📚 Mastering Integration by Parts for Edexcel A-Level Maths | Edexcel A-Level 数学：分部积分法精讲
Integration by parts is one of the most powerful techniques in the A-Level toolkit, allowing us to integrate products of functions that standard substitution cannot handle. Whether tackling exam questions on definite integrals or solving differential equations with integrating factors, a solid grasp of this method is essential. In this article, we break down the formula, sharpen your intuition for choosing u and dv, and explore recurring patterns like the 'boomerang' integral—all with the rigour needed for Edexcel success.
分部积分法是 A-Level 数学工具箱中最强大的技巧之一，它能帮我们处理标准换元法无法解决的两函数乘积的积分。无论是应对定积分考题，还是使用积分因子求解微分方程，扎实掌握这个方法都至关重要。在本文中，我们将拆解公式，强化你对选择 u 和 dv 的直觉，并探索像“回旋镖”积分这样的循环模式——全部以 Edexcel 所需的严谨性呈现。
1. The Formula and Its Origins | 公式及其来源
nging and integrating both sides gives the familiar formula: ∫ u (dv/dx) dx = uv − ∫ v (du/dx) dx, or more compactly ∫ u dv = uv − ∫ v du.
分部积分法来自微分乘法法则。若 u 和 v 都是 x 的函数，则 d(uv)/dx = u(dv/dx) + v(du/dx)。移项并对两边积分即得熟悉的公式：∫ u (dv/dx) dx = uv − ∫ v (du/dx) dx，或更紧凑地写为 ∫ u dv = uv − ∫ v du。
iginal. In the Edexcel specification, you are expected to recognise when to apply integration by parts and to execute it efficiently, sometimes more than once in a single problem.
该方法成功的关键在于明智地选择 u 和 dv：新积分 ∫ v du 应比原积分更简单。在 Edexcel 大纲中，考生需能识别何时使用分部积分法并高效执行，有时一道题需要多次运用该法。
∫ u dv = uv − ∫ v du
2. Choosing u: The LIATE Rule of Thumb | 选择 u：LIATE 经验法则
A helpful mnemonic for picking u is LIATE, which ranks function types from most preferable as u to least: Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential. The idea is that differentiating u should eventually simplify the integrand, while dv should be easy to integrate.
挑选 u 的一个实用助记法是 LIATE，它将函数类型按优先作为 u 的顺序排列：对数函数、反三角函数、代数函数、三角函数、指数函数。其思想是：对 u 求导最终应能化简被积函数，而 dv 则应易于积分。
For example, in ∫ x eˣ dx, Algebraic (x) comes before Exponential (eˣ), so let u = x and dv = eˣ c (ln x) comes before Algebraic (1), so set u = ln x, dv = 1 dx. LIATE is not a rigid rule—exceptions exist—but it significantly reduces guesswork in timed exam conditions.
例如，在 ∫ x eˣ dx 中，代数函数 (x) 在指数函数 (eˣ) 之前，故设 u = x，dv = eˣ dx。在 ∫ ln x dx 中，可视为 ∫ (ln x)·1 dx；对数函数 (ln x) 在代数函数 (1) 之前，故设 u = ln x，dv = 1 dx。LIATE 并非硬性规则——存在例外——但它能在限时考试中显著减少试错。
3. Integrating ln x: A Classic First Example | 积分 ln x：经典首例
pplying the formula: ∫ ln x dx = x ln x − ∫ x·(1/x) dx = x ln x − ∫ 1 dx = x ln x − x + C.
要积分 ln x，我们写成 ∫ ln x dx = ∫ (ln x)·1 dx。选择 u = ln x（求导得 1/x），dv = 1 dx（积分得 x）。套用公式：∫ ln x dx = x ln x − ∫ x·(1/x) dx = x ln x − ∫ 1 dx = x ln x − x + C。
This neat result is frequently tested, either directly or as part of a larger integral involving a logarithmic factor. Remember that the constant of integration is compulsory for indefinite integrals; in definite integrals, evaluate uv between the limits before subtracting the second integral.
