Integration by Parts (Edexcel C4) | 分部积分法（爱德思 C4）

📚 Integration by Parts (Edexcel C4) | 分部积分法（爱德思 C4）

Integration by parts is one of the essential analytical tools in A-Level Mathematics, particularly within the Edexcel C4 module. It allows us to integrate products of two functions by transforming the original integral into a simpler one. Mastering this technique is not just about memorising a formula; it requires a deep understanding of function types, the strategic choice of u and dv/dx, and the ability to recognise when repeated or cyclic use is needed.

分部积分法是A-Level数学（尤其是爱德思考局C4模块）中必不可少的分析工具之一。它能通过将原积分转化为更简单的积分，来解决两个函数乘积的积分问题。掌握这一技巧不仅仅是记住公式，还需要深刻理解函数类型、策略性地选择 u 与 dv/dx，并能识别何时需要反复使用或循环使用。

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1. Introduction to Integration by Parts | 分部积分法简介

Integration by parts is essentially the reverse of the product rule for differentiation. While the product rule deals with derivatives of products, integration by parts provides a way to integrate products by splitting them into two parts that are easier to handle.

分部积分本质上就是微分中乘积法则的逆运算。乘积法则处理的是乘积的导数，而分部积分通过将被积函数拆分成更容易处理的两部分，提供了对乘积进行积分的方法。

In many A-Level problems, you will face integrals like ∫ x eˣ dx or ∫ x sin x dx, where direct integration is not possible. A substitution may not work, but integration by parts will reduce the integral to a basic form.

在许多A-Level题目中，你会碰到像 ∫ x eˣ dx 或 ∫ x sin x dx 这样的积分，这些都无法直接积分。换元法可能行不通，但分部积分法能够将积分约简为一个基本形式。

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2. Deriving the Formula | 公式推导

Start with the product rule for differentiation: d/dx (u v) = u (dv/dx) + v (du/dx). If we rearrange and integrate both sides with respect to x, we obtain the integration by parts formula.

从微分的乘积法则出发：d/dx (u v) = u (dv/dx) + v (du/dx)。重新整理并对 x 积分，就得到了分部积分公式。

Integrating both sides gives ∫ d/dx (u v) dx = ∫ u (dv/dx) dx + ∫ v (du/dx) dx. The left-hand side is simply u v. Hence, ∫ u (dv/dx) dx = u v – ∫ v (du/dx) dx. This is the standard formula used in Edexcel C4.

两边积分得到 ∫ d/dx (u v) dx = ∫ u (dv/dx) dx + ∫ v (du/dx) dx。左边就是 u v。因此，∫ u (dv/dx) dx = u v – ∫ v (du/dx) dx。这就是爱德思C4中使用的标准公式。

∫ u (dv/dx) dx = u v – ∫ v (du/dx) dx

A more compact version is often written as ∫ u dv = u v – ∫ v du, which highlights the symmetry: we differentiate u and integrate dv/dx.

更紧凑的形式常写作 ∫ u dv = u v – ∫ v du，这凸显了对称性：对 u 求导，对 dv/dx 积分。

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3. The LIATE Rule for Choosing u | 选择 u 的 LIATE 法则

The key to successful integration by parts is selecting which part of the integrand to call u and which to call dv/dx. A helpful mnemonic is LIATE, which orders function types from highest priority for u to lowest:

成功进行分部积分的关键在于选择被积函数中的哪一部分作为 u，哪一部分作为 dv/dx。一个有用的记忆方法是 LIATE，它将函数类型从作为 u 的最高优先级排列到最低优先级：

Letter	Type	Examples
L	Logarithmic	ln x, ln(2x)
I	Inverse trigonometric	arcsin x, arctan x
A	Algebraic (polynomial)	x², 3x+1
T	Trigonometric	sin x, cos 3x
E	Exponential	eˣ, 2ˣ

This means if you have an integral containing ln x and a polynomial, choose u = ln x. For x eˣ, u = x (algebraic) since exponential is lower on the list. The rule works well for most Edexcel exam questions.

这意味着，如果被积函数包含 ln x 和多项式，选择 u = ln x。对于 x eˣ，u = x（代数函数），因为指数函数的优先级更低。该法则适用于大多数爱德思考试题目。

Sometimes the choice is forced; for example, ∫ ln x dx has only one obvious candidate for u. In other cases, you may need to experiment. LIATE is a guide, not an absolute law.

