A-Level数学 分部积分 选u 循环型 定积分
积分题里的乘积形式是 A-Level 考生绕不开的一类。看到 x 乘 e^x、x 乘 sin x，或者孤零零一个 ln x，靠基本公式和凑微分都做不动，能用的工具就是分部积分。它的公式来自乘积求导法则，考场上真正拉开差距的却不是公式，而是拿到题之后怎么摆放两个因子。
Pearson Edexcel、AQA 与 OCR 三家考试局的 A-Level 数学考纲都把分部积分放在纯数学的积分章节里，考纲用的词是 simple cases，只要求处理结构规整的乘积。下面按这个难度写：从公式来历讲到选择规则、四类常见题型、卷面结构与扣分点。
Integration by parts often looks like a formula to memorise, yet it is really the product rule read backwards. Every board in England asks for simple cases only, which in practice means products such as x e^x, x sin x or ln x. The marks are won or lost in the first five seconds, when you decide which factor becomes u and which becomes dv; the rest is routine algebra.
分部积分是乘积法则的逆过程
乘积求导法则写成 (uv)' = u'v + uv'，两边同时积分就得到分部积分公式。移项以后变成 ∫ u v' dx = uv - ∫ u' v dx，也就是课本上印的 ∫ u (dv/dx) dx = uv - ∫ v (du/dx) dx。Pearson Edexcel 9MA0 考纲在积分一章里写明，分部积分是乘积法则的逆过程，换元则是链式法则的逆过程。这句话解释了公式右边为什么会出现一项乘积再减一个积分：被减掉的那部分，正是你原本不想直接处理的因子。求导会把 u 变简单，积分要处理 dv。两个方向的目标刚好相反。
The product rule says that differentiating uv gives u'v plus uv'. Integrating both sides and rearranging produces the formula printed in your formula booklet, and the Pearson Edexcel 9MA0 specification describes the method as the inverse of the product rule; substitution plays the same role for the chain rule. The first term on the right is a product you can write down at once, while the second is a fresh integral that you are hoping is easier than the one you started with. Everything below is about making that hope realistic.
选 u 与 dv：LIATE 顺序的实际用法
拿到 ∫ f(x)g(x) dx 这类题，先决定谁进微分号。常用的顺序是 LIATE：对数、反三角、代数式、三角、指数，越靠前的越适合当 u，剩下的当 dv。这条口诀不是考纲内容，考试局也不会印出来，它只是把「谁求导之后变简单」写成一条可操作的顺序。另一条判断同样重要：dv 必须是你立刻能积出来的东西。对数放进 dv 就积不动，代数式放进 dv 却很简单。很多学生以为分部积分是先套公式再想细节，实际上一开始的判断错了，后面整题都做不对。
Faced with a product, you must first decide which factor becomes u and which becomes dv. LIATE puts logarithms and inverse trigonometric functions first, then algebraic terms, then trigonometric and exponential ones; the earlier a factor appears in that order, the better it serves as u. This mnemonic is a teaching aid rather than specification content, and no board prints it. The second test matters just as much; if dv cannot be integrated on sight, the choice is already wrong.
多项式乘指数或三角：一次分部就降次
∫ x e^x dx 是最标准的类型。取 u = x，于是 du = dx；取 dv = e^x dx，于是 v = e^x。代入公式得到 x e^x - ∫ e^x dx，答案是 x e^x - e^x + c，整理成 e^x(x - 1) + c。复核一遍：对 e^x(x - 1) 求导得到 e^x(x - 1) + e^x，化简正是 x e^x，和被积函数一致。同一套节奏用在 ∫ x cos x dx 上，取 u = x，dv = cos x dx，结果是 x sin x + cos x + c。多项式当成 u 的价值在于求导一次就降一次次，任何这类题最后都会落回简单积分。
The standard case is a polynomial multiplied by an exponential or a trigonometric function. Set u as the polynomial and dv as the other factor, and the new integral contains x to a lower power, so the process terminates after one round. For x e^x the answer e^x(x - 1) + c differentiates back to x e^x, and the same routine gives x sin x + cos x + c for x cos x. Differentiating your answer takes ten seconds and catches sign slips before they cost marks; many candidates skip that check and never notice the error.
ln x 这类看不出乘积的题
ln x 看着不是乘积，但可以写成 ln x 乘 1。取 u = ln x，dv = dx，于是 v = x，代入得到 x ln x - ∫ x 乘 (1/x) dx，化简是 x ln x - x + c。这里的关键动作是主动补一个因子 1，很多学生卡在这一步，因为他们只盯着题目里看得见的乘积。稍微长一点的题目是 x 乘 ln x：取 u = ln x，dv = x dx，得到 (x^2 / 2) ln x - ∫ (x / 2) dx，整理为 (x^2 / 2) ln x - x^2 / 4 + c。优先让对数当 u，是因为 ln x 求导后变成 1/x。乘进剩下的因子里，它反而变简单。
A logarithm does not look like a product until you write ln x as ln x times 1. Choose u as ln x and dv as dx, so v is x, and the new integral turns into 1 dx, which is trivial; the answer x ln x - x + c follows immediately. For x ln x, take u as ln x and dv as x dx, and the leftover integral is a simple power. Logs earn the priority because they differentiate into something simpler than themselves, which is exactly what the formula wants from u.
