A-Level Maths Integration by Parts 分部积分技巧
什么是分部积分法？
分部积分法（Integration by Parts）是A-Level数学微积分部分最核心的积分技巧之一。当被积函数是两个不同类型函数的乘积时：例如多项式乘以指数函数、三角函数乘以对数函数：普通换元法往往无从下手，这时分部积分法就派上了用场。它的本质是将一个复杂的积分转化为另一个更容易处理的积分，背后的数学原理是乘积法则（Product Rule）的逆向应用。对于准备A-Level Pure Mathematics考试的同学来说，分部积分不仅是必考内容，更是后续学习Further Mathematics中更高级积分技巧的基础。
Integration by Parts is one of the most essential integration techniques in A-Level Pure Mathematics. It is used when the integrand is a product of two different types of functions -- for example, a polynomial multiplied by an exponential function, or a trigonometric function multiplied by a logarithmic function. Ordinary substitution often fails in these cases, and that is where Integration by Parts comes in. Its essence is transforming a complicated integral into another integral that is easier to handle. The underlying mathematical principle is the reverse application of the Product Rule from differentiation. For students preparing for the A-Level Pure Mathematics exam, Integration by Parts is not only mandatory content but also the foundation for learning more advanced integration techniques in Further Mathematics.
分部积分公式的推导
分部积分法的公式来源于微分中的乘积法则。回想一下，对于两个可微函数 u(x) 和 v(x)，乘积法则告诉我们：d(uv)/dx = u(dv/dx) + v(du/dx)。对这个等式两边同时积分，得到：∫ d(uv)/dx dx = ∫ u(dv/dx) dx + ∫ v(du/dx) dx。左边就是 uv，移项后得到核心公式：∫ u(dv/dx) dx = uv − ∫ v(du/dx) dx。写成更简洁的形式就是 ∫ u dv = uv − ∫ v du。考试中通常会给公式表，但理解推导过程能帮助你在选择 u 和 dv 时做出更准确的判断。
The Integration by Parts formula is derived from the Product Rule in differentiation. Recall that for two differentiable functions u(x) and v(x), the Product Rule states: d(uv)/dx = u(dv/dx) + v(du/dx). Integrating both sides gives us: ∫ d(uv)/dx dx = ∫ u(dv/dx) dx + ∫ v(du/dx) dx. The left side simplifies to uv, and rearranging yields the core formula: ∫ u(dv/dx) dx = uv − ∫ v(du/dx) dx. In the more compact form, this is written as ∫ u dv = uv − ∫ v du. While formula sheets are typically provided in the exam, understanding the derivation helps you make better judgments when choosing u and dv.
LIATE规则：如何选择u和dv
分部积分法成功的关键在于正确选择哪个函数作为 u（求导的那部分），哪个函数作为 dv（积分的那部分）。LIATE规则是A-Level学生最常用的选择优先级指南：L代表Logarithmic（对数函数，如ln x），I代表Inverse trigonometric（反三角函数），A代表Algebraic（代数函数，如x的幂函数），T代表Trigonometric（三角函数，如sin x, cos x），E代表Exponential（指数函数，如e^x）。优先级从高到低，排在前面的函数更适合作为u。举个例子，计算∫ x e^x dx时，x是Algebraic，e^x是Exponential，A排在E前面，所以选择u = x，dv = e^x dx。这个规则覆盖了A-Level考纲中90%以上的分部积分场景。
The key to success with Integration by Parts lies in correctly choosing which function to assign as u (the part to differentiate) and which as dv (the part to integrate). The LIATE rule is the most commonly used selection priority guide for A-Level students: L stands for Logarithmic functions (e.g., ln x), I for Inverse trigonometric functions, A for Algebraic functions (e.g., powers of x), T for Trigonometric functions (e.g., sin x, cos x), and E for Exponential functions (e.g., e^x). The priority goes from high to low, meaning functions appearing earlier in this list are better choices for u. For example, when computing ∫ x e^x dx, x is Algebraic and e^x is Exponential. Since A comes before E, we choose u = x and dv = e^x dx. This rule covers over 90% of Integration by Parts scenarios in the A-Level syllabus.
基础例题详解
让我们通过几个典型例题来巩固分部积分法的应用。第一个例子：计算∫ x sin x dx。根据LIATE规则，Algebraic (x) 优先于Trigonometric (sin x)，所以令 u = x，dv = sin x dx。则 du = dx，v = −cos x。代入公式∫ u dv = uv − ∫ v du，得到：∫ x sin x dx = x(−cos x) − ∫ (−cos x) dx = −x cos x + ∫ cos x dx = −x cos x + sin x + C。验证方法：对结果求导，应得到 x sin x。第二个例子：计算∫ ln x dx。这里没有明显的乘积形式，但可以看作∫ 1 · ln x dx。Logarithmic在LIATE中排第一，所以令 u = ln x，dv = dx。则 du = (1/x) dx，v = x。代入公式：∫ ln x dx = x ln x − ∫ x · (1/x) dx = x ln x − ∫ 1 dx = x ln x − x + C。这个结果在A-Level考试中经常被直接使用，建议熟记。
Let us consolidate the application of Integration by Parts through several typical examples. First example: compute ∫ x sin x dx. According to the LIATE rule, Algebraic (x) takes priority over Trigonometric (sin x), so we set u = x and dv = sin x dx. Then du = dx and v = −cos x. Substituting into the formula ∫ u dv = uv − ∫ v du yields: ∫ x sin x dx = x(−cos x) − ∫ (−cos x) dx = −x cos x + ∫ cos x dx = −x cos x + sin x + C. Verification: differentiate the result and you should obtain x sin x. Second example: compute ∫ ln x dx. There is no obvious product form here, but we can treat it as ∫ 1 · ln x dx. Logarithmic comes first in LIATE, so we set u = ln x and dv = dx. Then du = (1/x) dx and v = x. Substituting: ∫ ln x dx = x ln x − ∫ x · (1/x) dx = x ln x − ∫ 1 dx = x ln x − x + C. This result is frequently used directly in A-Level exams and is worth memorising.