有时选择是必然的；比如 ∫ ln x dx 只有一个明显的 u 候选。其他情况下可能需要尝试。LIATE 是一个指导原则，并非绝对规律。

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4. Basic Example: Polynomial Times Exponential | 基本例题：多项式乘以指数函数

Consider ∫ x eˣ dx. Let u = x, so du/dx = 1. Then let dv/dx = eˣ, which integrates to v = eˣ. Substituting into the formula gives ∫ x eˣ dx = x eˣ – ∫ eˣ (1) dx = x eˣ – eˣ + C.

考虑 ∫ x eˣ dx。令 u = x，则 du/dx = 1。再令 dv/dx = eˣ，积分得 v = eˣ。代入公式得到 ∫ x eˣ dx = x eˣ – ∫ eˣ (1) dx = x eˣ – eˣ + C。

∫ x eˣ dx = eˣ (x – 1) + C

Note how the polynomial x was differentiated to 1, which made the new integral much simpler. This is the hallmark of a good u choice: du/dx should be simpler than u.

注意多项式 x 被微分成了 1，这使得新的积分简单了许多。这是良好选择 u 的标志：du/dx 应该比 u 更简单。

For a slightly harder example, ∫ x² eˣ dx would require two applications of integration by parts, because differentiating x² twice brings it to a constant.

稍难一点的例子，∫ x² eˣ dx 需要两次分部积分，因为对 x² 微分两次会得到常数。

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5. Basic Example: Polynomial Times Trigonometric Function | 基本例题：多项式乘以三角函数

Evaluate ∫ x cos x dx. Set u = x (du/dx = 1) and dv/dx = cos x, so v = sin x. Then ∫ x cos x dx = x sin x – ∫ sin x (1) dx = x sin x + cos x + C.

计算 ∫ x cos x dx。设 u = x (du/dx = 1)，dv/dx = cos x，所以 v = sin x。那么 ∫ x cos x dx = x sin x – ∫ sin x (1) dx = x sin x + cos x + C。

As always, check by differentiating: d/dx (x sin x + cos x) = sin x + x cos x – sin x = x cos x, which matches the original integrand.

一如既往，通过求导验证：d/dx (x sin x + cos x) = sin x + x cos x – sin x = x cos x，与原被积函数一致。

If the trigonometric function had a coefficient inside, such as cos 2x, remember to adjust v accordingly. For dv/dx = cos 2x, v = ½ sin 2x.

如果三角函数内部有系数，比如 cos 2x，记得相应调整 v。若 dv/dx = cos 2x，则 v = ½ sin 2x。

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6. Integrating the Natural Logarithm | 积分自然对数函数

The integral ∫ ln x dx is a classic case where LIATE suggests u = ln x, and the remaining part is simply 1 dx. Let u = ln x (du/dx = 1/x) and dv/dx = 1, so v = x.

积分 ∫ ln x dx 是一个经典案例，LIATE 建议取 u = ln x，剩下的部分就是 1 dx。令 u = ln x (du/dx = 1/x)，dv/dx = 1，则 v = x。

Applying the formula: ∫ ln x dx = x ln x – ∫ x (1/x) dx = x ln x – ∫ 1 dx = x ln x – x + C.

运用公式：∫ ln x dx = x ln x – ∫ x (1/x) dx = x ln x – ∫ 1 dx = x ln x – x + C。

∫ ln x dx = x ln x – x + C

This result is frequently tested in C4. A similar approach works for ∫ (ln x)² dx, requiring parts twice.

这个结果在 C4 中经常考到。类似的方法也可用于 ∫ (ln x)² dx，需要两次分部积分。

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7. Integrating Inverse Trigonometric Functions | 积分反三角函数

Integrals like ∫ arctan x dx also rely on integration by parts with u = arctan x and dv/dx = 1. Then du/dx = 1/(1+x²) and v = x.

像 ∫ arctan x dx 这样的积分同样依赖于分部积分，令 u = arctan x，dv/dx = 1。则 du/dx = 1/(1+x²)，v = x。

Thus ∫ arctan x dx = x arctan x – ∫ x/(1+x²) dx. The remaining integral can be tackled by substitution (let w = 1+x²), giving ½ ln|1+x²| + C. The final answer is x arctan x – ½ ln(1+x²) + C.

