Integration by Parts: Techniques and Applications for Edexcel A-Level Mathematics — 分部积分法：Edexcel A-Level 数学的技巧与应用

Introduction: Why Integration by Parts Matters — 引言：为什么分部积分法如此重要

Integration by parts is arguably the single most important integration technique in the Edexcel A-Level Mathematics syllabus. It is the natural counterpart to the product rule of differentiation and provides a systematic method for integrating products of functions that cannot be simplified through substitution or algebraic manipulation alone. In Edexcel Pure Mathematics Paper 2, integration by parts questions appear consistently, typically worth between five and twelve marks. Beyond the pure mathematics context, the technique also features prominently in Mechanics problems involving variable forces, work done by non-constant forces, and the derivation of equations of motion from acceleration functions. A thorough command of integration by parts is therefore essential for achieving a top grade.

分部积分法可以说是 Edexcel A-Level 数学大纲中最重要的积分技巧。它是微分乘法法则的自然对应，提供了一种系统性的方法来对无法通过代换或代数化简来处理的函数乘积进行积分。在 Edexcel 纯数学试卷二中，分部积分法题目稳定出现，通常价值 5 到 12 分。在纯数学范围之外，该技巧也频繁出现在涉及变力、非常力做功以及从加速度函数推导运动方程的力学问题中。因此，彻底掌握分部积分法对于取得高分至关重要。

The Derivation from the Product Rule — 从乘法法则推导

The integration by parts formula is derived directly from the product rule of differentiation. Recall that for two differentiable functions u(x) and v(x), the product rule states: d/dx(uv) = u(dv/dx) + v(du/dx). If we integrate both sides with respect to x, we obtain: ∫ d/dx(uv) dx = ∫ u(dv/dx) dx + ∫ v(du/dx) dx. The left side simplifies to uv, giving: uv = ∫ u(dv/dx) dx + ∫ v(du/dx) dx. Rearranging yields the standard formula: ∫ u(dv/dx) dx = uv − ∫ v(du/dx) dx. This derivation is worth memorising because it reveals the underlying logic: we are trading one integral for another, and the technique only works when the new integral is simpler than the original.

分部积分公式直接由微分的乘法法则推导而来。回顾一下，对于两个可微函数 u(x) 和 v(x)，乘法法则为：d/dx(uv) = u(dv/dx) + v(du/dx)。如果对两边关于 x 积分，我们得到：∫ d/dx(uv) dx = ∫ u(dv/dx) dx + ∫ v(du/dx) dx。左边简化为 uv，得到：uv = ∫ u(dv/dx) dx + ∫ v(du/dx) dx。重新排列得到标准公式：∫ u(dv/dx) dx = uv − ∫ v(du/dx) dx。这个推导值得记住，因为它揭示了底层逻辑：我们是在用一个积分交换另一个积分，只有当新积分比原积分更简单时，这个技巧才有效。

The LIATE Rule: A Systematic Approach to Choosing u — LIATE 法则：选择 u 的系统方法

The most critical decision in any integration by parts problem is the choice of u and dv. A poor choice leads to a more complicated integral and a dead end. The LIATE mnemonic provides a reliable priority order for selecting u. The acronym stands for Logarithmic functions (ln x, logₐ x), Inverse trigonometric functions (arcsin x, arccos x, arctan x), Algebraic functions (xⁿ, polynomial expressions), Trigonometric functions (sin x, cos x, tan x), and Exponential functions (eˣ, aˣ). The function type appearing earliest in LIATE should typically be chosen as u, because differentiating these functions generally simplifies them: the derivative of ln x is 1/x, which is algebraically simpler; the derivative of arcsin x is 1/√(1−x²), which opens up substitution possibilities; while differentiating an algebraic polynomial reduces its degree.

