ALevel数学 分部积分 Integration by Parts 技巧 | A-Level Maths Integration by Parts Techniques

ALevel数学 分部积分 Integration by Parts 技巧

分部积分是 A-Level 纯数学（Pure Mathematics）中最重要的积分技巧之一，几乎所有考试局的 Paper 2 或纯数卷都会考到。它的本质是乘积法则（product rule）的逆运算：当我们遇到两个不同类型函数相乘的积分，例如 x 乘 e 的 x 次方、x 乘 cos x、或者单独的 ln x 时，直接积分往往行不通，这时就要用分部积分把难题转化成更简单的积分。这篇文章会用中文和英文双语，从公式推导讲到选择技巧，再带大家完整做四类高频例题，最后总结考场上的常见错误与得分要点。

Integration by parts is one of the most important techniques in A-Level Pure Mathematics, appearing regularly in Paper 2 or the pure paper of every major exam board. It is simply the product rule in reverse: when you face the integral of a product of two different types of function, such as x times e to the x, x times cos x, or even a lone ln x, direct integration usually fails. Integration by parts converts such a difficult integral into a simpler one. This bilingual guide moves from the derivation of the formula, to the art of choosing u, through four families of classic exam questions, and finishes with common mistakes and mark-scoring strategies.

分部积分公式与乘积法则的联系

回忆乘积法则：d/dx [f(x)g(x)] = f'(x)g(x) + f(x)g'(x)。两边同时对 x 求积分，左边就是 f(x)g(x)，于是 f(x)g(x) = ∫ f'(x)g(x) dx + ∫ f(x)g'(x) dx。移项后得到分部积分公式。只要把公式中的两项重新安排，让其中一项变成容易处理的积分，问题就迎刃而解。

Recall the product rule: d/dx [f(x)g(x)] = f'(x)g(x) + f(x)g'(x). Integrating both sides with respect to x gives f(x)g(x) = ∫ f'(x)g(x) dx + ∫ f(x)g'(x) dx. Rearranging produces the integration by parts formula. If we arrange the two terms so that one of the resulting integrals is easy, the whole problem solves itself.

标准公式写作：∫ u dv = uv - ∫ v du。其中 u 是一个函数，dv 是另一个函数的微分。使用时分三步：第一步选 u，第二步把剩下的部分当作 dv 并求出 v，第三步代入公式计算新积分 ∫ v du。新积分通常比原来的积分简单，因为我们对 u 求了导。

The standard form is ∫ u dv = uv - ∫ v du, where u is one function and dv is the differential of the other. The method runs in three steps: choose u first, treat whatever remains as dv and integrate it to find v, then substitute into the formula and evaluate the new integral ∫ v du. The new integral is usually simpler than the old one, because we have differentiated u.

选 u 的 LIATE 优先序

分部积分成败的关键在于选 u。经验法则叫 LIATE，按优先级从高到低排列：L 是对数函数 ln x，I 是反三角函数如 arctan x，A 是代数函数（多项式）如 x 或 x 的平方，T 是三角函数如 sin x，E 是指数函数如 e 的 x 次方。规则是：优先让列表中靠前的函数当 u，因为对它们求导会越变越简单。

The whole method stands or falls on the choice of u. The rule of thumb is LIATE, ranked from highest to lowest priority: L for Logarithmic functions such as ln x, I for Inverse trigonometric functions such as arctan x, A for Algebraic or polynomial functions such as x or x squared, T for Trigonometric functions such as sin x, and E for Exponential functions such as e to the x. Functions earlier in the list should become u, because differentiating them makes them simpler.

为什么这样排？ln x 求导后变成 1/x，形式立刻简化；x 求导一次变成常数；而 e 的 x 次方无论求导多少次都是自己，三角函数求导两次会循环。所以指数和三角函数适合放进 dv，让它们被积分而不是被求导。记住 LIATE 五个字母，大部分选择题就能直接写对。

Why this order? Differentiating ln x turns it into 1/x and instantly simplifies it; differentiating x once yields a constant; but e to the x differentiates to itself forever, and trigonometric functions cycle after two derivatives. Exponentials and trig functions therefore belong in dv, where they get integrated rather than differentiated. Memorise the five letters LIATE and most exam choices become automatic.