因此 ∫ arctan x dx = x arctan x – ∫ x/(1+x²) dx。剩下的积分可用代换法（令 w = 1+x²）求得 ½ ln|1+x²| + C。最终答案是 x arctan x – ½ ln(1+x²) + C。

For arcsin x, a similar pattern emerges. These integrals are less common in Edexcel but appear occasionally in extension questions. Recognising them as parts problems is crucial.

对于 arcsin x，情况类似。这些积分在爱德思考试中较少见，但偶尔会出现在拓展题中。将其识别为分部积分问题至关重要。

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8. Repeated Integration by Parts | 反复使用分部积分法

When you have an integral like ∫ x² eˣ dx, one application reduces the power of x to 1, but still leaves an x in the integral. You then apply integration by parts again.

当你遇到像 ∫ x² eˣ dx 这样的积分时，一次分部积分将 x 的幂次降为 1，但积分中仍留有 x。这时需要再次使用分部积分。

First, let u = x², dv/dx = eˣ, giving ∫ x² eˣ dx = x² eˣ – ∫ 2x eˣ dx. Then for ∫ 2x eˣ dx, let u = 2x, dv/dx = eˣ, yielding 2x eˣ – ∫ 2eˣ dx = 2x eˣ – 2eˣ. Combine: ∫ x² eˣ dx = x² eˣ – 2x eˣ + 2eˣ + C = eˣ(x² – 2x + 2) + C.

首先令 u = x², dv/dx = eˣ，得 ∫ x² eˣ dx = x² eˣ – ∫ 2x eˣ dx。然后对 ∫ 2x eˣ dx，令 u = 2x, dv/dx = eˣ，得 2x eˣ – ∫ 2eˣ dx = 2x eˣ – 2eˣ。合并：∫ x² eˣ dx = x² eˣ – 2x eˣ + 2eˣ + C = eˣ(x² – 2x + 2) + C。

The pattern is consistent: the polynomial part eventually differentiates to zero after enough steps. This technique is sometimes called tabular integration and can be organised in a table to save time.

模式是一致的：多项式部分在足够多步后最终会微分至零。这种技巧有时称为表格积分法，可以整理成表格以节省时间。

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9. Cyclic Integration by Parts: Exponential and Trigonometric | 循环分部积分：指数与三角函数

A particularly interesting case arises with integrals like ∫ eˣ sin x dx. Here neither function reduces to a constant upon differentiation; instead, the integral cycles back to itself after two applications.

一个特别有趣的例子是 ∫ eˣ sin x dx 这类积分。在这里，两个函数都不会因求导而缩减为常数；相反，两次分部积分后积分会循环回到自身。

Let I = ∫ eˣ sin x dx. First, choose u = sin x, dv/dx = eˣ, so du/dx = cos x, v = eˣ. Then I = eˣ sin x – ∫ eˣ cos x dx. Now apply parts again to the second integral with u = cos x, dv/dx = eˣ: ∫ eˣ cos x dx = eˣ cos x + ∫ eˣ sin x dx = eˣ cos x + I.

令 I = ∫ eˣ sin x dx。首先选择 u = sin x, dv/dx = eˣ，则 du/dx = cos x, v = eˣ。那么 I = eˣ sin x – ∫ eˣ cos x dx。现在对第二个积分再次分部，令 u = cos x, dv/dx = eˣ：∫ eˣ cos x dx = eˣ cos x + ∫ eˣ sin x dx = eˣ cos x + I。

Substituting back: I = eˣ sin x – (eˣ cos x + I). So I = eˣ sin x – eˣ cos x – I. Solving for I gives 2I = eˣ(sin x – cos x), therefore I = ½ eˣ(sin x – cos x) + C.

代回：I = eˣ sin x – (eˣ cos x + I)。所以 I = eˣ sin x – eˣ cos x – I。解出 I 得 2I = eˣ(sin x – cos x)，因此 I = ½ eˣ(sin x – cos x) + C。

∫ eˣ sin x dx = ½ eˣ (sin x – cos x) + C

This cyclic nature is a standard C4 trick. Students must be careful to keep track of signs and remember to add the constant only at the end.