在任何分部积分问题中，最关键的决定是 u 和 dv 的选择。糟糕的选择会导致积分变得更加复杂，走进死胡同。LIATE 口诀提供了一个可靠的选择 u 的优先级顺序。该缩写代表对数函数、反三角函数、代数函数（多项式）、三角函数和指数函数。LIATE 中出现最早的函数类型通常应该被选为 u，因为对这些函数求导通常会简化它们：ln x 的导数是 1/x，代数上更简单；arcsin x 的导数是 1/√(1−x²)，为代换法打开了可能性；而对代数多项式求导会降低其次数。

Worked Example 1: Basic Polynomial times Exponential — 例题一：基本多项式乘以指数函数

Evaluate the indefinite integral ∫ x eˣ dx. Following LIATE, we note that x is Algebraic (third position) and eˣ is Exponential (fifth position). Algebraic appears earlier, so we set u = x and dv/dx = eˣ. Then du/dx = 1, so du = dx. To find v, we integrate dv/dx: v = ∫ eˣ dx = eˣ. Substituting into the formula: ∫ x eˣ dx = x·eˣ − ∫ eˣ·1 dx = x eˣ − ∫ eˣ dx = x eˣ − eˣ + C. Factorising: ∫ x eˣ dx = eˣ(x − 1) + C. This is a foundational result; any polynomial multiplied by eˣ can be handled by repeated application of this approach.

计算不定积分 ∫ x eˣ dx。依照 LIATE，注意 x 是代数函数（第三位），eˣ 是指数函数（第五位）。代数函数出现更早，所以我们设 u = x，dv/dx = eˣ。那么 du/dx = 1，所以 du = dx。要求 v，我们对 dv/dx 积分：v = ∫ eˣ dx = eˣ。代入公式：∫ x eˣ dx = x·eˣ − ∫ eˣ·1 dx = x eˣ − ∫ eˣ dx = x eˣ − eˣ + C。因式分解：∫ x eˣ dx = eˣ(x − 1) + C。这是一个基础结果；任何多项式乘以 eˣ 都可以通过重复应用这个方法来解决。

Worked Example 2: Polynomial times Trigonometric Function — 例题二：多项式乘以三角函数

Evaluate ∫ x sin x dx. Here x is Algebraic and sin x is Trigonometric. According to LIATE, Algebraic precedes Trigonometric, so we let u = x and dv/dx = sin x. Then du/dx = 1, giving du = dx, and v = ∫ sin x dx = −cos x. Applying the formula: ∫ x sin x dx = x(−cos x) − ∫ (−cos x)·1 dx = −x cos x + ∫ cos x dx = −x cos x + sin x + C. To verify, differentiate: d/dx(−x cos x + sin x) = −cos x + x sin x + cos x = x sin x, which matches the original integrand.

计算 ∫ x sin x dx。这里 x 是代数函数，sin x 是三角函数。按 LIATE，代数函数在三角函数之前，所以我们令 u = x，dv/dx = sin x。那么 du/dx = 1，得 du = dx，而 v = ∫ sin x dx = −cos x。应用公式：∫ x sin x dx = x(−cos x) − ∫ (−cos x)·1 dx = −x cos x + ∫ cos x dx = −x cos x + sin x + C。验证：求导 d/dx(−x cos x + sin x) = −cos x + x sin x + cos x = x sin x，与原被积函数一致。

Worked Example 3: The Logarithm Trick — 例题三：对数函数的技巧

Evaluate ∫ ln x dx. At first glance, this appears to be a single function, not a product. However, we can always multiply by 1 without changing the value: ∫ ln x dx = ∫ 1·ln x dx. Now we have a product. Following LIATE, Logarithmic functions come first, so u = ln x and dv/dx = 1. Then du/dx = 1/x, giving du = (1/x)dx, and v = ∫ 1 dx = x. Substituting: ∫ ln x dx = x ln x − ∫ x·(1/x) dx = x ln x − ∫ 1 dx = x ln x − x + C. This is a classic result that every A-Level student should know by heart. The same trick works for inverse trigonometric functions: treat ∫ arctan x dx as ∫ 1·arctan x dx with u = arctan x.

