Exercise 18D: Integration by Parts | 练习18D：分部积分法

📚 Exercise 18D: Integration by Parts | 练习18D：分部积分法

In many IB Analysis & Approaches HL courses, Exercise 18D is the point where students move beyond simple anti‑differentiation and tackle integrals involving products of algebraic, exponential, logarithmic and trigonometric functions. This article consolidates the integration‑by‑parts technique, unpacks the reasoning behind the LIATE rule and walks through the types of problems you will encounter in that exercise set.

在许多IB数学分析与方法（HL）课程中，练习18D 是学生从简单反导数走向处理含代数、指数、对数和三角函数的乘积积分的关键节点。本文整合分部积分法，解析LIATE法则背后的逻辑，并带领大家一步步完成该练习中会遇到的各类问题。

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1. What Makes Exercise 18D Different | 练习18D的特殊之处

Exercise 18D usually appears after you have mastered basic antiderivatives, reverse chain rule and simple substitution. Its sole focus is integration by parts, the technique that reverses the product rule for differentiation. The questions progress from straightforward products like x e^x to more involved cases requiring repeated application or clever algebraic manipulation.

练习18D通常出现在你已经掌握了基本反导数、反向链式法则和简单换元法之后。它的唯一重点就是分部积分法，这种方法是微分乘法法则的逆运算。题目从简单的乘积如 x e^x 开始，逐步过渡到需要多次应用或巧妙代数处理的复杂情形。

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2. The Integration by Parts Formula | 分部积分公式

From the product rule (f g)' = f ' g + f g ' we obtain the fundamental identity used in Exercise 18D:

∫ f(x) g '(x) dx = f(x) g(x) – ∫ f '(x) g(x) dx

If we let u = f(x) and dv = g '(x) dx, the formula is more compactly written as ∫ u dv = u v – ∫ v du.

由乘法法则 (f g)' = f ' g + f g ' 可以得到练习18D使用的核心恒等式：

∫ f(x) g '(x) dx = f(x) g(x) – ∫ f '(x) g(x) dx

若令 u = f(x)，dv = g '(x) dx，则公式可更紧凑地写为 ∫ u dv = u v – ∫ v du。

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3. Choosing u and dv: The LIATE Rule | 选择 u 和 dv：LIATE法则

Success in Exercise 18D hinges on making the right choice for u and dv. A widely used priority list is LIATE: Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential. The function higher in the list is typically chosen as u, because its derivative is simpler than its integral.

在练习18D中取得成功的关键在于为 u 和 dv 做出正确选择。广泛使用的优先级列表是 LIATE：对数函数、反三角函数、代数函数、三角函数、指数函数。列表中位置越靠前的函数通常被选为 u，因为它的导数比它的积分更简单。

• L – Logarithmic: ln x, logₐ x
• I – Inverse trig: arcsin x, arctan x
• A – Algebraic: xⁿ (polynomials)
• T – Trigonometric: sin x, cos x
• E – Exponential: e^x, a^x

ays test your choice mentally: the new integral ∫ v du should be easier than the original.

尽管LIATE是一个有用的指导，但每次都要在脑子里检验你的选择：新的积分 ∫ v du 应该比原来的积分更简单。

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4. Basic Examples: Polynomial Times Exponential | 基本例子：多项式乘指数函数

Consider ∫ x e^2x dx. Here we have an algebraic function x and an exponential e^2x. Following LIATE, we let u = x (Algebraic) and dv = e^2x dx. Then du = dx and v = (1/2) e^2x.

考虑 ∫ x e^2x dx。这里有一个代数函数 x 和一个指数函数 e^2x。按照LIATE，我们设 u = x（代数函数），dv = e^2x dx。那么 du = dx，v = (1/2) e^2x。

∫ x e^2x dx = x·(1/2)e^2x – ∫ (1/2)e^2x dx = (1/2)x e^2x – (1/4)e^2x + C

This type of question frequently appears in the early parts of Exercise 18D to build confidence.

这类题目经常出现在练习18D的前半部分，以帮助建立信心。

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5. Integrating Products with Trigonometric Functions | 含三角函数的乘积积分

When facing ∫ x sin x dx, LIATE suggests u = x (Algebraic) and dv = sin x dx. Then v = –cos x, giving:

∫ x sin x dx = –x cos x – ∫ (–cos x) dx = –x cos x + sin x + C

当面对 ∫ x sin x dx 时，LIATE建议设 u = x（代数），dv = sin x dx。那么 v = –cos x，得到：

∫ x sin x dx = –x cos x – ∫ (–cos x) dx = –x cos x + sin x + C

integral. Exercise 18D may also mix trigonometric functions with exponentials, such as ∫ e^x cos x dx, which requires a special cyclic approach.