这个简洁的结果常被考查，无论是直接出现还是作为一个包含对数因式的较大积分的一部分。记住，对于不定积分，积分常数是必需的；在定积分中，先计算 uv 在上下限之间的值，再减去第二个积分。
4. Integrating x eˣ: Product of Algebraic and Exponential | 积分 x eˣ：代数函数与指数函数的乘积
Consider ∫ x eˣ dx. By LIATE, Algebraic (x) takes priority for u. Let u = x, so du = dx. Then dv = eˣ dx gives v = eˣ. Substituting: ∫ x eˣ dx = x eˣ − ∫ eˣ dx = x eˣ − eˣ + C = eˣ(x − 1) + C.
考虑 ∫ x eˣ dx。根据 LIATE，代数函数 (x) 优先作为 u。设 u = x，则 du = dx。然后 dv = eˣ dx 得 v = eˣ。代入得：∫ x eˣ dx = x eˣ − ∫ eˣ dx = x eˣ − eˣ + C = eˣ(x − 1) + C。
If the algebraic part is of higher degree, say x² eˣ, we apply integration by parts twice, each time reducing the power of x. This 'repeated integration by parts' pattern is a common exam theme, where systematic working and clear notation are vital to avoid errors.
若代数部分为更高次数，比如 x² eˣ，则需两次运用分部积分法，每次降低 x 的幂次。这种“重复分部积分”模式是常见考点，系统化的解题步骤和清晰的符号对避免错误至关重要。
5. Integrating x sin x and x cos x: Trigonometric Encounters | 积分 x sin x 与 x cos x：遇见三角函数
For ∫ x sin x dx, LIATE still places Algebraic (x) before Trigonometric (sin x). Set u = x ⇒ du = dx, and dv = sin x dx ⇒ v = −cos x. Thus ∫ x sin x dx = −x cos x − ∫ (−cos x) dx = −x cos x + sin x + C.
对于 ∫ x sin x dx，LIATE 仍将代数函数 (x) 排在三角函数 (sin x) 之前。设 u = x ⇒ du = dx，dv = sin x dx ⇒ v = −cos x。于是 ∫ x sin x dx = −x cos x − ∫ (−cos x) dx = −x cos x + sin x + C。
For ∫ x cos x dx, set u = x, dv = cos x dx ⇒ efficiency, but always derive them from first principles in an exam to show full working.
对于 ∫ x cos x dx，设 u = x，dv = cos x dx ⇒ v = sin x。于是 ∫ x cos x dx = x sin x − ∫ sin x dx = x sin x + cos x + C。为效率起见，这两个结果都值得记住，但在考试中务必从基本原理推导以展示完整步骤。
6. The Boomerang Integral: eˣ sin x and eˣ cos x | “回旋镖”积分：eˣ sin x 与 eˣ cos x
Integrals of the form ∫ eˣ sin x dx or ∫ eˣ ause after two applications of integration by parts, the original integral reappears. This 'boomerang' effect allows us to solve for the integral algebraically.
形如 ∫ eˣ sin x dx 或 ∫ eˣ cos x dx 的积分很特别，因为经过两次分部积分后，原积分会重新出现。这种“回旋镖”效应使我们能用代数方法解出积分。
Let I = ∫ eˣ sin x dx. Choose u = sin x, dv = eˣ dx. Then du = cos x dx, v = eˣ. So I = eˣ sin x − ∫ eˣ cos x dx. Now let J = ∫ eˣ cos x dx. For J, set u = cos x, dv = eˣ dx ⇒ du = −sin x dx, v = eˣ. Then J = eˣ cos x − ∫ eˣ(−sin x) 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) ⇒ I = ½ eˣ(sin x − cos x) + C.