计算 ∫ ln x dx。乍一看这像个单一函数，不是乘积。然而，我们总是可以乘以 1 而不改变值：∫ ln x dx = ∫ 1·ln x dx。现在我们有了一个乘积。按 LIATE，对数函数排在最前面，所以 u = ln x，dv/dx = 1。那么 du/dx = 1/x，得 du = (1/x)dx，而 v = ∫ 1 dx = x。代入：∫ ln x dx = x ln x − ∫ x·(1/x) dx = x ln x − ∫ 1 dx = x ln x − x + C。这是每个 A-Level 学生都应该熟记于心的经典结果。同样的技巧适用于反三角函数：将 ∫ arctan x dx 视为 ∫ 1·arctan x dx，设 u = arctan x。

Worked Example 4: Repeated Integration by Parts — 例题四：重复分部积分

Evaluate ∫ x² eˣ dx. We set u = x² (Algebraic) and dv/dx = eˣ (Exponential). Then du/dx = 2x, giving du = 2x dx, and v = eˣ. First application: ∫ x² eˣ dx = x² eˣ − ∫ eˣ·2x dx = x² eˣ − 2∫ x eˣ dx. The new integral ∫ x eˣ dx still requires integration by parts. We apply the technique again with u = x, dv/dx = eˣ, giving du = dx, v = eˣ. Then: ∫ x eˣ dx = x eˣ − ∫ eˣ dx = x eˣ − eˣ + C₁. Substituting this back into the original expression: ∫ x² eˣ dx = x² eˣ − 2(x eˣ − eˣ) + C = x² eˣ − 2x eˣ + 2eˣ + C. Factorising: ∫ x² eˣ dx = eˣ(x² − 2x + 2) + C. Notice the emerging pattern: for ∫ xⁿ eˣ dx, the result is eˣ times a polynomial of degree n with alternating signs.

计算 ∫ x² eˣ dx。我们设 u = x²（代数函数），dv/dx = eˣ（指数函数）。那么 du/dx = 2x，得 du = 2x dx，而 v = eˣ。第一次应用：∫ x² eˣ dx = x² eˣ − ∫ eˣ·2x dx = x² eˣ − 2∫ x eˣ dx。新的积分 ∫ x eˣ dx 仍需要分部积分。我们再次应用该技巧，设 u = x，dv/dx = eˣ，得 du = dx，v = eˣ。那么：∫ x eˣ dx = x eˣ − ∫ eˣ dx = x eˣ − eˣ + C₁。将其代回原表达式：∫ x² eˣ dx = x² eˣ − 2(x eˣ − eˣ) + C = x² eˣ − 2x eˣ + 2eˣ + C。因式分解：∫ x² eˣ dx = eˣ(x² − 2x + 2) + C。注意其中显现的模式：对于 ∫ xⁿ eˣ dx，结果是 eˣ 乘以一个带有交替符号的 n 次多项式。

Worked Example 5: The Circular Integral Pattern — 例题五：循环积分模式

Evaluate ∫ eˣ sin x dx. This is a famous case where integration by parts appears to lead in circles, but this circularity is exactly what gives us the answer. Let u = sin x (Trigonometric) and dv/dx = eˣ (Exponential). Although LIATE would suggest Trigonometric before Exponential, in practice both choices work, but one may be more convenient. With our choice, du/dx = cos x, giving du = cos x dx, and v = eˣ. First application: I = ∫ eˣ sin x dx = eˣ sin x − ∫ eˣ cos x dx. Now apply integration by parts to the new integral ∫ eˣ cos x dx. Let u = cos x, dv/dx = eˣ. Then du/dx = −sin x, giving du = −sin x dx, and v = eˣ. This gives: ∫ eˣ cos x dx = eˣ cos x − ∫ eˣ(−sin x) dx = eˣ cos x + ∫ eˣ sin x dx = eˣ cos x + I. Substituting back into the first equation: I = eˣ sin x − (eˣ cos x + I) = eˣ sin x − eˣ cos x − I. Adding I to both sides: 2I = eˣ sin x − eˣ cos x. Therefore: I = (1/2)eˣ(sin x − cos x) + C. This circular approach also works for ∫ eˣ cos x dx and for integrals involving products of trigonometric and exponential functions.