重复分部积分法
有些积分需要多次应用分部积分法才能得到最终结果。最典型的例子是∫ x² e^x dx。第一次分部积分：令 u = x²，dv = e^x dx，得到 du = 2x dx，v = e^x。代入得：∫ x² e^x dx = x² e^x − ∫ e^x · 2x dx = x² e^x − 2∫ x e^x dx。此时新积分∫ x e^x dx仍需要再次使用分部积分。令 u = x，dv = e^x dx，则 du = dx，v = e^x。得到：∫ x e^x dx = x e^x − ∫ e^x dx = x e^x − e^x + C。代回原式：∫ x² e^x dx = x² e^x − 2(x e^x − e^x) + C = e^x(x² − 2x + 2) + C。这种"套娃"式的重复操作在考试中很常见，注意每次都要正确追踪符号和系数。
Some integrals require multiple applications of Integration by Parts to reach the final result. The classic example is ∫ x² e^x dx. First application: set u = x² and dv = e^x dx, giving du = 2x dx and v = e^x. Substituting: ∫ x² e^x dx = x² e^x − ∫ e^x · 2x dx = x² e^x − 2∫ x e^x dx. Now the new integral ∫ x e^x dx itself requires Integration by Parts. Set u = x and dv = e^x dx, then du = dx and v = e^x. We obtain: ∫ x e^x dx = x e^x − ∫ e^x dx = x e^x − e^x + C. Substituting back: ∫ x² e^x dx = x² e^x − 2(x e^x − e^x) + C = e^x(x² − 2x + 2) + C. This nesting-style repeated application is very common in exams -- always track signs and coefficients carefully at each step.
表格法：高效处理多次分部积分
当遇到如∫ x³ sin x dx这类需要三次分部积分的题目时，手动一步步写会非常繁琐且容易出错。这时表格法（Tabular Method，也称DI Method）可以大幅提高效率和准确率。方法如下：创建两列，左列依次写下u及其各阶导数（直到导数为0），右列依次写下dv的反复积分结果。第一列：x³ → 3x² → 6x → 6 → 0；第二列：sin x → −cos x → −sin x → cos x → sin x。然后按对角线交替加减：+x³(−cos x) − 3x²(−sin x) + 6x(cos x) − 6(sin x)。整理得到最终结果。表格法在Further Mathematics中也经常出现，熟练掌握可以在考试中节省大量时间。
When faced with integrals requiring three rounds of Integration by Parts, such as ∫ x³ sin x dx, manual step-by-step working becomes tedious and error-prone. The Tabular Method (also called the DI Method) can dramatically improve both efficiency and accuracy. The procedure: create two columns. In the left column, write u and its successive derivatives (until reaching zero). In the right column, write the repeated integrals of dv. Left column: x³ → 3x² → 6x → 6 → 0. Right column: sin x → −cos x → −sin x → cos x → sin x. Then alternate signs along the diagonals: +x³(−cos x) − 3x²(−sin x) + 6x(cos x) − 6(sin x). Simplify to obtain the final result. The Tabular Method also appears frequently in Further Mathematics. Mastering it can save substantial time in exams.
特殊情况与技巧
有些特殊形式的积分需要灵活运用分部积分法。第一类是循环积分：当被积函数中两种函数的导数具有周期性时，分部积分后会产生与原积分相同形式的项，从而可以通过解方程求得结果。例如∫ e^x sin x dx，经过两次分部积分后，会重新出现∫ e^x sin x dx，把它当作未知数解出来即可。第二类是定积分的分部积分：公式变为∫[a,b] u dv = [uv]|[a,b] − ∫[a,b] v du，需要代入上下限计算。第三类是结合换元法使用：先用换元化简表达式，再用分部积分处理。这些综合题型经常出现在A-Level试卷的最后几题，分数权重较高。
Certain special forms of integrals require flexible application of Integration by Parts. The first type is cyclic integrals: when the derivatives of both functions in the integrand are periodic, applying Integration by Parts will regenerate a term with the same form as the original integral, allowing us to solve for it algebraically. For example, ∫ e^x sin x dx: after two rounds of Integration by Parts, ∫ e^x sin x dx reappears on the right side. Treat it as an unknown and solve the resulting equation. The second type is definite Integration by Parts: the formula becomes ∫[a,b] u dv = [uv]|[a,b] − ∫[a,b] v du, requiring substitution of upper and lower limits. The third type combines substitution with Integration by Parts: first simplify the expression via substitution, then apply Integration by Parts. These comprehensive question types often appear in the final problems of A-Level papers and carry significant marks.
