2) − e⁰(0 −0 +2) = e·1 − 2 = e − 2.
另一个常见的定积分需重复分部积分，如 ∫₀¹ x² eˣ dx。代入之前的不定积分结果：[eˣ (x² − 2x + 2)]₀¹ = e¹(1 −2 +2) − e⁰(0 −0 +2) = e − 2。
7. Integration by Parts with Logarithmic Functions | 涉及对数函数的分部积分
its derivative 1/x simplifies the integrand. Example: ∫ x ln x dx. Set u = ln x, dv = x dx ⇒ n ∫ x ln x dx = ½ x² ln x − ∫ ½ x²·(1/x) dx = ½ x² ln x − ½ ∫ x dx = ½ x² ln x − ¼ x² + C.
含有 ln x 的积分几乎总是令 u = ln x，因为其导数 1/x 可简化被积函数。例：∫ x ln x dx。设 u = ln x, dv = x dx ⇒ du = 1/x dx, v = ½ x²。则 ∫ x ln x dx = ½ x² ln x − ∫ ½ x²·(1/x) dx = ½ x² ln x − ½ ∫ x dx = ½ x² ln x − ¼ x² + C。
∫ x ln x dx = ½ x² ln x − ¼ x² + C
n x, dv = dx, giving x ln x − x + C. This is a standard result frequently quoted in exams.
对于 ∫ ln x dx 本身，可写为 ∫ 1·ln x dx 并选 u = ln x, dv = dx，得 x ln x − x + C。这是考试中常直接引用的标准结果。
8. Integration by Parts with Inverse Trigonometric Functions | 涉及反三角函数的分部积分
dx. Write as ∫ 1·arctan x dx, let u = arctan x, dv = dx → du = 1/(1 + x²) dx, v = x. Hence ∫ arctan x dx = x arctan x − ∫ x/(1 + x²) dx. The remaining integral can be solved via substitution (w = 1 + x²) or recognition, giving −½ ln |1 + x²|. Final answer: x arctan x − ½ ln(1 + x²) + C.
反三角函数如 arctan x 或 arcsin x 同样适用 LIATE 规则，u 选用反三角函数。例：∫ arctan x dx。写为 ∫ 1·arctan x dx，令 u = arctan x, dv = dx → du = 1/(1 + x²) dx, v = x。因此 ∫ arctan x dx = x arctan x − ∫ x/(1 + x²) dx。剩余积分可用代换 w = 1 + x² 或直接识别，得到 −½ ln |1 + x²|。最终答案为 x arctan x − ½ ln(1 + x²) + C。
∫ arctan x dx = x arctan x − ½ ln(1 + x²) + C
9. Common Mistakes and Tips | 常见错误与技巧
mplifies the product. When integrating dv, include the constant of integration only in the final step for indefinite integrals. For definite integrals, evaluate uv at the limits before subtracting the new integral. Also, be careful with signs when integrating trigonometric functions; a common slip is missing a minus sign when integrating sin x (v = −cos x).
考试常见错误包括 u 选择不当、遗漏 v du 项或丢失常数。务必检查 u 求导后是否确实简化乘积。积分不定积分时，仅在最后步骤加上积分常数。对于定积分，先计算 uv 在上下限的值再减去新积分。另外，处理三角函数的积分要注意符号；常见失误是积分 sin x 时漏掉负号（v = −cos x）。
Tip: : If the new integral looks more complicated than the original, swap your choice of u and dv. Sometimes the non‑LIATE choice works when LIATE fails – for instance, ∫ x³ ex² dx is better handled with substitution, but if integration by parts is forced, an unconventional choice may help.
技巧：若新积分比原积分更复杂，交换 u 和 dv 的选择。有时非 LIATE 选择反而奏效——例如 ∫ x³ ex² dx 用代换法更佳，但如果必须使用分部积分，非常规选择可能有效。
10. Practice Problem Set | 练习题集
Apply the techniques above to the following exam‑style integrals. Answers are provided for self‑checking.
用上述技巧求解下列考试风格积分题，附答案供自查。
- 1. ∫ x e²ˣ dx
Answer: (½ x e²ˣ − ¼ e²ˣ) + C - 2. ∫ x² e⁻ˣ dx
Answer: −e⁻ˣ (x² + 2x + 2) + C - 3. ∫ x sin 2x dx
Answer: −½ x cos 2x + ¼ sin 2x + C - 4. ∫ ln(2x) dx
Answer: x ln(2x) − x + C - 5. ∫ eˣ cos x dx
Answer: ½ eˣ (cos x + sin x) + C - 6. ∫₀¹ x eˣ dx
Answer: 1 - 7. ∫ x arctan x dx
Answer: ½ (x² + 1) arctan x − ½ x + C - 8. ∫ (ln x)/x² dx
Answer: −(ln x + 1)/x + C
Work through each problem by clearly stating u and dv, writing the intermediate steps, and verifying by differentiation. Repetition builds the pattern recognition essential for the Pure 4 exam.
逐一练习时，明确写出 u 和 dv，记录中间步骤，并通过求导验证。重复训练有助于形成 Pure 4 考试所必需的模式识别能力。
Published by TutorHao | Pure Mathematics 4 Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图