设 I = ∫ eˣ sin x dx。选择 u = sin x，dv = eˣ dx。则 du = cos x dx，v = eˣ。于是 I = eˣ sin x − ∫ eˣ cos x dx。现在设 J = ∫ eˣ cos x dx。对 J，设 u = cos x，dv = eˣ dx ⇒ du = −sin x dx，v = eˣ。则 J = eˣ cos x − ∫ eˣ(−sin x) dx = eˣ cos x + ∫ eˣ sin x dx = eˣ cos x + I。代回：I = eˣ sin x − (eˣ cos x + I) ⇒ I = eˣ sin x − eˣ cos x − I ⇒ 2I = eˣ(sin x − cos x) ⇒ I = ½ eˣ(sin x − cos x) + C。
the original integral saves time and avoids infinite loops. The same method works for ∫ eᵃˣ sin bx dx and ∫ eᵃˣ cos bx dx, giving more general results.
这种技巧极可能被考查；知道何时停下并解出原积分可以节省时间并避免无限循环。同样的方法适用于 ∫ eᵃˣ sin bx dx 和 ∫ eᵃˣ cos bx dx，得到更一般的结果。
7. Integration by Parts with Definite Integrals | 定积分的分部积分法
ated at the upper and lower limits, and the remaining integral is also evaluated over the same limits. It is often safer to compute the indefinite integral first (without limits) then substitute limits at the end, but the direct method is quicker if you track signs carefully.
当对定积分使用分部积分法时，公式变为 ∫ₐᵇ u dv = [uv]ₐᵇ − ∫ₐᵇ v du。uv 项在上下限处求值，剩余积分也在相同区间上求值。通常更安全的做法是先计算不定积分（不带上下限），最后再代入，但若仔细跟踪符号，直接法更快。
For example, evaluate ∫₀¹ x eˣ dx. Using u = x, dv = eˣ dx ⇒ du = dx, v = eˣ. Then ∫₀¹ x eˣ dx = [x eˣ]₀¹ − ∫₀¹ eˣ dx = (1·e¹ − 0·e⁰) − [eˣ]₀¹ = e − (e¹ − e⁰) = e − (e − 1) = 1. Always check that your final answer is reasonable—here the area under x eˣ from 0 to 1 is exactly 1.
例如，计算 ∫₀¹ x eˣ dx。利用 u = x，dv = eˣ dx ⇒ du = dx，v = eˣ。则 ∫₀¹ x eˣ dx = [x eˣ]₀¹ − ∫₀¹ eˣ dx = (1·e¹ − 0·e⁰) − [eˣ]₀¹ = e − (e¹ − e⁰) = e − (e − 1) = 1。务必检查最终答案是否合理——此处 x eˣ 从 0 到 1 下方的面积恰为 1。
8. Repeated Integration by Parts and Tabular Method | 重复分部积分与表格法
When facing ∫ xⁿ eᵃˣ ss: differentiate the power function repeatedly until it becomes zero, while integrating the exponential/trigonometric part the same number of times. Then combine terms with alternating signs.
当面对 ∫ xⁿ eᵃˣ dx 或 ∫ xⁿ sin(ax) dx 时，需要重复分部积分直到代数项消失。表格法（或 DI 法）可以简化过程：对幂函数重复求导直至为零，同时对指数/三角函数部分积分相同次数。然后用交替符号组合各项。
For ∫ x² eˣ dx, the table for u = x², dv = eˣ dx is:
| Differentiate u | Integrate dv |
|---|---|
| x² | eˣ |
| 2x | eˣ |
| 2 | eˣ |
| 0 | eˣ |
对于 ∫ x² eˣ dx，u = x²，dv = eˣ dx 的表格如下：对角线相乘并附上交替符号：+ x²·eˣ − 2x·eˣ + 2·eˣ，得 eˣ(x² − 2x + 2) + C。该方法可减少代数疏漏，但若被要求需加以证明；在 Edexcel 考试中，最好先展示至少一步完整的分部积分再使用它。
9. Integration by Parts for Inverse Trigonometric Functions | 反三角函数的分部积分
)·1 dx and set u = the inverse trig function, dv = 1 dx. The derivative of u will be algebraic, making the new integral manageable.