计算 ∫ eˣ sin x dx。这是一个著名的例子，分部积分法看似在原地绕圈，但正是这种循环性给出了答案。设 u = sin x（三角函数），dv/dx = eˣ（指数函数）。虽然 LIATE 会建议三角函数在指数函数之前，但实际上两种选择都可行，但其中一种可能更方便。按我们的选择，du/dx = cos x，得 du = cos x dx，v = eˣ。第一次应用：I = ∫ eˣ sin x dx = eˣ sin x − ∫ eˣ cos x dx。现在对新积分 ∫ eˣ cos x dx 应用分部积分法。设 u = cos x，dv/dx = eˣ。那么 du/dx = −sin x，得 du = −sin x dx，v = eˣ。得到：∫ eˣ cos x dx = eˣ cos x − ∫ eˣ(−sin x) dx = eˣ cos x + ∫ eˣ sin x dx = eˣ cos x + I。代回第一个方程：I = eˣ sin x − (eˣ cos x + I) = eˣ sin x − eˣ cos x − I。两边加 I：2I = eˣ sin x − eˣ cos x。因此：I = (1/2)eˣ(sin x − cos x) + C。这种循环方法也适用于 ∫ eˣ cos x dx 以及涉及三角函数和指数函数乘积的积分。

Worked Example 6: Definite Integration by Parts — 例题六：定积分的分部积分法

For definite integrals, the formula becomes: ∫ₐᵇ u (dv/dx) dx = [uv]ₐᵇ − ∫ₐᵇ v (du/dx) dx. The key difference is that the uv term is evaluated at the limits before subtracting the remaining integral. Consider ∫₀¹ x eˣ dx. From our earlier indefinite result, we know ∫ x eˣ dx = eˣ(x − 1). Evaluating at the limits: F(1) = e¹(1 − 1) = 0, F(0) = e⁰(0 − 1) = −1. Therefore ∫₀¹ x eˣ dx = 0 − (−1) = 1. Alternatively, applying the definite formula directly: ∫₀¹ x eˣ dx = [x eˣ]₀¹ − ∫₀¹ eˣ dx = (1·e¹ − 0·e⁰) − [eˣ]₀¹ = e − (e − 1) = 1. Both methods yield the same result.

对于定积分，公式变为：∫ₐᵇ u (dv/dx) dx = [uv]ₐᵇ − ∫ₐᵇ v (du/dx) dx。关键区别在于 uv 项在减去剩余积分之前需要在上下限处求值。考虑 ∫₀¹ x eˣ dx。从我们之前的不定积分结果可知 ∫ x eˣ dx = eˣ(x − 1)。在上下限处求值：F(1) = e¹(1 − 1) = 0，F(0) = e⁰(0 − 1) = −1。因此 ∫₀¹ x eˣ dx = 0 − (−1) = 1。另一种方法，直接应用定积分公式：∫₀¹ x eˣ dx = [x eˣ]₀¹ − ∫₀¹ eˣ dx = (1·e¹ − 0·e⁰) − [eˣ]₀¹ = e − (e − 1) = 1。两种方法得出相同的结果。