注意 u 的导数（为1）使余下的积分大大简化。练习18D还可能将三角函数与指数函数混合，例如 ∫ e^x cos x dx，这需要一种特殊的循环处理方式。

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6. Integrating the Natural Logarithm (ln x) | 自然对数 ln x 的积分

the logarithmic function must be u, so let u = ln x, dv = 1 dx. Then du = (1/x) dx, v = x.

乍一看 ∫ ln x dx 不像乘积，但它可以写作 ∫ 1·ln x dx。按照LIATE，对数函数必须作为 u，因此设 u = ln x，dv = 1 dx。那么 du = (1/x) dx，v = x。

∫ ln x dx = x ln x – ∫ x · (1/x) dx = x ln x – x + C

This trick is tested regularly in IB exams and is a highlight of Exercise 18D because it forces you to think about the hidden “1”.

这个技巧在IB考试中经常出现，也是练习18D的一个亮点，因为它迫使你去思考那个隐藏的“1”。

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7. Repeated Integration by Parts | 重复分部积分

Some integrals, like ∫ x² e^x g integration by parts twice because the algebraic factor has degree 2. Set u = x², dv = e^x dx; then the new integral contains x e^x, which needs a second treatment.

某些积分，如 ∫ x² e^x dx，需要应用两次分部积分，因为代数因子的幂次为2。设 u = x²，dv = e^x dx；那么新的积分包含 x e^x，需要第二次处理。

First application: ∫ x² e^x dx = x² e^x – ∫ 2x e^x dx. Second application on ∫ 2x e^x dx gives 2x e^x – 2e^x. Collecting terms yields:

∫ x² e^x dx = x² e^x – 2x e^x + 2e^x + C

第一次应用：∫ x² e^x dx = x² e^x – ∫ 2x e^x dx。再对 ∫ 2x e^x dx 应用分部积分，得到 2x e^x – 2e^x。合并各项后：

∫ x² e^x dx = x² e^x – 2x e^x + 2e^x + C

Exercise 18D often arranges questions so that you systematically reduce the polynomial degree.

练习18D通常会安排题目让你系统地降低多项式的次数。

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8. The Cyclic Case: When the Integral Reappears | 循环情形：当原积分再次出现时

For ∫ e^x sin x dx, both factors are ‘self‑’ under differentiation and integration. Apply integration by parts with u = sin x, dv = e^x dx, giving:

对于 ∫ e^x sin x dx，两个因子在微分和积分下都是“自我再生”的。设 u = sin x，dv = e^x dx 进行分部积分，得到：

∫ e^x sin x dx = e^x sin x – ∫ e^x cos x dx. Now apply parts again to the new integral, this time with u = cos x, dv = e^x dx. The result contains the original integral ∫ e^x sin x dx once more.

∫ e^x sin x dx = e^x sin x – ∫ e^x cos x dx。现在对新积分再次使用分部积分，这次设 u = cos x，dv = e^x dx。结果中会再次出现原积分 ∫ e^x sin x dx。

Let I = ∫ e^x sin x dx. After two steps we obtain I = e^x sin x – e^x cos x – I, so 2I = e^x (sin x – cos x) and therefore I = (1/2) e^x (sin x – cos x) + C. This cyclic reasoning is a classic in Exercise 18D.

令 I = ∫ e^x sin x dx。经过两步后我们得到 I = e^x sin x – e^x cos x – I，因此 2I = e^x (sin x – cos x)，于是 I = (1/2) e^x (sin x – cos x) + C。这种循环推导是练习18D中的经典题型。

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9. Definite Integrals and Integration by Parts | 定积分的分部积分

When a definite integral is given, such as ∫₀¹ x e^–x dx, the integration‑by‑parts formula is applied to the antiderivative first, then the limits are substituted. Equivalently, we can use the definite version:

∫ₐᵇ u dv = [u v]ₐᵇ – ∫ₐᵇ v du

当给出定积分，如 ∫₀¹ x e^–x dx，先对不定积分使用分部积分，再代入上下限。等效地，我们可以使用定积分版本：

∫ₐᵇ u dv = [u v]ₐᵇ – ∫ₐᵇ v du

For example, ∫₀¹ x e^–x dx = [–x e^–x]₀¹ + ∫₀¹ e^–x dx = (–1·e⁻¹ + 0) + [–e^–x]₀¹ = –e⁻¹ – (e⁻¹ – 1) = 1 – 2e⁻¹. Many Exercise 18D problems ask for exact answers in terms of e.

