📚 Mastering Integration by Parts (Exercise 13C) | 掌握分部积分法（练习13C）
Integration by parts is one of the most powerful techniques in an IB Mathematics AA HL student's toolkit. In Exercise 13C, you are challenged to apply this method to a variety of integrals, ranging from straightforward products of polynomials and exponentials to more intricate combinations involving logarithms and trigonometric functions. This article will guide you through the essential concepts, step‑by‑step strategies, and common pitfalls, helping you turn integration by parts into a reliable skill for both your homework and the final exam.
分部积分法是IB数学AA HL学生工具箱中最强大的技巧之一。在练习13C中，你将面对一系列应用该方法的积分题目，从简单的多项式与指数函数的乘积，到包含对数函数和三角函数的更复杂组合。本文会带你梳理核心概念、分步策略以及常见陷阱，帮你把分部积分法变成作业和期末大考中一项可靠的技能。
1. The Core Idea of Integration by Parts | 分部积分法的核心思想
At its heart, integration by parts is the integration counterpart of the product rule for differentiation. Whenever you have an integral of a product of two functions, the product rule in reverse can often break it down into simpler parts. The technique transforms a difficult integral into an easier one (or sometimes into an equation that can be solved for the original integral).
分部积分法本质上是求导乘法法则的积分对应形式。当你遇到两个函数乘积的积分时，反过来运用乘法法则往往能把它拆分成更简单的部分。这个技巧会把一个难积分的式子转化成一个更容易的积分（或者有时会变成一个关于原积分的方程）。
2. The Formula and Its Derivation | 公式及推导
Start with the product rule for differentiation: d/dx [u(x) v(x)] = u'(x)v(x) + u(x)v'x). By rearranging and integrating both sides, we obtain the integration by parts formula: ∫ u dv = uv − ∫ v du. Here we choose one part of the original integrand to be u, whose derivative du = u' dx we can easily compute, and the rest to be dv, which we can integrate to obtain v.
从乘法的求导法则出发：d/dx [u(x) v(x)] = u'(x)v(x) + u(x)v'(x)。移项并对两边积分，我们就得到了分部积分公式：∫ u dv = uv − ∫ v du。在这里，我们从原被积函数中选择一部分作为u（要容易求导得到du = u' dx），其余部分作为dv（要容易积分得到v）。
In practice, the formula is often written as ∫ f(x)g(x) dx = f(x)G(x) − ∫ f'(x)G(x) dx, where f is the function chosen for differentiation and g is the function chosen for integration. However, the u · dv notation is more compact and widely used in IB textbooks.
实际使用时，公式常常写成 ∫ f(x)g(x) dx = f(x)G(x) − ∫ f'(x)G(x) dx，其中 f 是被选择用于求导的函数，g 是被选择用于积分的函数。不过 u · dv 的记号更简洁，也是IB教材中通用的形式。
3. Choosing u and dv: The LIATE Rule | 选择u和dv：LIATE规则
A correct choice of u and dv is crucial. The LIATE rule gives a priority order for selecting u: Logarithmic, Inverse trigonometric, Algebraic (polynomials), Trigonometric, Exponential. The function that appears earlier in this list is usually chosen as u, while the rest (including dx) becomes dv.
正确选择u和dv至关重要。LIATE规则为选择u提供了一个优先顺序：Logarithmic（对数函数）、Inverse trigonometric（反三角函数）、Algebraic（代数函数，如多项式）、Trigonometric（三角函数）、Exponential（指数函数）。列表中越靠前的函数通常选作u，其余部分（包括dx）作为dv。
| Priority | Function Type | Examples |
|---|---|---|
| 1 (highest priority as u) | Logarithmic | ln x, log₂ x |
| 2 | Inverse trigonometric | arcsin x, arctan x |
| 3 | Algebraic (polynomials) | x², 3x + 1 |
| 4 | Trigonometric | sin x, cos 2x |
| 5 (lowest priority) | Exponential | eˣ, 2ˣ |
For the integral ∫ x eˣ dx, x is algebraic (A) and eˣ al (E). According to LIATE, algebraic comes before exponential, so we set u = x and dv = eˣ dx. This gives du = dx and v = eˣ, leading to a straightforward result.
对于积分 ∫ x eˣ dx，x 是代数函数（A），eˣ 是指数函数（E）。按照LIATE，代数优先于指数，所以我们设 u = x，dv = eˣ dx。这样 du = dx，v = eˣ，从而得到简洁的结果。
4. Basic Example: ∫ x eˣ dx | 基本例题：∫ x eˣ dx
Let u = x, so du = dx; dv = eˣ dx, so v = eˣ. Applying the formula: ∫ x eˣ dx = u v − ∫ v du = x eˣ − ∫ eˣ dx = x eˣ − eˣ + C = eˣ e product was transformed into a simple exponential integral minus a basic antiderivative.