需要两轮分部积分的题
多项式次数高一点，分部积分就要做两轮。第一轮的取法是让 u 等于 x^2，dv 等于 e^x dx，得到 x^2 e^x - 2 ∫ x e^x dx；括号里的积分正是上一节处理过的那一个，代入结果是 e^x(x^2 - 2x + 2) + c。同一节奏也适用于 x^2 乘 sin x：第一轮把 x^2 降成 x，第二轮把 x 降成常数。每做一轮，多项式次数掉一次。这个下降保证了过程会停。第二轮要重新指定 u 与 dv，忘了重新指定是漏分的常见来源之一。
When the polynomial is a higher power, the method has to run twice. Taking u as x^2 for the first round leaves the integral of x e^x, which is the case already solved above; for x^2 sin x the first round lowers the power and the second finishes the job. Each round reduces the degree by one, which is why the process must stop. Re-declaring u and dv at the start of the second round matters, because examiners award method marks for that working and many candidates simply drop it.
指数乘三角的循环型积分
∫ e^x sin x dx 属于另一个类型，无论怎么选，被积函数的结构都不会自己变简单。把原积分记作 I，取 u = sin x，dv = e^x dx，于是 I = e^x sin x - ∫ e^x cos x dx。对右边这个积分再做一次分部积分，得到 e^x cos x + ∫ e^x sin x dx，代回去就有 I = e^x sin x - e^x cos x - I。把 I 移到等式左边，2I = e^x(sin x - cos x)，答案是 e^x(sin x - cos x) / 2 + c。两轮的选择要保持同类：第一轮用三角函数当 u，第二轮也要用三角函数。中途换边会把 I 消掉。
Products of an exponential and a trigonometric function behave differently, because no amount of integrating makes the structure simpler. Label the original integral I, apply the formula once, then apply it again to the integral that appears on the right; the original I reappears, which is the point. Collecting the I terms on one side gives 2I and the answer follows directly. Keep the same type of factor as u in both rounds, since swapping the roles in the middle makes the two I terms cancel and leaves nothing behind.
定积分的分部积分与上下限处理
带上限下限时，公式写成 ∫(a 到 b) u dv = [uv](a 到 b) - ∫(a 到 b) v du。方括号那部分直接代入上下限，结果是一个数字，不用再写常数。从 0 到 1 的定积分用刚才的原函数 e^x(x - 1)：代入上限 1 得 0，代入下限 0 得 -1，相减得到 1。另一种算法是按 [x e^x] 减 ∫ e^x dx 逐项处理，结果一样。经典例子还有 1 到 e 上的 ln x，原函数 x ln x - x，代入得到 (e - e) 减 (0 - 1)，答案也是 1。定积分算出负数不代表做错，它说明曲线落在坐标轴下方。面积题通常要取绝对值。
With limits, the formula becomes the bracket of uv between a and b, minus the integral of v du over the same interval. Evaluate the bracket first and you have a number, so no constant of integration appears at the end. For the integral of x e^x from 0 to 1 the value is 1, and integrating ln x from 1 to e also gives 1. A negative answer is not automatically an error; it means the curve sits below the axis, and area questions usually expect the size of that region instead.
什么时候换元比分部积分更快
看到乘积不必马上想到分部积分。以 2x 乘 e^(x^2) 为例，前面的 2x 恰好是 x^2 的导数，换元 u = x^2 一步就出结果。对数除以 x 的积分同理，因为 1/x 正好是 ln x 的导数。判别方法是看两个因子之间有没有导数关系，如果有，换元通常更快。另一种情况是分部积分之后积分变得更复杂，例如把 ln x 放进 dv 的位置，第二轮反而更难处理。动笔前花十秒判断结构。写完两行再划掉，代价要大得多。
Not every product should be attacked with this method. When one factor is the derivative of the other, as in 2x e^(x^2) or (ln x) / x, a substitution finishes the question in one line. Look for a derivative relationship before you write the formula down, because the wrong method can double the length of your working. Ten seconds spent reading the structure is cheaper than two lines crossed out.