常见错误与避坑指南
在A-Level考试中，分部积分相关的失分点主要集中在以下几方面。第一，选错u和dv导致积分越算越复杂：如果发现新积分∫ v du比原积分更复杂，说明选反了，应立刻调换。第二，忘记常数C：不定积分必须加上积分常数，这在CAIE和Edexcel的评分标准中都会被扣分。第三，定积分代入时符号错误：特别是在处理负号时容易搞混。第四，遇到ln x时忘记将其作为u（因为ln x的积分不容易直接写出，但求导很简单）。第五，被积函数可以展开时却直接用分部积分：先检查是否可以通过代数化简避免分部积分。养成"先化简、再判断、后动手"的解题习惯，可以大幅减少考试中的无谓失分。
In A-Level exams, marks are commonly lost on Integration by Parts in the following areas. First, choosing the wrong u and dv, which makes the resulting integral more complicated than the original -- if you find ∫ v du harder than the original integral, swap your choices immediately. Second, forgetting the constant C -- indefinite integrals must include the constant of integration; this is penalised in both CAIE and Edexcel mark schemes. Third, sign errors when substituting limits in definite integrals, particularly with negative signs. Fourth, forgetting to set ln x as u when it appears, since ln x is hard to integrate directly but trivial to differentiate. Fifth, jumping straight to Integration by Parts when the integrand can be expanded algebraically -- always check whether algebraic simplification can avoid the need for Integration by Parts entirely. Developing the habit of simplifying first, judging second, and executing third can substantially reduce unnecessary mark losses in exams.
考前速查：核心公式与解题模板
为帮助你在考前快速回顾，这里整理分部积分法的核心要点。公式：∫ u(dv/dx) dx = uv − ∫ v(du/dx) dx，简写为∫ u dv = uv − ∫ v du。LIATE选择规则：Log → Inverse Trig → Algebraic → Trig → Exponential，排在前面的选为u。重复分部积分：直到某一项导数为0或出现循环。表格法：适合多项式×三角/指数函数，多项式放在左列求导直到0，另一函数放在右列反复积分，对角线交替加减。循环积分：两次分部后解方程。定积分：记得在uv项代入上下限。考前把这些要点过一遍，配合几道真题练习，分部积分相关题目就能稳稳拿分。
Here is a quick-reference summary of the core points for Integration by Parts to review before your exam. Formula: ∫ u(dv/dx) dx = uv − ∫ v(du/dx) dx, compactly written as ∫ u dv = uv − ∫ v du. LIATE selection rule: Log → Inverse Trig → Algebraic → Trig → Exponential -- choose the one appearing earlier as u. Repeated Integration by Parts: continue until a derivative reaches zero or a cyclic pattern emerges. Tabular Method: ideal for polynomial × trig/exponential functions; place the polynomial in the left column, differentiate down to zero; place the other function in the right column, integrate repeatedly; alternate signs along the diagonals. Cyclic integrals: apply Integration by Parts twice, then solve the resulting equation. Definite integrals: remember to evaluate the uv term at the upper and lower limits. Review these key points before your exam, practise with a few past paper questions, and you will confidently secure full marks on Integration by Parts problems.
进阶延伸：Further Mathematics中的分部积分
对于计划继续学习Further Mathematics的同学，分部积分法在更高级的背景下还有广泛应用。在FP2/FP3中，你会遇到分部积分用于推导递推公式（Reduction Formulae），例如通过分部积分建立I_n = ∫ sin^n x dx的递推关系。在求解微分方程时，积分因子法中也会用到分部积分。此外，在大学一年级的数学课程中，分部积分是理解Laplace变换和Fourier分析等高级主题的必要前置技能。A-Level阶段把分部积分练到炉火纯青，不仅是为了眼前的考试，更是为未来的数学学习铺设坚实的基础。建议在熟练掌握基本题型后，尝试一些FP1/FP2级别的综合题，提前感受更高层次的积分应用。
For students planning to continue with Further Mathematics, Integration by Parts has extensive applications in more advanced contexts. In FP2 and FP3, you will encounter Integration by Parts used to derive Reduction Formulae -- for instance, establishing a recurrence relation for I_n = ∫ sin^n x dx via Integration by Parts. When solving differential equations, the integrating factor method also involves Integration by Parts. Moreover, in first-year university mathematics courses, Integration by Parts is an essential prerequisite for understanding advanced topics such as Laplace transforms and Fourier analysis. Mastering Integration by Parts thoroughly at the A-Level stage is not only for the immediate exam but also for laying a solid foundation for future mathematical studies. It is recommended that, after becoming proficient with basic question types, you attempt some FP1/FP2-level comprehensive problems to experience integration applications at a higher level in advance.
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