设 u = x，则 du = dx；dv = eˣ dx，则 v = eˣ。代入公式：∫ x eˣ dx = u v − ∫ v du = x eˣ − ∫ eˣ dx = x eˣ − eˣ + C = eˣ(x − 1) + C。可以看到乘积被转化成了一个简单的指数积分减去基本原函数的形式。
l ∫ v du must be easier to evaluate than the original. If we had swapped the choice (u = eˣ, dv = x dx), the new integral would become ∫ (x²/2) eˣ dx, which is even more complicated.
这个例子展示了该方法的精髓：新的积分 ∫ v du 必须比原积分更容易计算。如果我们反过来选择（u = eˣ，dv = x dx），新的积分会变成 ∫ (x²/2) eˣ dx，这反而更复杂了。
5. Example with a Logarithm: ∫ ln x dx | 对数的例子：∫ ln x dx
x and v = x. Thus, ∫ ln x dx = x ln x − ∫ x · (1/x) dx = x ln x − ∫ 1 dx = x ln x − x + C.
自然对数的积分是分部积分法大放异彩的经典情形，尽管被积函数看起来只有一个函数。我们可以把它视为 ln x 与 1 dx 的乘积。设 u = ln x（LIATE中的L），dv = 1 dx，则 du = (1/x) dx，v = x。因此，∫ ln x dx = x ln x − ∫ x · (1/x) dx = x ln x − ∫ 1 dx = x ln x − x + C。
ks for inverse trigonometric functions like ∫ arctan x dx, where u = arctan x and dv = dx.
这种利用 dv = dx 的小技巧在IB考试中经常出现。同样的方法也适用于反三角函数，例如 ∫ arctan x dx，设 u = arctan x，dv = dx。
6. Repeated Integration by Parts | 重复分部积分
x, giving ∫ x² cos x dx = x² sin x − ∫ 2x sin x dx. The remaining integral ∫ 2x sin x dx is then handled with a second integration by parts (u = 2x, dv = sin x dx).
当被积函数是 xⁿ 乘以指数函数或三角函数的形式时，一次分部积分会把 x 的幂次降低一次。反复应用就可以完全消去多项式部分。例如，∫ x² cos x dx 需要两轮。首先设 u = x²，dv = cos x dx，得到 ∫ x² cos x dx = x² sin x − ∫ 2x sin x dx。剩下的积分 ∫ 2x sin x dx 再通过第二次分部积分处理（u = 2x，dv = sin x dx）。
In Exercise 13C, you will often encounter integrals that demand two or even three iterations. Patience and careful bookkeeping of signs and constants are essential to avoid careless mistakes.
在练习13C中，你经常会遇到需要进行两次甚至三次迭代的积分。耐心以及仔细处理符号和常数对于避免粗心错误至关重要。
7. Cyclic Integrals with Trigonometric and Exponential Functions | 三角函数与指数函数的循环积分
Some integrals, like ∫ eˣ sin x dx or ∫ eˣ re the original integral reappears after two applications of integration by parts. For ∫ eˣ sin x dx, set u = sin x, dv = eˣ dx (or the other way around – both routes work). After the first round you get an expression involving ∫ eˣ cos x dx. After a second integration by parts on that integral, the original ∫ eˣ s an unknown I and solving the resulting linear equation, you can obtain the final answer.
某些积分，如 ∫ eˣ sin x dx 或 ∫ eˣ cos x dx，在两次分部积分之后会出现原积分复现的情况。对于 ∫ eˣ sin x dx，可以设 u = sin x，dv = eˣ dx（或者反过来——两种方式都行）。第一轮之后你会得到一个包含 ∫ eˣ cos x dx 的表达式；对这个积分做第二次分部积分后，原来的 ∫ eˣ sin x dx 又以不同的系数重新出现。把原积分看作未知数 I 并求解得到的线性方程，就能得出最终答案。
This cyclic method is a favourite in IB exams because it tests your algebraic manipulation as well as your integration skills. Always remember to add the constant of integration only at the very end.
这种循环方法是IB考试中最爱考察的类型之一，因为它既考验积分能力，又考验代数运算。一定要记住只在最后才加上积分常数。
8. Tabular Integration (Shortcut for Repeated Parts) | 表格积分法（重复分部的捷径）
need to integrate a product of a polynomial and a repeatedly integrable function (e.g., eˣ rite the other function and its repeated integrals in a second column. Draw diagonal arrows with alternating signs starting with '+' gonals and sum the results. This method is especially efficient for integrals like ∫ x³ e²ˣ dx.
