Integration by Parts Made Easy for Edexcel A-Level Maths | 分部积分法精讲

📚 Integration by Parts Made Easy for Edexcel A-Level Maths | 分部积分法精讲

Integration by parts is one of the most important techniques in the Edexcel A-Level Pure Mathematics specification, especially in Paper 2 and Paper 3. It allows you to integrate products of functions such as x eˣ, x sin x and ln x. Many students find the method tricky at first because it changes the way we think about integration: instead of reversing a derivative directly, we transform one integral into another. This revision guide breaks down the formula, shows how to choose u and dv/dx, and works through typical exam-style examples. By the end you will be able to handle both indefinite and definite integrals with confidence.

分部积分法是 Edexcel A-Level 纯数学考纲中最重要的技巧之一，尤其在 Paper 2 和 Paper 3 中经常出现。它可以用来积分 x eˣ、x sin x 和 ln x 这类函数乘积。许多学生一开始觉得这个方法棘手，因为它改变了我们对积分的思考方式：不是直接逆转导数，而是把一个积分转换成另一个积分。本复习指南将拆解公式，说明如何选择 u 和 dv/dx，并演练典型考试例题。学完后你将能自信地处理不定积分和定积分。

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

For the product of two functions, we use the formula ∫ u (dv/dx) dx = uv − ∫ v (du/dx) dx. Here u is a function we differentiate, and dv/dx is a function we integrate to obtain v. The formula comes from the product rule for differentiation: d(uv)/dx = u dv/dx + v du/dx. Rearranging and integrating both sides gives the result above. In practice we write the four parts u, du/dx, v and dv/dx in a table before substituting into the formula.

对于两个函数的乘积，我们使用公式 ∫ u (dv/dx) dx = uv − ∫ v (du/dx) dx。这里 u 是我们要求导的函数，dv/dx 是我们要求积分的函数并得到 v。该公式来自微分的乘法法则：d(uv)/dx = u dv/dx + v du/dx。重新整理并对两边积分就得到上面的结果。实际解题时，我们先把 u、du/dx、v 和 dv/dx 四部分写成表格，再代入公式。

∫ u (dv/dx) dx = uv − ∫ v (du/dx) dx

2. Choosing u and dv/dx | 选择 u 和 dv/dx

Choosing the correct u is the key to making integration by parts work. A useful rule is LIATE: Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential. Pick u as the function that appears earliest in this list. For example, in ∫ x eˣ dx, choose u = x because Algebraic comes before Exponential, and set dv/dx = eˣ dx. In ∫ ln x dx, choose u = ln x because Logarithmic comes first. This often reduces the new integral to something simpler.

正确选择 u 是分部积分法成功的关键。一个有用的法则是 LIATE：对数函数、反三角函数、代数函数、三角函数、指数函数。选择列表中排在最前面的函数作为 u。例如，在 ∫ x eˣ dx 中，选 u = x，因为代数函数排在指数函数之前，并令 dv/dx = eˣ dx。在 ∫ ln x dx 中，选 u = ln x，因为对数函数排在最前。这通常能把新积分化简为更简单的形式。

LIATE order: Logarithmic → Inverse trigonometric → Algebraic → Trigonometric → Exponential. For Edexcel papers, the most common choices are u = ln x, u = x, u = x², or u = arctan x, with dv/dx taking the remaining factor.

LIATE 顺序：对数函数 → 反三角函数 → 代数函数 → 三角函数 → 指数函数。在 Edexcel 考试中，最常见的 u 选择是 u = ln x、u = x、u = x² 或 u = arctan x，而 dv/dx 则取剩余因子。

3. Worked Example 1: ∫ x eˣ dx | 例题 1：∫ x eˣ dx

Let u = x and dv/dx = eˣ. Then du/dx = 1 and v = eˣ. Substituting into the formula gives ∫ x eˣ dx = x eˣ − ∫ eˣ dx. The remaining integral is simply eˣ, so we get x eˣ − eˣ + C. Factorising gives eˣ(x − 1) + C. Always include the constant of integration for indefinite integrals.

令 u = x，dv/dx = eˣ。则 du/dx = 1，v = eˣ。代入公式得到 ∫ x eˣ dx = x eˣ − ∫ eˣ dx。剩下的积分就是 eˣ，因此得到 x eˣ − eˣ + C。提取公因式得到 eˣ(x − 1) + C。不定积分一定要加上积分常数。

∫ x eˣ dx = eˣ(x − 1) + C

4. Worked Example 2: ∫ x sin x dx | 例题 2：∫ 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 + ∫ cos x dx. Since ∫ cos x dx = sin x, the final answer is −x cos x + sin x + C. Notice how the negative sign is a common source of error.

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 + ∫ cos x dx。因为 ∫ cos x dx = sin x，最终答案是 −x cos x + sin x + C。注意负号是常见的出错点。

∫ x sin x dx = −x cos x + sin x + C

5. Worked Example 3: ∫ ln x dx | 例题 3：∫ ln x dx

This example surprises many students because there is only one obvious function. We set u = ln x and dv/dx = 1. Then du/dx = 1/x and v = x. Substitution gives ∫ ln x dx = x ln x − ∫ x(1/x) dx = x ln x − ∫ 1 dx. Therefore the result is x ln x − x + C, often written as x(ln x − 1) + C.

这个例题让许多学生感到意外，因为被积函数看似只有一个函数。我们令 u = ln x，dv/dx = 1。则 du/dx = 1/x，v = x。代入得到 ∫ ln x dx = x ln x − ∫ x(1/x) dx = x ln x − ∫ 1 dx。因此结果是 x ln x − x + C，通常写作 x(ln x − 1) + C。

∫ ln x dx = x(ln x − 1) + C

6. Repeated Integration by Parts | 多次分部积分

Some integrals such as ∫ x² eˣ dx require applying integration by parts more than once. First choose u = x² and dv/dx = eˣ, giving du/dx = 2x and v = eˣ. This yields ∫ x² eˣ dx = x² eˣ − ∫ 2x eˣ dx. Now apply the method again to ∫ 2x eˣ dx with u = 2x and dv/dx = eˣ. The result is x² eˣ − 2(x eˣ − eˣ) + C, which simplifies to eˣ(x² − 2x + 2) + C. Keep your working organised to avoid sign errors.

有些积分如 ∫ x² eˣ dx 需要多次使用分部积分法。首先选 u = x²，dv/dx = eˣ，得到 du/dx = 2x，v = eˣ。于是 ∫ x² eˣ dx = x² eˣ − ∫ 2x eˣ dx。再对 ∫ 2x eˣ dx 使用同样方法，令 u = 2x，dv/dx = eˣ。结果是 x² eˣ − 2(x eˣ − eˣ) + C，化简为 eˣ(x² − 2x + 2) + C。保持计算过程清晰，以避免符号错误。

∫ x² eˣ dx = eˣ(x² − 2x + 2) + C

7. Definite Integrals | 定积分应用

For definite integrals, evaluate the uv term between the limits first, then subtract the integral of v du/dx. For example, ∫₀¹ x eˣ dx = [x eˣ − eˣ]₀¹ = [eˣ(x − 1)]₀¹. Substituting the upper limit gives e¹(1 − 1) = 0, and the lower limit gives e⁰(0 − 1) = −1. Therefore the value is 0 − (−1) = 1. Remember that the limits apply to the whole expression,

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