阅卷扣分点：符号与漏项
阅卷时，这类题后半部分的分数往往丢在符号上：公式里是减号，写成加号之后后面全部跟着错。另一种丢分是忘记把 dv 积出来，直接写成 uv 减 ∫ u' v' dx，公式只用了一半。两个因子的配对要写在卷面上，只给结果不写中间步骤，方法分是拿不到的。不定积分的答案最后要加常数 c，定积分不加。常见错误是把 ln x 当成多项式处理，写出 (x^2 / 2) 这种答案。写完之后对被积函数求导复核，是最省时间的自检方式。
Most marks lost in this topic come from a single sign. The formula contains a subtraction, and writing a plus there spoils every line that follows, even when the rest of the algebra is fine. A second loss comes from half-applying the formula: v must actually be integrated, not copied, and u and dv should be stated on the page for the method mark. Answers to indefinite integrals need the constant c, definite ones do not, and logs are never handled by a power rule.
公式册与图形计算器怎么用
公式册提供的是结果。它不会替你做出选择。OCR H240 的公式页把分部积分公式 ∫ u (dv/dx) dx = uv - ∫ v (du/dx) dx 印了出来，Pearson Edexcel 的 Mathematical Formulae and Statistical Tables 也收录了标准积分形式，所以背公式本身不是难点。真正要练的是拿到题之后几秒钟内决定 u 与 dv。图形计算器可以在做完之后帮你验证：把同一个定积分输入计算器得到一个数值，再和自己的答案比对，差得多就说明符号或者系数出了问题。Edexcel、AQA 与 OCR 的 A-Level 数学都允许在三份卷子上使用科学计算器或图形计算器。
Formula booklets give you results, not judgement. The OCR H240 formula page prints the integration by parts formula itself, and the Pearson Edexcel booklet lists the standard integrals, so recalling the formula is rarely the difficulty. What needs practice is the faster decision underneath it, namely which factor becomes u. A graphical calculator is useful after the fact; enter the same definite integral, compare the number with your answer, and a large gap points to a sign or coefficient slip.
Edexcel AQA OCR 卷面结构对照
三家考试局的卷面结构不一样，安排复习时要跟着改。Pearson Edexcel 9MA0 是三份两小时、各一百分的卷子，前两份考纯数学，第三份统计与力学各占五十分。AQA 7357 同样是三份两小时一百分的卷子，第一份是纯数学，第二份是纯数学加力学，第三份是纯数学加统计。OCR H240 的 Paper 1 是纯数学，Paper 2 是纯数学与统计，Paper 3 是纯数学与力学，每份两小时一百分。纯数学内容在三份卷子上都可能出现，所以分部积分落在哪一份都不奇怪。按两小时一百分计算，平均每分的写作时间是一点二分钟。
The three boards structure their papers differently, which changes how you should plan revision. Pearson Edexcel 9MA0 uses three two-hour papers of 100 marks each; the first two are pure and the third splits statistics and mechanics fifty-fifty. AQA 7357 also runs three two-hour papers of 100 marks, with pure content on the first and mixed pure content on the other two. OCR H240 keeps pure mathematics in every paper as well, so a by parts question can appear anywhere, and 120 minutes for 100 marks works out at about 1.2 minutes per mark.
按被积函数结构分类刷真题
把近几年的真题按被积函数分类，你会发现分部积分只有几种固定长相。多项式乘指数或者三角函数，靠降次做完；单纯的 ln x 或 x 乘 ln x，靠补一个因子 1；指数乘三角，靠解方程把 I 提出来；定积分加面积，考的是上下限处理。四类各做五道题，比按年份刷整卷更快形成手感。每道题做完都写一行复核。用求导或计算器确认答案，比重新读一遍笔记更快。错题只记错在哪一步，不抄整题，第二遍复习时能直接看到自己的模式。
Sorting past paper questions by the shape of the integrand is more efficient than working through whole papers by year. Four patterns cover almost everything: a polynomial times an exponential or a trigonometric function, a lone logarithm, an exponential times a trigonometric function, and a definite integral attached to an area. Five questions of each type build recognition quickly. Record only the step where you went wrong rather than copying the whole question, so the second pass shows your own pattern of errors.
自测方式与错题分类
自测的方式很简单：合上笔记，从乘积求导法则重新推出公式。再默写四类题型的处理顺序。很多学生以为公式记住了就等于会做题，真正的门槛在第一步的选择，选错方向后面无从下手。另一个信号是做题时反复试错，这通常说明结构判断还没有形成条件反射。常见的错误是两轮分部积分中途换了 u 的类型，循环型积分里的 I 于是被消掉。模拟考之后把错题按结构分类，下一次遇到同类题先写下 u 与 dv 再动笔。
A useful self test starts from the product rule: close your notes, rebuild the formula, then write out the handling order for each of the four patterns. Knowing the formula is not the same as being able to choose u, and repeated trial and error during practice usually means that choice is still slow. The most common structural error is switching the type of u between rounds in a cyclic integral, which makes the I terms cancel. File your mistakes by pattern rather than by paper.
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