当需要积分一个多项式与一个可反复积分的函数（如 eˣ、sin x、cos x）的乘积时，表格积分法可以加速计算。在一列中写下多项式及其逐次导数，直到导数为零；在另一列中写下另一个函数及其反复积分。画斜箭头，符号从 '+' 开始交替。沿斜线相乘并求和。这种方法对于诸如 ∫ x³ e²ˣ dx 的积分特别高效。
IB markschemes do not require you to use tabular integration, but it is a popular time‑saving tool among high‑scoring students. Just be sure you understand the underlying integration by parts logic before relying on it.
IB评分方案并不要求你使用表格积分法，但它是高分考生中很流行的时间节省工具。不过在依赖它之前，务必确保你理解了底层的分部积分逻辑。
9. Common Mistakes and Pitfalls | 常见错误与陷阱
Choosing u and dv poorly: Ignoring LIATE is the number one reason students get stuck. Always check if the new integral is genuinely simpler. Forgetting the minus sign: The formula is uv − ∫ v du, not plus. A misplaced sign can ruin the entire solution. Dropping the constant of integration: Even if intermediate steps omit +C for brevity, the final answer must include it. Mis‑computing du or v: A derivative mistake (especially with chain rule) or an integration mistake (e.g., missing a factor) compounds quickly. Double‑check each derivative and antiderivative.
选择u和dv不当：忽略LIATE规则是学生卡壳的首要原因。务必检查新积分是否真的更简单。遗忘负号：公式是 uv − ∫ v du，不是加号。放错一个符号会毁掉整个解答。丢掉积分常数：尽管中间步骤为简洁可能省略 +C，最终答案必须包含。算错 du 或 v：求导出错（尤其是链式法则）或积分出错（例如漏掉系数）会迅速叠加。务必复查每一个导数和原函数。
10. Connection to IB Exam Questions | 与IB考试题的联系
In IB Mathematics Analysis and Approaches HL, integration by parts appears in Paper 1 (without calculator) and Paper 2 (with calculator). Questions are often structured in parts: first they may ask you to find a derivative, then use that result to integrate a related function. Other typical tasks include finding definite integrals using integration by parts, evaluating areas under curves, or solving differential equations where the integrating factor method leads to a by‑parts integral.
在IB数学分析与方法HL中，分部积分法出现在试卷一（无计算器）和试卷二（有计算器）中。题目通常分小问：先让你求一个导数，然后利用该结果积分一个相关函数。其他典型任务包括用分部积分法求定积分、计算曲线下方面积，或者解微分方程时利用积分因子导致的分部积分。
Exercise 13C is a direct preparation for these scenarios. Mastering the varied types of problems in this exercise will give you the fluency needed to tackle any exam question on this topic.
练习13C正是对这些情境的直接准备。掌握该练习中各种类型的问题，会给你带来足够的熟练度，以应对考试中任何有关该主题的题目。
11. Practice Tips for Exercise 13C | 练习13C的技巧
Work systematically: For each problem, write down u, dv, du, v explicitly before applying the formula. Verify by differentiation: Once you find an antiderivative, differentiate it mentally (or on scratch paper) to confirm it gives the original integrand. Check your boundaries: For definite integrals, remember to evaluate uv at both endpoints before subtracting the new integral. Mix and match: Some integrals require a combination of substitution and integration by parts – finish one technique first, then switch.
有条理地解题：对每道题，先明确写下 u、dv、du、v，再代入公式。用求导验证：求出原函数后，在心里（或草稿纸上）对其求导，确认是否得到原来的被积函数。检查定积分的上下限：对定积分，记住先在各端点计算 uv 的值，再减去新积分。混合使用技巧：有些积分需要换元积分法和分部积分法的组合——先完成一种，再转为另一种。
12. Final Thoughts | 总结
Integration by parts is not just another item on the syllabus; it is a gateway to deeper calculus thinking. By understanding when and how to apply it – and by practicing thoroughly with Exercise 13C – you build the analytical versatility that distinguishes an excellent IB mathematician. Every time you successfully choose u and dv, track signs, and recognise a cyclic pattern, you are sharpening skills that will serve you well in university mathematics and beyond.
分部积分法不仅仅是教学大纲上的又一个条目；它是通往更深层微积分思维的大门。通过理解何时以及如何应用它——并通过练习13C的充分训练——你将练就分析上的多面性，这正是出色IB数学家的标志。每一次你成功地选择 u 和 dv、追踪符号，并识别出循环结构，你都是在磨练那些会对大学数学乃至更远的学习大有裨益的技能。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图