📚 Integration by Parts | 分部积分法
Integration by parts is a powerful technique for evaluating integrals of products of functions. It is derived from the product rule for differentiation and often transforms a difficult integral into a simpler one. This article covers the formula, strategic choices, worked examples, and exam tips for the Edexcel A-Level Maths syllabus.
分部积分法是计算函数乘积积分的强大技巧。它源自微分的乘积法则，常能将困难积分转化为较简单的积分。本文涵盖公式、策略性选择、详细范例以及针对 Edexcel A-Level 数学大纲的考试技巧。
1. Introduction to Integration by Parts | 分部积分法简介
The product rule for differentiation states that (uv)' = u'v + uv'. Integrating both sides with respect to x gives ∫ uv' dx = uv - ∫ u'v dx. Replacing v' dx with dv and u' dx with du leads to the integration by parts formula.
微分的乘积法则表明 (uv)' = u'v + uv'。对两边关于 x 积分得 ∫ uv' dx = uv - ∫ u'v dx。将 v' dx 换作 dv，u' dx 换作 du，即得分部积分公式。
l × trigonometric. Instead of expanding, we assign parts to u and dv to simplify the integration process.
当被积函数是两种不同类型函数的乘积时，该方法尤其有用，例如多项式 × 指数函数或多项式 × 三角函数。我们通过指定 u 和 dv 的部分来简化积分过程，而不是展开。
2. The Integration by Parts Formula | 分部积分公式
dx. To apply the formula we must identify u and dv from the integrand, then compute du = u' dx and v = ∫ dv, and finally evaluate the resulting integral.
标准形式为 ∫ u dv = uv - ∫ v du。其中 u 和 v 是 x 的函数，微分 dv 包含 dx。应用该公式须从被积函数中确定 u 和 dv，然后计算 du = u' dx 以及 v = ∫ dv，最后求出产生的新积分。
∫ u dv = uv - ∫ v du
The success of the technique depends crucially on making the right choices for u and dv. A poor choice can make the new integral even more complicated than the original one.
该技巧的成败关键在于正确选择 u 和 dv。不当的选择会使新积分比原积分更复杂。
3. Choosing u and dv: The LIATE Rule | 选择 u 和 dv：LIATE 法则
A systematic guideline is the LIATE rule, which ranks candidate functions for u in the following order: Logarithmic functions, Inverse trigonometric functions, Algebraic functions, Trigonometric functions, Exponential functions. Choose u as the function appearing earliest in this list, and let dv be the remaining part of the integrand, including dx.
一个系统性的指导是 LIATE 法则，它将候选项 u 的函数按以下顺序排列：对数函数 (Logarithmic)、反三角函数 (Inverse trigonometric)、代数函数 (Algebraic)、三角函数 (Trigonometric)、指数函数 (Exponential)。将列表最靠前的函数选作 u，被积函数的剩余部分（包括 dx）作为 dv。
For example, in ∫ x² ln x dx, u = ln x (Logarithmic) and dv = x² dx. In ∫ x eˣ dx, u = x (Algebraic) and dv = eˣ dx. In ∫ eˣ cos x dx, u could be cos x (Trigonometric) or eˣ (Exponential) — either may work because they lead to a cycle, but LIATE suggests u = cos x.
例如，在 ∫ x² ln x dx 中，u = ln x (对数)，dv = x² dx。在 ∫ x eˣ dx 中，u = x (代数)，dv = eˣ dx。在 ∫ eˣ cos x dx 中，u 可选 cos x (三角)或 eˣ (指数) —— 两者皆可因产生循环，而 LIATE 建议 u = cos x。
4. Basic Examples | 基本范例
Example 1: ∫ x eˣ dx. Let u = x, dv = eˣ dx. Then du = dx, v = eˣ. Applying the formula: ∫ x eˣ dx = x eˣ - ∫ eˣ dx = x eˣ - eˣ + C = eˣ(x - 1) + C.
例1：∫ x eˣ dx。设 u = x，dv = eˣ dx，则 du = dx，v = eˣ。代入公式：∫ x eˣ dx = x eˣ - ∫ eˣ dx = x eˣ - eˣ + C = eˣ(x - 1) + C。
en du = dx, v = -cos x. So ∫ x sin x dx = -x cos x - ∫ (-cos x) dx = -x cos x + sin x + C.
例2：∫ x sin x dx。设 u = x，dv = sin x dx，du = dx，v = -cos x。因此 ∫ x sin x dx = -x cos x - ∫ (-cos x) dx = -x cos x + sin x + C。
These simple cases illustrate how the algebraic part is chosen as u, thereby reducing the power of x after differentiation, while dv integrates to a function of similar complexity.
这些简单情形说明代数部分被选为 u，通过求导降低 x 的幂次，而 dv 积分后复杂度不变。
5. Repeated Integration by Parts | 多次分部积分
When the algebraic factor has degree 2 or higher, such as ∫ x² eˣ dx, we need to apply integration by parts twice. First set u = x², dv = eˣ dx, giving du = 2x dx, v = eˣ. Then ∫ x² eˣ dx = x² eˣ - ∫ 2x eˣ dx.
当代数因子的次数为2或更高时，例如 ∫ x² eˣ dx，需要两次分部积分。首次设 u = x²，dv = eˣ dx，得 du = 2x dx，v = eˣ。则 ∫ x² eˣ dx = x² eˣ - ∫ 2x eˣ dx。
The remaining integral ∫ 2x eˣ dx again requires parts: let u = 2x, dv = eˣ dx, so du = 2 dx, v = eˣ. This yields 2x eˣ - ∫ 2eˣ dx = 2x eˣ - 2eˣ + C. Substituting back: x² eˣ - (2x eˣ - 2eˣ) + C = eˣ(x² - 2x + 2) + C.
剩余积分 ∫ 2x eˣ dx 仍需分部积分：令 u = 2x，dv = eˣ dx，
Published by TutorHao | A-Level Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图