例1：∫ x e的x次方 dx 的标准做法

看第一类高频题 ∫ x ex dx。按 LIATE，x 是代数函数，e 的 x 次方是指数函数，所以 u = x，dv = ex dx。求导得 du = dx，积分得 v = ex。代入公式：∫ x ex dx = x ex - ∫ ex dx = x ex - ex + C。最后别忘了加积分常数 C。

Take the classic ∫ x ex dx. By LIATE, x is algebraic and e to the x is exponential, so set u = x and dv = ex dx. Then du = dx and v = ex. Substituting: ∫ x ex dx = x ex - ∫ ex dx = x ex - ex + C. Always add the constant of integration C at the end.

这个例子展示分部积分的典型收益：原来要积 x 乘 e 的 x 次方，代入公式后只需要积 e 的 x 次方，一步完成。如果反过来选 u = e 的 x 次方，dv = x dx，那么 v = x 平方除以 2，新积分 ∫ e 的 x 次方乘 x 平方除以 2 dx 反而更复杂，题目越做越难。这就是 LIATE 的实战意义。

This example shows the typical payoff: instead of integrating x times e to the x, the formula leaves only e to the x to integrate, which is done in one line. If you reverse the choice, u = e to the x and dv = x dx, then v = x squared over 2 and the new integral is worse than the original. That is LIATE in action.

例2：∫ x cos x dx 与三角函数搭配

第二类高频题是多项式乘三角函数，如 ∫ x cos x dx。u = x（代数函数），dv = cos x dx，于是 du = dx，v = sin x。代入：∫ x cos x dx = x sin x - ∫ sin x dx = x sin x + cos x + C。注意最后一个符号变化：负的 sin x 积分得负负得正。

The second common family is polynomial times trig, for example ∫ x cos x dx. Choose u = x, dv = cos x dx, so du = dx and v = sin x. Substitution gives ∫ x cos x dx = x sin x - ∫ sin x dx = x sin x + cos x + C. Watch the final sign change: integrating negative sin x gives a double negative.

考试中这类题常以定积分出现，例如求 0 到 π/2 区间上 x cos x 的积分。处理方式与不定积分完全相同，先算原函数 x sin x + cos x，再代入上限 π/2 减去代入下限 0 的结果。代入上限得 π/2 乘 sin(π/2) 加 cos(π/2)，即 π/2；代入下限得 0 加 1。答案就是 π/2 - 1。

In exams this family often appears as a definite integral, such as ∫ x cos x dx from 0 to π/2. Work exactly as for an indefinite integral, find the antiderivative x sin x + cos x, then substitute the upper limit π/2 and subtract the value at the lower limit 0. The upper end gives π/2 times sin(π/2) plus cos(π/2), which is π/2; the lower end gives 0 plus 1. The answer is π/2 - 1.

例3：∫ ln x dx 单独一个函数也能分部

很多同学想不到，单独一个 ln x 也可以（也必须）用分部积分。技巧是把被积函数看成 1 乘 ln x：u = ln x，dv = 1 dx = dx。求导得 du = (1/x) dx，积分得 v = x。代入公式：∫ ln x dx = x ln x - ∫ x 乘 (1/x) dx = x ln x - ∫ 1 dx = x ln x - x + C。

Many students do not realise that a lone ln x can, and must, be handled by integration by parts. The trick is to view the integrand as 1 times ln x: let u = ln x and dv = 1 dx = dx. Then du = (1/x) dx and v = x. Substituting: ∫ ln x dx = x ln x - ∫ x times (1/x) dx = x ln x - ∫ 1 dx = x ln x - x + C.

关键点：如果 ln 或反三角里面套着别的函数，例如 ∫ ln(3x) dx 或 ∫ arctan x dx，一律把整个外层函数当 u，把 1 当 dv。凡是求导后能变成有理式的函数，例如 ln、arctan，都天然适合当 u，因为它们的导数让新积分变成纯有理函数积分。

The general principle: whenever ln or an inverse trig function wraps something else, such as ∫ ln(3x) dx or ∫ arctan x dx, make the whole outer function u and let dv be 1 dx. Any function whose derivative becomes a rational expression, such as ln or arctan, is a natural candidate for u because its derivative leaves a purely rational integral.

