📚 Integration by Parts | 分部积分法
Integration by parts is a fundamental technique in calculus that allows us to integrate products of functions. It is derived directly from the product rule for differentiation and is a core topic in the Edexcel A-Level Mathematics syllabus. Students are expected not only to apply the formula correctly but also to make strategic choices for u and dv in order to simplify complex integrals. This article provides a thorough review of the method, complete with step-by-step examples and exam-focused advice.
分部积分法是微积分中一种基本技巧，可用于求解函数乘积的积分。它直接源自微分的乘法法则，是Edexcel A-Level数学大纲中的核心主题。考生不仅要能正确套用公式，还需在解题时对 u 和 dv 做出策略性选择，以便将复杂积分化简。本文全面回顾该方法，辅以逐步示例和针对考试的实用建议。
1. The Formula for Integration by Parts | 分部积分公式
The integration by parts formula is expressed as ∫ u dv = uv − ∫ v du to a hopefully easier integral ∫ v du. Effective use of the formula therefore relies on a clever choice of u and dv.
分部积分公式可表示为 ∫ u dv = uv − ∫ v du。式中 u 与 v 均为积分变量（通常为 x）的函数。整个思路是把一个较难的积分 ∫ u dv 转化成期望中更简单的积分 ∫ v du。因此，有效运用该公式的关键在于巧妙地选择 u 和 dv。
2. Derivation from the Product Rule | 从乘积法则推导
v du. Rearranging this equation yields the integration by parts formula: ∫ u dv = uv − ∫ v du. This derivation highlights why the method works and reinforces the connection between differentiation and integration.
从乘积的微分法则出发：d(uv) = u dv + v du。对两边积分得 ∫ d(uv) = ∫ u dv + ∫ v du，化简后为 uv = ∫ u dv + ∫ v du。整理该等式即得到分部积分公式：∫ u dv = uv − ∫ v du。这一推导过程揭示了该方法成立的原因，并巩固了微分与积分之间的联系。
d(uv) = u dv + v du → ∫ u dv = uv − ∫ v du
3. Choosing u and dv: The LIATE Rule | 选择 u 和 dv：LIATE 法则
Trigonometric, Exponential. The function that appears earlier in the list should generally be chosen as u, while the remaining factor (along with dx) becomes dv. This helps ensure that the resulting ∫ v du is simpler than the original integral.
系统选择 u 的一个经验法则是 LIATE 规则，按优先级依次为：对数函数、反三角函数、代数函数、三角函数、指数函数。通常应选择列表中排序靠前的函数作为 u，而剩余因子（连同 dx）则作为 dv。这样有助于保证得到的 ∫ v du 比原积分更为简单。
- L – Logarithmic functions (e.g., ln x) | 对数函数（如 ln x）
- I – Inverse trigonometric functions (e.g., arctan x) | 反三角函数（如 arctan x）
- A – Algebraic functions (e.g., xⁿ) | 代数函数（如 xⁿ）
- T – Trigonometric functions (e.g., sin x, cos x) | 三角函数（如 sin x, cos x）
- E – Exponential functions (e.g., ex) | 指数函数（如 ex）
Although this rule covers most A-Level problems, there are exceptional cases (such as ∫ ex sin x dx) where either choice works and repeated application is required. Flexibility and practice are key.
虽然这一规则涵盖了大部分 A-Level 试题，但也存在特殊情况（如 ∫ ex sin x dx），此时任一种选择均可，但需重复应用分部积分。灵活性与练习至关重要。
4. Example 1: Basic Polynomial-Exponential Integration | 例1：基本多项式-指数积分
Evaluate ∫ x ex dx. Following LIATE, the algebraic part x is chosen as u, and ex dx as dv. Then du = dx and v = ex. Substituting into the formula:
计算 ∫ x ex dx。根据 LIATE 法则，选择代数部分 x 作为 u，ex dx 作为 dv。则 du = dx，v = ex。代入公式：
∫ x ex dx = x ex − ∫ ex dx = x ex − ex + C
This example shows how choosing the algebraic part for u reduces the new integral to a standard form. The method turns a product of two functions into a simple exponential integral.
