📚 Integration by Parts for Edexcel A-Level Maths | Edexcel A-Level 数学：分部积分法
Integration by parts is a core integration technique in Edexcel A-Level Mathematics, especially in Pure Mathematics 3 and 4. It reverses the product rule and is used when the integrand is a product of two different types of functions, such as algebraic and exponential, algebraic and trigonometric, or logarithmic alone. Edexcel exam questions frequently ask students to integrate expressions like x sin x, x² eˣ, ln x, and eˣ cos x, and may also combine the method with definite integrals or repeated applications.
分部积分法是 Edexcel A-Level 数学（尤其是纯数学 3 和 4）中的核心积分技巧。它反用乘积法则，适用于被积函数为两类不同函数乘积的情况，例如代数函数与指数函数、代数函数与三角函数，或单独的对数函数。Edexcel 考试题经常要求学生积分 x sin x、x² eˣ、ln x、eˣ cos x 等表达式，也可能将分部积分与定积分或多次应用相结合。
1. The Formula and Its Origin | 公式与来源
The formula comes from the product rule. For two differentiable functions u(x) and v(x), the derivative of the product uv is d(uv)/dx = u dv/dx + v du/dx. Rearranging gives u dv/dx = d(uv)/dx − v du/dx. Integrating both sides with respect to x produces the integration by parts formula.
该公式来源于乘积法则。对于两个可导函数 u(x) 和 v(x)，乘积 uv 的导数为 d(uv)/dx = u dv/dx + v du/dx。移项可得 u dv/dx = d(uv)/dx − v du/dx。两边同时对 x 积分，就得到了分部积分公式。
∫ u (dv/dx) dx = uv − ∫ v (du/dx) dx
2. Choosing u and dv | 选择 u 和 dv
Choosing u and dv/dx is the most important decision. A useful priority order for u is LIATE: Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential. Functions higher in this list generally become simpler when differentiated, so they are usually chosen as u. The remaining part dv/dx must be easy to integrate to find v.
选择 u 和 dv/dx 是最重要的步骤。一个实用的 u 优先顺序是 LIATE：对数函数、反三角函数、代数函数、三角函数、指数函数。排列越靠前的函数求导后通常越简单，因此一般被选为 u。剩下的部分 dv/dx 必须容易积分以求出 v。
3. Basic Polynomial Times Exponential | 基础：多项式乘以指数函数
Example: Find ∫ 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 = x eˣ − eˣ + C.
例题：求 ∫ x eˣ dx。设 u = x，则 du/dx = 1。设 dv/dx = eˣ，则 v = eˣ。代入公式得到 x eˣ − ∫ eˣ dx = x eˣ − eˣ + C。
The polynomial x becomes 1 after differentiation, so the new integral is simpler. This is exactly why we choose the algebraic function as u.
多项式 x 求导后变为 1，因此新积分更简单。这正是选择代数函数作为 u 的原因。
4. Integrating ln x | 对 ln x 积分
ln x does not look like a product, but we can write it as 1 × ln x. Set u = ln x and dv/dx = 1. Then du/dx = 1/x and v = x. The formula gives x ln x − ∫ x × (1/x) dx = x ln x − ∫ 1 dx = x ln x − x + C.
ln x 看起来不像乘积，但我们可以将其写成 1 × ln x。设 u = ln x，dv/dx = 1。则 du/dx = 1/x，v = x。代入公式得 x ln x − ∫ x × (1/x) dx = x ln x − ∫ 1 dx = x ln x − x + C。
5. Polynomial Times Trigonometric Functions | 多项式乘以三角函数
For ∫ x sin x dx, choose u = x and dv/dx = sin x. Then du/dx = 1 and v = −cos x. Applying the formula: ∫ x sin x dx = −x cos x − ∫ (−cos x) dx = −x cos x + sin x + C. The minus sign must be handled carefully.