例4：∫ x平方 e的x次方 dx 需要两次分部

当多项式的次数高于一次，例如 ∫ x2 ex dx，一次分部不够。第一步 u = x2，dv = ex dx，得到 x2 ex - ∫ 2x ex dx。新积分里 2x ex 与例1完全同型，再分部一次：∫ 2x ex dx = 2x ex - 2ex。合并得到完整答案。

When the polynomial degree exceeds one, such as ∫ x2 ex dx, one round is not enough. First set u = x2 and dv = ex dx to obtain x2 ex - ∫ 2x ex dx. The new integrand 2x ex has exactly the same shape as Example 1, so apply parts again: ∫ 2x ex dx = 2x ex - 2ex. Combining gives the full answer x2 ex - 2x ex + 2ex + C.

每分部一次，x 的次数就降一：从二次降到一次，再一次降到常数。所以含 x 的 n 次方乘 e 的 x 次方或 sin x 的题，通常需要做 n 次分部。A-Level 真题常见到二次，进阶数学（Further Maths）才考到三次。做题时每步把新积分写清楚，避免漏项。

Each application reduces the power of x by one: from quadratic to linear, then to a constant. An integral of x to the n times e to the x or sin x therefore usually needs n rounds of parts. A-Level papers commonly test n = 2, while Further Maths may reach n = 3. Write out the new integral clearly at every step so nothing is dropped.

循环型积分 ∫ e的x次方 sin x dx 的经典套路

最特别的一类是 e 的 x 次方乘 sin x 或 cos x：无论怎么选 u，新积分都和原题同型。设 u = ex，dv = sin x dx，分部一次后得到含 ex cos x 的积分；对它再分部一次，结果中重新出现 ∫ ex sin x dx。此时把原积分当作未知数 I，把两轮结果写成方程，移项解出 I 即可。

The most distinctive family is e to the x times sin x or cos x: whichever function you pick for u, the new integral looks just like the original. Set u = ex and dv = sin x dx; one round of parts leaves an integral involving ex cos x. Apply parts to that and the original ∫ ex sin x dx reappears. Treat that integral as an unknown I, write the two rounds as an equation, and solve for I.

推导结果：∫ ex sin x dx = (1/2) ex (sin x - cos x) + C。这种循环型题目在 A-Level 纯数卷属于高分题（通常 6 到 8 分），步骤完整、符号正确就能拿满分。进阶数学还可能出现 ∫ e 的 x 次方 cos(2x) dx 这类变体，方法完全一样，注意系数即可。

The derivation yields ∫ ex sin x dx = (1/2) ex (sin x - cos x) + C. Cyclic questions are high-value items in A-Level pure papers, typically worth 6 to 8 marks, and full method marks are available if every line and sign is correct. Further Maths variants such as ∫ ex cos(2x) dx work identically, with extra care over the coefficients.

定积分代入上下限的两个细节

分部积分用于定积分时，公式变为：∫(a 到 b) u dv = [uv](a 到 b) - ∫(a 到 b) v du。中括号 [uv] 表示先代入上限 b 减去代入下限 a。两个细节最容易丢分：第一，如果题目给出积分上下限，分部后另一侧的新积分必须保留同样的上下限；第二，代入时小心符号，尤其是下限处括号整体取负。

For definite integrals the formula becomes ∫ from a to b of u dv = [uv] from a to b - ∫ from a to b of v du. The bracket [uv] means evaluate at the upper limit b and subtract the value at the lower limit a. Two details lose the most marks: first, the new integral on the right must keep the same limits a and b; second, when substituting, mind the sign, because the whole bracket at the lower limit is subtracted.