例如，∫₀¹ x e^–x dx = [–x e^–x]₀¹ + ∫₀¹ e^–x dx = (–1·e⁻¹ + 0) + [–e^–x]₀¹ = –e⁻¹ – (e⁻¹ – 1) = 1 – 2e⁻¹。练习18D中的许多题目都要求用 e 表示的精确值。

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10. Common Pitfalls and How to Avoid Them | 常见错误及避免方法

Swapping u and dv unwisely: If you choose u = e^x in ∫ x e^x dx, you will end up with a more complicated integral. Always run a quick “derivative test” on your chosen u.

不明智地交换 u 和 dv：如果在 ∫ x e^x dx 中选择 u = e^x，最终会得到一个更复杂的积分。务必对你选择的 u 进行一次快速的“导数测试”。

Forgetting the constant of integration: In definite integrals this is not an issue, but in indefinite integrals leaving out + C loses a mark in IB examinations.

忘记积分常数：在定积分中这不是问题，但在不定积分中漏掉 +C 会在IB考试中扣分。

Mishandling signs: When dv = sin x dx, v = –cos x. A sign error in v spoils the entire solution. Write a small sign chart if needed.

符号处理错误：当 dv = sin x dx 时，v = –cos x。v 中的一个符号错误就会毁掉整个解答。如有必要，可以画一个小符号表。

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11. Exercise 18D Practice Questions Walkthrough | 练习18D典型题目分析

Let us examine three questions representative of the set. (1) ∫ x² ln x dx be u. Set u = ln x, dv = x² dx → du = (1/x) dx, v = x³/3. Then ∫ x² ln x dx = (x³/3) ln x – ∫ (x³/3)(1/x) dx = (x³/3) ln x – (1/3)∫ x² dx = (x³/3) ln x – x³/9 + C.

让我们分析该练习中三道有代表性的题目。(1) ∫ x² ln x dx。LIATE告诉我们 ln x 必须作为 u。设 u = ln x，dv = x² dx → du = (1/x) dx，v = x³/3。那么 ∫ x² ln x dx = (x³/3) ln x – ∫ (x³/3)(1/x) dx = (x³/3) ln x – (1/3)∫ x² dx = (x³/3) ln x – x³/9 + C。

(2) ∫ arctan x dx ∫ 1·arctan x dx. Let u = arctan x, dv = 1 dx → du = 1/(1+x²) dx, v = x. Result: x arctan x – dx. The remaining integral can be solved by substitution (let w = 1+x²), giving x arctan x – (1/2) ln(1+x²) + C.

(2) ∫ arctan x dx。写作 ∫ 1·arctan x dx。设 u = arctan x，dv = 1 dx → du = 1/(1+x²) dx，v = x。结果为：x arctan x – ∫ x/(1+x²) dx。剩下的积分可用换元法（令 w = 1+x²），得到 x arctan x – (1/2) ln(1+x²) + C。

(3) ∫ e^–x cos(2x) dx. This is a cyclic integral. Let I = ∫ e^–x cos(2x) dx. Choose u = cos(2x), dv = e^–x dx. Then I = –e^–x cos(2x) – ∫ 2e^–x sin(2x) dx. Apply parts again to the new integral: u = sin(2x), dv = e^–x dx leads to an equation involving I. Solving gives I = (e^–x/5)(2 sin(2x) – cos(2x)) + C.

(3) ∫ e^–x cos(2x) dx。这是一个循环积分。令 I = ∫ e^–x cos(2x) dx。选择 u = cos(2x)，dv = e^–x dx。则 I = –e^–x cos(2x) – ∫ 2e^–x sin(2x) dx。再对新积分使用分部积分：u = sin(2x)，dv = e^–x dx 会得到一个包含 I 的方程。解出 I = (e^–x/5)(2 sin(2x) – cos(2x)) + C。

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12. Summarising the Strategy for Exercise 18D | 练习18D策略总结

The core message of Exercise 18D is: identify u and dv wisely, apply the formula meticulously, and don’t hesitate to use repeated or cyclic applications. Before moving on to definite integrals, ensure your indefinite integration is flawless. Double‑check your v for sign errors and always verify by differentiating your final answer.

练习18D的核心信息是：明智地确定 u 和 dv，一丝不苟地应用公式，并且不要犹豫使用重复或循环应用。在进行定积分之前，确保你的不定积分准确无误。反复检查 v 的符号，并始终通过求导来验证你的最终答案。

By working through the full set of Exercise 18D, you internalise a technique that is essential not only for the IB exam but also for further study in calculus, physics and engineering.

通过完成练习18D的全部题目，你将内化这项技巧，这不仅对IB考试至关重要，对未来的微积分、物理和工程学习也同样重要。

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