积分反三角函数如 arcsin x、arccos x 或 arctan x 的技巧与 ln x 相同：将被积函数写成 (反三角函数)·1 dx，并设 u = 反三角函数，dv = 1 dx。u 的导数将是代数函数，使得新积分易于处理。
For ∫ arctan x dx, let u = arctan x ⇒ du = 1/(1+x²) dx, dv = 1 dx ⇒ integral can be solved by substitution (let w = 1+x²) or recognition: ∫ x/(1+x²) dx = ½ ln|1+x²| + C. Hence the result is x arctan x − ½ ln(1+x²) + C.
对于 ∫ arctan x dx，设 u = arctan x ⇒ du = 1/(1+x²) dx，dv = 1 dx ⇒ v = x。则 ∫ arctan x dx = x arctan x − ∫ x/(1+x²) dx。剩余积分可用换元法（设 w = 1+x²）或识别：∫ x/(1+x²) dx = ½ ln|1+x²| + C。因此结果为 x arctan x − ½ ln(1+x²) + C。
10. Connecting to Integrating Factors in Differential Equations | 联系微分方程中的积分因子
In the Edexcel A-Level, integration by parts often appears when solving first-order linear differential equations using an integrating factor. After multiplying through by the integrating factor, the left-hand side becomes the derivative of a product, but evaluating the right-hand side integral may require parts.
在 Edexcel A-Level 中，当使用积分因子求解一阶线性微分方程时，常出现分部积分。乘以积分因子后，左边变成乘积的导数，但求解右边积分时可能需要分部积分。
2x sec x dx, which simplifies and might need parts. Recognising these cross-topic links strengthens your overall problem-solving ability and is a favourite of examiners testing synergy between calculus and differential equations.
例如，求解 dy/dx + y tan x = sin 2x 可能导出类似 ∫ sin 2x sec x dx 的积分，它化简后可能需要分部积分。识别这些跨主题联系能增强整体解题能力，这也是出题人喜爱考查微积分与微分方程协同作用的点。
11. Common Pitfalls and How to Avoid Them | 常见陷阱及如何避开
an the original, swap your choices. Another is forgetting the constant of integration or mishandling signs, especially when v involves a negative function like −cos x.
一个常见错误是错误识别 u 和 dv，导致积分更复杂。若 ∫ v du 看起来比原积分更糟，则交换选择。另一个是忘记积分常数或符号处理出错，特别是当 v 涉及负函数如 −cos x 时。
Also, in definite integrals, don't forget to apply limits to the uv term as well as the new integral. And when using the boomerang technique, be careful with algebraic manipulation: always re-check that you have correctly isolated the original integral. Finally, never assume integration by parts is the only method—substitution might be simpler for some products.
此外，在定积分中，别忘了对 uv 项和新积分都应用上下限。使用回旋镖技巧时，注意代数操作：始终重新检查你是否正确地分离出原积分。最后，切勿假设分部积分是唯一方法——对某些乘积而言，换元法可能更简单。
12. Exam Strategy and Practice Tips | 考试策略与练习建议
In Edexcel exams, integration by parts questions are often worth 4–7 marks and may appear within larger multi-step problems. Clearly show your choice of u and dv/dx, write the formula, and present each substitution line by line. Even if the final answer is wrong, method marks are substantial.
在 Edexcel 考试中，分部积分题通常值 4-7 分，并可能出现在较大的多步问题中。清楚地写出你对 u 和 dv/dx 的选择，展示公式，并逐行写出每次代入。即使最终答案错误，方法分也很可观。
Practise a wide variety: products of polynomials with exponentials, trig functions, logarithms, inverse trig, and the boomerang type. Time yourself to build speed while maintaining accuracy. Finally, after obtaining an answer, differentiate it mentally to check that you recover the original integrand—this quick verification can save marks lost to sign errors.
练习多种题型：多项式与指数函数、三角函数、对数函数、反三角函数以及回旋镖型乘积的积分。计时练习以提高速度同时保持准确性。最后，得到答案后在脑中对其求导，检查是否复原了原被积函数——这个快速验证可以挽救因符号错误丢失的分数。
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