本例表明，将代数部分选为 u 可以把新的积分化归为标准形式。该方法将两个函数的乘积转化为简单的指数积分。
5. Example 2: Integrating the Natural Logarithm | 例2：积分自然对数
)·ln x dx. Let u = ln x and dv = 1 dx. Then du = (1/x) dx and v = x. Applying the formula:
计算 ∫ ln x dx 时，并无明显的第二个函数，因此将积分写为 ∫ (1)·ln x dx。设 u = ln x，dv = 1 dx。则 du = (1/x) dx，v = 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
The trick of inserting 1 as a factor is extremely useful whenever the integrand consists of a single 'difficult' function, such as inverse trigonometric functions or logarithms.
当被积函数仅由一个“困难”函数构成时，插入因子 1 的技巧极其有用，例如反三角函数或对数函数的情形。
6. Example 3: Trigonometric Integral with Polynomial | 例3：多项式与三角函数的积分
ic) and dv = sin x dx (trigonometric). Then du = 2x dx and v = −cos x. The formula yields:
考虑 ∫ x² sin x dx。根据 LIATE，令 u = x²（代数），dv = sin x dx（三角）。则 du = 2x dx，v = −cos x。公式得出：
∫ x² sin x dx = −x² cos x − ∫ (−cos x)·2x dx = −x² cos x + 2 ∫ x cos x dx
ires another integration by parts, but the power of x has decreased from 2 to 1, showing how the method systematically breaks down the polynomial degree.
新的积分 ∫ x cos x dx 仍需再使用一次分部积分，但 x 的幂次已从 2 降至 1，这表明该方法可以逐步降低多项式的次数。
7. Example 4: Reversing the Roles – ∫ ex sin x dx | 例4：角色互换 – ∫ ex sin x dx
For ∫ ex sin x dx, LIATE does not impose a strict choice because both ex and sin x can be swapped. Let u = sin x, dv = ex dx. Then du = cos x dx, v = ex. We obtain: ∫ ex sin x dx = ex sin x − ∫ ex cos x dx. Apply parts again on the new integral with u = cos x, dv = ex dx:
对于 ∫ ex sin x dx，LIATE 法则并无严格限定，因为 ex 与 sin x 可以互换。设 u = sin x，dv = ex dx。则 du = cos x dx，v = ex。得到：∫ ex sin x dx = ex sin x − ∫ ex cos x dx。再对新积分使用分部积分，设 u = cos x，dv = ex dx：
∫ ex sin x dx = ex sin x − ex cos x − ∫ ex sin x dx
Notice the original integral appears on both sides. Letting I = ∫ ex sin x dx, we have I = ex sin x − ex cos x − I, so 2I = ex(sin x − cos x) and finally I = ½ ex(sin x − cos x) + C. This looping technique is a classic exam problem.
注意原积分出现在等号两边。设 I = ∫ ex sin x dx，则有 I = ex sin x − ex cos x − I，故 2I = ex(sin x − cos x)，最终 I = ½ ex(sin x − cos x) + C。这种循环技巧是经典的考试题型。
8. Repeated Integration by Parts | 重复分部积分
When the integrand includes xⁿ times a trigonometric or exponential function, integration by parts may need to be applied n times, each time reducing the power of x by one. A tabular method can speed up the process: list derivatives of u down a column and integrals of dv in a second column, then multiply diagonally with alternating signs. This is especially efficient for large values of n.
当被积函数包含 xⁿ 与三角函数或指数函数的乘积时，可能需应用 n 次分部积分，每次将 x 的幂次降低 1。列表法可加速这一过程：在一列中依次写下 u 的各阶导数，在另一列中写下 dv 的反复积分，然后沿对角线相乘并交替冠以正负号。对于较大的 n，该方法尤为高效。
| Derivatives of u | u 的导数 | Integrals of dv | dv 的积分 | Sign | 符号 |
| x² | sin x | + |
| 2x | −cos x | − |
| 2 | −sin x | + |
| 0 | cos x |
x) + 2(cos x) + C, which simplifies correctly. This organised approach reduces algebraic errors in examinations.
于是 ∫ x² sin x dx = x²(−cos x) − 2x(−sin x) + 2(cos x) + C，化简后正确无误。这种条理化方法可减少考试中的代数运算错误。
9. Integration by Parts for Definite Integrals | 定积分的分部积分
re a and b are the limits of integration. Simply evaluate the uv term at the boundaries and subtract the definite integral of v du. No need to re-substitute limits at each step; just carry the limits throughout.