实操建议：先完全不写下限做完整的不定积分，求出原函数后再代入上下限，这样能减少符号混乱。不少评分标准按步骤给分，方法分（M 分）占大头，即使最后答案算错，只要原函数正确也能拿到大部分分数，所以过程必须工整。

A practical tip: do the whole integration without limits first, find the antiderivative, then substitute the limits at the very end to reduce sign confusion. Mark schemes award method marks generously, so even a wrong final answer earns most of the credit if the antiderivative is correct. Neat working matters.

与换元法配合的混合题型

有些题要分部积分和换元法（substitution）先后使用。典型例子是 ∫ ln x 除以 x 平方 dx 或含根号的乘积。面对这类题，先判断外层结构：如果能看出复合函数，优先换元；如果明显是两个独立函数相乘，优先分部。部分题目两步都需要，顺序要灵活，先做能简化形式的一步。

Some questions require substitution and parts in sequence, for example ∫ ln x over x squared dx or products involving square roots. Look first at the outer structure: if a composite function is visible, substitute first; if two independent functions are multiplied, start with parts. Some problems need both, and the order should be the step that simplifies the form more.

经典示范：∫ x 乘 (x + 1) 的 7 次方 dx 用换元 t = x + 1 更直接；而 ∫ x 乘 ln x dx 用分部更快，u = ln x，dv = x dx，得到 (1/2) x2 ln x - (1/4) x2 + C。考试中先花 30 秒观察结构再动手，比盲目套公式更能拿分。

A quick comparison: ∫ x (x + 1)7 dx is handled better by the substitution t = x + 1, while ∫ x ln x dx is faster by parts with u = ln x and dv = x dx, giving (1/2) x2 ln x - (1/4) x2 + C. Spending thirty seconds studying the structure before touching the pen earns more marks than blindly applying a formula.

考场高频错误清单

阅卷报告反复指出的错误有六类：第一，u 和 dv 选反，导致新积分更复杂；第二，忘记在最后加积分常数 C；第三，v 的符号出错，例如积 sin x 时把负号丢掉；第四，换元后的新积分忘记回代或忘记处理上下限；第五，两次分部之间抄错系数；第六，把分部积分与换元法的适用场景混淆。

Examiner reports repeatedly flag six error families: choosing u and dv the wrong way round so the new integral grows worse; forgetting the constant C; losing a sign while integrating, such as dropping the minus in front of cos x; failing to back-substitute after a change of variable or mishandling limits; copying coefficients wrongly between two rounds of parts; and confusing when to use parts versus substitution.

自查方法：做完后对答案求导验证，原函数求导应当回到被积函数。验证只需要十秒钟，却是最有用的检查手段。另外养成先写 u、du、v、dv 四行的习惯，让每一步有据可查，评卷人也更容易给方法分。

Self-checking: differentiate your final answer. The derivative must return to the original integrand, and this ten-second check is the most useful habit you can build. Also form the habit of writing the four lines u, du, v, dv before substituting, which keeps every step auditable and helps examiners award method marks.

冲刺建议与练习资源

备考策略分三步。第一步，把本文四类例题独立重做一遍，不看答案；第二步，按考试局刷分类真题，Edexcel 的 Paper 2、AQA 的 Paper 1、CAIE 的 Pure 2 与 3 中分部积分是固定考点，用近五年真题练习手感；第三步，整理自己的错题本，把每道错题标注错误类型，考前一周集中回看。

A three-step revision plan works best. First, redo the four example families in this article from scratch without peeking. Second, drill topic-sorted past paper questions from your board, since integration by parts is a guaranteed topic in Edexcel Paper 2, AQA Paper 1, and CAIE Pure 2 and Pure 3, and five years of papers build real fluency. Third, keep an error log that labels every mistake by type and review it in the final week.

记住三条主线：选 u 看 LIATE，做 n 次分部对应 x 的 n 次方，循环型题列方程解 I。这三条能覆盖 90% 的考点。熟练之后，分部积分会成为你最有把握的得分工具之一，遇到任何乘积型积分都不会慌。

Keep three threads in mind: choose u by LIATE, expect n rounds of parts for a power of x to the n, and solve cyclic integrals by writing an equation for I. These three rules cover about 90 percent of what examiners ask. Once fluent, integration by parts becomes one of your most reliable scoring tools, and no product integrand will frighten you again.

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