📚 Integration by Parts for Edexcel A-Level Maths | 爱德思A-Level数学：分部积分法
Integration by parts is one of the core A2 techniques in the Edexcel A-Level Pure Mathematics specification. It allows you to integrate products of two functions where standard substitution or inspection will not work, such as x eˣ, x sin x, or ln x. Mastering the method depends not only on remembering the formula but also on choosing which part to differentiate and which part to integrate.
分部积分法是爱德思 A-Level 纯数学 A2 阶段的核心技巧之一。当标准换元法或观察法无法处理两个函数的乘积时，例如 x eˣ、x sin x 或 ln x，就需要使用分部积分法。掌握这一方法不仅在于记住公式，还在于正确选择哪一部分求导、哪一部分积分。
1. The Formula and Its Origins | 公式及其来源
The integration by parts formula comes from reversing the product rule for differentiation. If u and v are both functions of x, then the product rule states that the derivative of uv is u dv/dx + v du/dx. Rearranging and integrating both sides gives the standard formula.
分部积分公式来源于乘积法则的逆运算。如果 u 和 v 都是 x 的函数，那么乘积法则说明 uv 的导数是 u dv/dx + v du/dx。整理两边并积分，就得到标准公式。
∫ u (dv/dx) dx = u v – ∫ v (du/dx) dx
In words, the integral of u times dv/dx is equal to u times v minus the integral of v times du/dx. This is often remembered as ‘u v minus integral of v du’. The key idea is to transfer the difficulty from one integral to another that should be easier to evaluate.
简单来说，u 乘以 dv/dx 的积分等于 u 乘以 v 减去 v 乘以 du/dx 的积分。常被记作 ‘u v 减去 v du 的积分’。其核心思想是把一个较难的积分转移成另一个更容易计算的积分。
2. Choosing u and dv: The LIATE Strategy | 选择 u 和 dv：LIATE 策略
A common mistake is choosing u and dv/dx poorly, which can make the new integral more complicated than the original. A useful guideline is the acronym LIATE, which gives the priority order for choosing u: Logarithmic functions, Inverse trigonometric functions, Algebraic functions such as polynomials, Trigonometric functions, and Exponential functions.
一个常见错误是 u 和 dv/dx 选择不当，导致新的积分比原来的更复杂。一个实用的指导原则是首字母缩写 LIATE，它给出了选择 u 的优先顺序：对数函数、反三角函数、代数函数（如多项式）、三角函数和指数函数。
- L – Logarithmic: ln x, ln(2x)
- I – Inverse trig: arcsin x, arctan x
- A – Algebraic: x, x², 3x³
- T – Trigonometric: sin x, cos 2x
- E – Exponential: eˣ, e⁻²ˣ
For example, in ∫ x eˣ dx, x is algebraic and eˣ is exponential. Since A comes before E in LIATE, choose u = x and dv/dx = eˣ. For ∫ x ln x dx, choose u = ln x because L has higher priority than A.
例如，在 ∫ x eˣ dx 中，x 是代数函数，eˣ 是指数函数。由于 LIATE 中 A 排在 E 之前，因此选择 u = x，dv/dx = eˣ。对于 ∫ x ln x dx，应选择 u = ln x，因为 L 的优先级高于 A。
3. Basic Polynomial × Exponential Example | 基本多项式乘指数例题
Consider the integral ∫ x eˣ dx. Let u = x, so du/dx = 1. Let dv/dx = eˣ, so v = eˣ. Substituting into the formula gives x eˣ – ∫ eˣ dx, which simplifies to x eˣ – eˣ + C.
考虑积分 ∫ x eˣ dx。令 u = x，则 du/dx = 1。令 dv/dx = eˣ，则 v = eˣ。代入公式得到 x eˣ – ∫ eˣ dx，化简为 x eˣ – eˣ + C。
∫ x eˣ dx = x eˣ – eˣ + C
Always check your result by differentiating. The derivative of x eˣ is eˣ + x eˣ by the product rule, and the derivative of -eˣ is -eˣ, leaving x eˣ exactly as required.
始终通过求导来检验结果。x eˣ 的导数是 eˣ + x eˣ（乘积法则），而 -eˣ 的导数是 -eˣ，剩下正好是 x eˣ，符合要求。
4. Polynomial × Trigonometric Example | 多项式乘三角函数例题
Integrate ∫ x cos x dx. According to LIATE, choose u = x and dv/dx = cos x. Then du/dx = 1 and v = sin x. The formula gives x sin x – ∫ sin x dx, which equals x sin x + cos x + C because the integral of sin x is -cos x.