对于定积分，公式变为 ∫ₐᵇ u dv = [uv]ₐᵇ − ∫ₐᵇ v du，其中 a 和 b 是积分上下限。只需在边界处计算 uv 项，并减去 v du 的定积分。无需在每一步重新代入上下限，直接始终保留积分限即可。
Example: ∫₀¹ x ex dx. Using u = x, dv = ex dx, we have [x ex]₀¹ − ∫₀¹ ex dx = (1·e¹ − 0·e⁰) − [ex 1) = 1. This neat result can be verified by simple differentiation of the antiderivative.
例：∫₀¹ x ex dx。设 u = x，dv = ex dx，得 [x ex]₀¹ − ∫₀¹ ex dx = (1·e¹ − 0·e⁰) − [ex]₀¹ = e − (e − 1) = 1。这一整洁的结果可通过求原函数的导数简单验证。
10. Handling Inverse Trigonometric Functions | 处理反三角函数
es: ∫ arctan x dx = x arctan x − ∫ x/(1+x²) dx. The remaining integral can be tackled by the substitution w = 1+x², leading to ½ ln(1+x²).
对于 ∫ arctan x dx 这类积分，由于缺少代数因子，可设 u = arctan x，dv = 1 dx。则 du = 1/(1+x²) dx，v = x。代入公式得：∫ arctan x dx = x arctan x − ∫ x/(1+x²) dx。余下的积分可通过代换 w = 1+x² 解决，结果为 ½ ln(1+x²)。
x²) + C. Similar strategies apply to arcsin x and arccos x. Edexcel particularly favours such combinations of techniques in exam questions.
因此 ∫ arctan x dx = x arctan x − ½ ln(1+x²) + C。类似的策略也适用于 arcsin x 和 arccos x。Edexcel 在考试题中尤为偏爱这种多种技巧的组合。
11. Common Pitfalls and How to Avoid Them | 常见陷阱及避免方法
Mistake 1: Incorrect identification of u and dv. Students sometimes choose u as the function that looks easier to integrate, which makes the new integral more complicated. Always use LIATE and check that ∫ v du is genuinely simpler.
错误1： u 和 dv 选择不当。部分学生有时会选看起来好积分的函数作为 u，结果使新积分变得更复杂。务必使用 LIATE 法则，并检查 ∫ v du 是否确实更简单。
Mistake 2: Forgetting the constant of integration C for indefinite integrals. Although the process involves several expressions, the final answer must always include +C.
错误2： 不定积分忘记加积分常数 C。尽管过程中包含多个表达式，最终答案必须总是包含 +C。
Mistake 3: In definite integrals, misplacing the limits when computing [uv] or when evaluating the final integral. Write out [uv] with limits clearly before substituting numbers.
错误3： 在定积分中，计算 [uv] 或最终积分时代错上下限。应先清晰地写出带上下限的 [uv]，再代入数值。
Mistake 4: Algebraic slips when differentiating u or integrating dv. Double-check the derivative of u and the antiderivative of dv rapidly by mental differentiation.
错误4： 对 u 求导或对 dv 积分时出现代数失误。可通过心算快速复核 u 的导数与 dv 的原函数。
12. Summary and Exam Tips | 总结与考试提示
nd the LIATE guide. Practice recognising when the technique is needed: products of mixed function types, logarithms, inverse trigonometric functions, and products where standard substitutions fail. In exam conditions, stay calm and methodical; sometimes a second application of parts will reveal a repeating pattern or a simpler integrand. With consistent practice, you will be able to tackle integration by parts questions with confidence and accuracy.
分部积分法是 A-Level 数学不可或缺的技能。牢记公式 ∫ u dv = uv − ∫ v du 以及 LIATE 指引。多加练习，识别何时需要运用该技巧：不同函数类型的乘积、对数、反三角函数，以及标准代换法失效的情形。考试时保持冷静、条理清晰；有时第二次分部积分后会出现循环模式或更简单的被积函数。通过持续练习，你将能够自信且准确地解答分部积分问题。
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