这种循环特性是 C4 中的标准技巧。学生必须小心处理符号，并记得只在最后加上常数。

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10. Definite Integration by Parts | 定积分的分部积分法

For definite integrals, the integration by parts formula includes the limits directly: ∫ₐᵇ u (dv/dx) dx = [u v]ₐᵇ – ∫ₐᵇ v (du/dx) dx.

对于定积分，分部积分公式直接包含上下限：∫ₐᵇ u (dv/dx) dx = [u v]ₐᵇ – ∫ₐᵇ v (du/dx) dx。

This means you evaluate the product u v at the upper and lower limits first, and then subtract the integral of v du/dx within those limits. It saves steps because you don’t need to revert to the indefinite integral first.

这意味着你先计算乘积 u v 在上下限处的值，再减去 v du/dx 在同一区间内的积分。这省去了先求不定积分的步骤。

For example, evaluate ∫₀¹ x eˣ dx. Take u = x, dv/dx = eˣ. Then [x eˣ]₀¹ = (1·e¹ – 0) = e. Next, – ∫₀¹ eˣ dx = – [eˣ]₀¹ = -(e – 1) = 1 – e. So the total is e + 1 – e = 1.

例如，计算 ∫₀¹ x eˣ dx。令 u = x，dv/dx = eˣ。则 [x eˣ]₀¹ = (1·e¹ – 0) = e。然后，- ∫₀¹ eˣ dx = – [eˣ]₀¹ = -(e – 1) = 1 – e。因此总和为 e + 1 – e = 1。

∫₀¹ x eˣ dx = 1

Always check the consistency of limits; they must be applied after finding v and throughout the remaining integral.

始终检查上下限的一致性；在求出 v 之后以及后续积分中都必须应用这些限值。

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11. Common Pitfalls and Tips | 常见误区与技巧

A frequent mistake is choosing the wrong function for u, which makes the new integral more complicated rather than simpler. Always think: does differentiating u give something simpler?

一个常见错误是选错了 u，导致新的积分反而更复杂。始终要思考：对 u 求导是否能得到更简单的表达式？

Another pitfall is forgetting to include the constant of integration in indefinite integrals, or mishandling signs when inserting into the formula. Double-checking by differentiation is highly recommended.

另一个误区是在不定积分中忘记加上积分常数，或者在代入公式时处理符号出错。强烈建议通过求导验算。

When limits are involved, students sometimes mistakenly apply the formula as if it were a product evaluation without the minus integral. Writing out each step clearly can prevent this.

涉及上下限时，学生有时会错误地应用公式，好像只需计算乘积而忘了减去积分。清晰地写出每一步可以防止此类错误。

Finally, for cyclic integrals, algebraic errors in solving for I are easy to make. Always isolate I systematically and verify by differentiating the final answer.

最后，对于循环积分，解 I 时容易发生代数错误。务必系统地将 I 分离出来，并通过求导验证最终答案。

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12. Practice Questions and Exam Advice | 练习题及考试建议

To build fluency, practise a mix of problems: ∫ x² ln x dx, ∫ e²ˣ cos 3x dx, and definite integrals like ∫₁ᵉ ln x dx. Edexcel often asks for exact values, so simplify fully.

为了熟练掌握，要练习各种题目：∫ x² ln x dx，∫ e²ˣ cos 3x dx，以及像 ∫₁ᵉ ln x dx 这样的定积分。爱德思常要求精确值，因此要彻底化简。

In the exam, show every step: state your choice of u and dv/dx, write du/dx and v, substitute into the formula, and simplify. Marks are awarded for correct method even if a small slip occurs.

考试中，要展示每一步：说明 u 与 dv/dx 的选择，写出 du/dx 和 v，代入公式并化简。即使有小失误，正确的方法也能获得步骤分。

It is also useful to recognise when integration by parts is not needed. For instance, ∫ sin x cos x dx is better tackled with a simple substitution. Subject knowledge helps you pick the right tool.

此外，识别何时不需要分部积分也很有用。例如 ∫ sin x cos x dx 用简单的代换法处理更好。扎实的学科知识能帮助你选择合适的工具。

Finally, time management is key. If you get stuck, move on and return later. A clear structured approach to integration by parts will save you time and earn you the marks.

最后，时间管理很关键。如果卡住了，就继续前进，稍后再回来。对分部积分采取条理清晰的方法将为你节省时间并赢得分数。

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