计算 ∫ x cos x dx。根据 LIATE，选择 u = x，dv/dx = cos x。则 du/dx = 1，v = sin x。公式给出 x sin x – ∫ sin x dx，等于 x sin x + cos x + C，因为 sin x 的积分是 -cos x。
∫ x cos x dx = x sin x + cos x + C
Note that if you had chosen u = cos x and dv/dx = x, the resulting integral would involve x² sin x, which is more difficult. This shows why the LIATE order matters in practice.
注意，如果选择了 u = cos x、dv/dx = x，得到的积分将涉及 x² sin x，更加困难。这说明了为什么 LIATE 顺序在实践中很重要。
5. Integration of Natural Logarithm | 自然对数的积分
The integral of ln x is a classic application where no product is visible. You can treat ln x as 1 × ln x and choose u = ln x, dv/dx = 1. Then du/dx = 1/x and v = x. The formula gives x ln x – ∫ x (1/x) dx, which simplifies to x ln x – ∫ 1 dx = x ln x – x + C.
ln x 的积分是一个经典应用，表面上没有乘积。你可以把 ln x 看作 1 × ln x，选择 u = ln x，dv/dx = 1。则 du/dx = 1/x，v = x。公式给出 x ln x – ∫ x (1/x) dx，化简为 x ln x – ∫ 1 dx = x ln x – x + C。
∫ ln x dx = x ln x – x + C
This result is frequently used in exam questions, especially when finding areas under logarithmic curves. It is worth memorising the final form, but you must also be able to derive it using integration by parts.
这一结果在考试题中经常使用，尤其是在求对数曲线下的面积时。最终形式值得记忆，但你还必须能够使用分部积分法推导它。
6. Repeated Integration by Parts | 反复分部积分
Some integrals require the formula more than once. For example, ∫ x² eˣ dx needs two applications. First set u = x² and dv/dx = eˣ, giving v = eˣ and du/dx = 2x. The result is x² eˣ – ∫ 2x eˣ dx. Then apply integration by parts again to ∫ 2x eˣ dx with u = 2x and dv/dx = eˣ.
有些积分需要多次使用公式。例如 ∫ x² eˣ dx 需要两次。首先令 u = x²，dv/dx = eˣ，得到 v = eˣ，du/dx = 2x。结果是 x² eˣ – ∫ 2x eˣ dx。然后再对 ∫ 2x eˣ dx 使用分部积分，令 u = 2x，dv/dx = eˣ。
∫ x² eˣ dx = x² eˣ – 2x eˣ + 2eˣ + C
Each application reduces the power of x by one, so a term with xⁿ will generally require n applications of integration by parts. Keeping work organised in a table or clear steps is essential to avoid sign errors.
每使用一次，x 的幂就减少一次，因此含有 xⁿ 的项通常需要 n 次分部积分。保持步骤清晰或使用表格，对于避免符号错误至关重要。
7. Special Case: Boomerang Integrals | 特例：回旋积分
Sometimes integration by parts brings back the original integral. A standard Edexcel example is ∫ eˣ sin x dx. Let I = ∫ eˣ sin x dx. Choose u = sin x and dv/dx = eˣ, so v = eˣ and du/dx = cos x. This gives I = eˣ sin x – ∫ eˣ cos x dx. Now apply integration by parts again to the new integral with u = cos x and dv/dx = eˣ.
有时分部积分会带回原来的积分。一个标准的爱德思例子是 ∫ eˣ sin x dx。令 I = ∫ eˣ sin x dx。选择 u = sin x，dv/dx = eˣ，则 v = eˣ，du/dx = cos x。得到 I = eˣ sin x – ∫ eˣ cos x dx。现在对新的积分再次使用分部积分，令 u = cos x，dv/dx = eˣ。
After the second application you get ∫ eˣ cos x dx = eˣ cos x + ∫ eˣ sin x dx, so I = eˣ sin x – eˣ cos x – I. Collecting I terms gives 2I = eˣ sin x – eˣ cos x, and therefore I = ½ eˣ (sin x – cos x) + C.