对于 ∫ x sin x dx，选择 u = x，dv/dx = sin x。则 du/dx = 1，v = −cos x。应用公式：∫ x sin x dx = −x cos x − ∫ (−cos x) dx = −x cos x + sin x + C。负号必须小心处理。
6. Repeating Integration by Parts | 重复使用分部积分法
When the algebraic part has degree 2 or higher, repeat the process. For ∫ x² eˣ dx, first set u = x² and dv/dx = eˣ, giving x² eˣ − ∫ 2x eˣ dx. Then apply parts to ∫ 2x eˣ dx with u = 2x, dv/dx = eˣ. This yields 2x eˣ − ∫ 2 eˣ dx = 2x eˣ − 2 eˣ + C. Combining gives x² eˣ − 2x eˣ + 2 eˣ + C.
当代数部分的次数为 2 或更高时，需要重复使用分部积分。对于 ∫ x² eˣ dx，先设 u = x²，dv/dx = eˣ，得到 x² eˣ − ∫ 2x eˣ dx。再对 ∫ 2x eˣ dx 使用分部积分，设 u = 2x，dv/dx = eˣ，得到 2x eˣ − ∫ 2 eˣ dx = 2x eˣ − 2 eˣ + C。合并得到 x² eˣ − 2x eˣ + 2 eˣ + C。
7. Circular Integrals | 循环积分法
Some integrals return to the original expression after two applications. Let I = ∫ eˣ sin x dx. First, choose u = sin x, dv/dx = eˣ; then v = eˣ and du/dx = cos x, so I = eˣ sin x − ∫ eˣ cos x dx. For the second integral, choose u = cos x, dv/dx = eˣ; then v = eˣ and du/dx = −sin x, giving ∫ eˣ cos x dx = eˣ cos x + ∫ eˣ sin x dx = eˣ cos x + I. Substitute back: I = eˣ sin x − (eˣ cos x + I), so I = eˣ sin x − eˣ cos x − I. Rearranging gives 2I = eˣ sin x − eˣ cos x + C, hence I = ½ eˣ(sin x − cos x) + C.
有些积分在两次分部积分后回到原式。设 I = ∫ eˣ sin x dx。第一次设 u = sin x，dv/dx = eˣ；则 v = eˣ，du/dx = cos x，所以 I = eˣ sin x − ∫ eˣ cos x dx。第二个积分设 u = cos x，dv/dx = eˣ；则 v = eˣ，du/dx = −sin x，得到 ∫ eˣ cos x dx = eˣ cos x + ∫ eˣ sin x dx = eˣ cos x + I。代回得 I = eˣ sin x − (eˣ cos x + I)，即 I = eˣ sin x − eˣ cos x − I。移项得 2I = eˣ sin x − eˣ cos x + C，因此 I = ½ eˣ(sin x − cos x) + C。
8. Definite Integrals | 定积分的分部积分
For definite integrals, the formula is ∫ₐᵇ u (dv/dx) dx = [uv]ₐᵇ − ∫ₐᵇ v (du/dx) dx. Evaluate uv at b and a first, then subtract the remaining definite integral. Example: ∫₀¹ x eˣ dx = [x eˣ − eˣ]₀¹ = (1 e¹ − e¹) − (0 e⁰ − e⁰) = 0 − (−1) = 1.
对于定积分，公式为 ∫ₐᵇ u (dv/dx) dx = [uv]ₐᵇ − ∫ₐᵇ v (du/dx) dx。先计算 uv 在 b 和 a 处的值，再减去剩余的定积分。例：∫₀¹ x eˣ dx = [x eˣ − eˣ]₀¹ = (1 e¹ − e¹) − (0 e⁰ − e⁰) = 0 − (−1) = 1。
Make sure the limits stay attached to all terms if you write each part separately.
如果分开书写每一项，请确保积分限始终伴随所有项。
9. Common Mistakes | 常见错误
Common errors include forgetting the minus sign, choosing u and dv/dx the wrong way round, failing to simplify du/dx before the second integration, losing the constant of integration, and making sign errors when v is a negative cosine or sine. Also, do not differentiate v by mistake; v must be integrated correctly.
常见错误包括忘记负号、u 和 dv/dx 选反、在第二次积分前没有化简 du/dx、丢失积分常数，以及当 v 为负余弦或正弦时出现符号错误。此外，不要把 v 错误地求导；v 必须正确地积分。
10. Exam Tips and Summary | 考试技巧与总结
In the exam, always show the substitutions
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