第二次应用后得到 ∫ eˣ cos x dx = eˣ cos x + ∫ eˣ sin x dx，于是 I = eˣ sin x – eˣ cos x – I。合并 I 项得到 2I = eˣ sin x – eˣ cos x，因此 I = ½ eˣ (sin x – cos x) + C。
∫ eˣ sin x dx = ½ eˣ (sin x – cos x) + C
This ‘boomerang’ method is common for products of exponential and trigonometric functions, and it is important to add the constant of integration only at the very end.
这种 ‘回旋’ 方法常见于指数函数与三角函数的乘积，并且一定要在最后才加上积分常数。
8. Definite Integrals and Bounds | 定积分与上下限
For a definite integral, you can either evaluate the indefinite integral first and then substitute the limits, or apply the limits directly to the uv term. The direct method often saves time: ∫ₐᵇ u dv = [uv]ₐᵇ – ∫ₐᵇ v du.
对于定积分，你可以先求不定积分再代入上下限，也可以直接将上下限应用于 uv 项。直接方法通常更省时：∫ₐᵇ u dv = [uv]ₐᵇ – ∫ₐᵇ v du。
∫ₐᵇ u (dv/dx) dx = [u v]ₐᵇ – ∫ₐᵇ v (du/dx) dx
For example, to evaluate ∫₀¹ x eˣ dx, use u = x and dv/dx = eˣ. Then [x eˣ]₀¹ = e – 0 = e, and ∫₀¹ eˣ dx = e – 1. Therefore the exact value is e – (e – 1) = 1.
例如，计算 ∫₀¹ x eˣ dx，令 u = x，dv/dx = eˣ。则 [x eˣ]₀¹ = e – 0 = e，且 ∫₀¹ eˣ dx = e – 1。因此精确值为 e – (e – 1) = 1。
9. Common Mistakes and Exam Tips | 常见错误与考试技巧
One typical error is differentiating v instead of integrating when finding v from dv/dx. If dv/dx = eˣ, then v = eˣ, not the derivative. Another common mistake is forgetting the minus sign in the formula, especially when using the compact form ∫ u dv = uv – ∫ v du.
一个典型错误是在由 dv/dx 求 v 时进行了求导而不是积分。如果 dv/dx = eˣ，那么 v = eˣ，而不是它的导数。另一个常见错误是忘记公式中的负号，尤其是在使用紧凑形式 ∫ u dv = uv – ∫ v du 时。
- Always write down u, dv/dx, du/dx and v before substituting.
- Choose u using LIATE, but be flexible if one choice clearly makes the new integral simpler.
- Check the final answer by differentiating.
- Do not forget + C for indefinite integrals.
- With definite integrals, evaluate the uv term at both limits carefully.
考试时，先写出 u、dv/dx、du/dx 和 v，再代入。根据 LIATE 选择 u，但如果某个选择明显使新积分更简单，也要灵活处理。最后通过求导检查答案。不定积分不要忘记 + C。定积分要小心计算 uv 项在两个限值处的差。
10. Practice Checklist | 练习清单
Before your Edexcel exam, make sure you can confidently integrate products of polynomials with eˣ and trigonometric functions, integrate ln x, handle repeated integration by parts, and solve boomerang integrals such as ∫ eˣ sin x dx. You should also be able to apply the method to definite integrals and combine it with other techniques like trigonometric identities or partial fractions.
在参加爱德思考试之前，请确保你能自信地计算多项式与 eˣ、三角函数的乘积积分，能积分 ln x，能处理反复分部积分，也能求解如 ∫ eˣ sin x dx 的回旋积分。你还应能将分部积分法应用于定积分，并与三角恒等式或有理分式分解等其他技巧结合使用。
| Integral | Method summary |
| ∫ x e²ˣ dx | u = x, dv/dx = e²ˣ, then v = ½ e²ˣ |
| ∫ x sin 3x dx | u = x, dv/dx = sin 3x, then v = -⅓ cos 3x |
| ∫ ln 2x dx | u = ln 2x, dv/dx = 1, then simplify using log rules |
| ∫ x² cos x dx | Two applications, reducing x² to x then to a constant |
| ∫ eˣ cos 2x dx | Boomerang method, collect the original integral I |
Use this table as a quick revision card. Practise writing out full solutions, not just short answers, because Edexcel examiners award method marks for correct structure even if a small arithmetic error occurs later.
把这张表作为快速复习卡片使用。练习写出完整解题过程，而不仅仅是简略答案，因为爱德思考官会根据正确的方法步骤给分，即便之后出现小的计算错误也能获得方法分。
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