The Tabular Method: A Shortcut for Repeated Applications — 表格法：重复应用的捷径

When the integrand takes the form xⁿ eᵃˣ or xⁿ sin(ax) with a large value of n, performing integration by parts n times becomes tedious and error-prone. The tabular method, sometimes called the DI method or the rapid repeated integration by parts method, organises the computation into a simple table. Create two columns. In the left column, write u and repeatedly differentiate until you reach zero. In the right column, write dv and repeatedly integrate the same number of times. Then draw diagonal arrows from each left entry to the right entry one row below, alternating signs starting with positive. Multiply along each diagonal and sum the results. For ∫ x³ eˣ dx: differentiate x³ down the left column (x³, 3x², 6x, 6, 0); integrate eˣ down the right column (eˣ, eˣ, eˣ, eˣ, eˣ). The result is x³eˣ − 3x²eˣ + 6xeˣ − 6eˣ + C = eˣ(x³ − 3x² + 6x − 6) + C. This method is not examinable as a separate technique in Edexcel A-Level, but it provides a reliable verification tool.

当被积函数的形式为 xⁿ eᵃˣ 或 xⁿ sin(ax) 且 n 较大时，执行 n 次分部积分法变得繁琐且容易出错。表格法，有时称为 DI 法或快速重复分部积分法，将计算组织成一个简单的表格。创建两列。在左列中，写下 u 并重复求导直到变为零。在右列中，写下 dv 并重复积分相同次数。然后从每个左列条目向下一行的右列条目画对角线箭头，从正号开始交替符号。沿每条对角线相乘并求和。对于 ∫ x³ eˣ dx：在左列对 x³ 向下求导（x³, 3x², 6x, 6, 0）；在右列对 eˣ 向下积分（eˣ, eˣ, eˣ, eˣ, eˣ）。结果为 x³eˣ − 3x²eˣ + 6xeˣ − 6eˣ + C = eˣ(x³ − 3x² + 6x − 6) + C。这种方法在 Edexcel A-Level 中不作为独立的考试技巧，但它提供了可靠的验证工具。

Integration by Parts with Inverse Trigonometric Functions — 反三角函数的分部积分

Evaluate ∫ arctan x dx. Following LIATE, Inverse trigonometric functions are second in priority, so we let u = arctan x and dv/dx = 1. Then du/dx = 1/(1 + x²), giving du = dx/(1 + x²), and v = x. Applying the formula: ∫ arctan x dx = x arctan x − ∫ x/(1 + x²) dx. The remaining integral can be solved by substitution. Let t = 1 + x², then dt = 2x dx, so x dx = dt/2. Thus ∫ x/(1 + x²) dx = ∫ (1/t)·(dt/2) = (1/2) ln|t| + C = (1/2) ln(1 + x²) + C. Therefore ∫ arctan x dx = x arctan x − (1/2) ln(1 + x²) + C. This combination of integration by parts and substitution is a common pattern in Edexcel A-Level questions.

计算 ∫ arctan x dx。按 LIATE，反三角函数排在第二位，所以我们令 u = arctan x，dv/dx = 1。那么 du/dx = 1/(1 + x²)，得 du = dx/(1 + x²)，而 v = x。应用公式：∫ arctan x dx = x arctan x − ∫ x/(1 + x²) dx。剩余的积分可以通过代换法求解。令 t = 1 + x²，则 dt = 2x dx，所以 x dx = dt/2。因此 ∫ x/(1 + x²) dx = ∫ (1/t)·(dt/2) = (1/2) ln|t| + C = (1/2) ln(1 + x²) + C。因此 ∫ arctan x dx = x arctan x − (1/2) ln(1 + x²) + C。这种分部积分法与代换法的组合是 Edexcel A-Level 考题中的常见模式。

Applications in Mechanics: Work Done by Variable Forces — 力学中的应用：变力做功

Integration by parts is indispensable in Edexcel A-Level Mechanics. Consider a particle moving along the x-axis under the influence of a variable force F(x) = x e⁻ˣ. The work done by this force as the particle moves from x = 0 to x = a is given by W = ∫₀ᵃ F(x) dx = ∫₀ᵃ x e⁻ˣ dx. Let u = x, dv/dx = e⁻ˣ, so du = dx, v = −e⁻ˣ. Then W = [−x e⁻ˣ]₀ᵃ − ∫₀ᵃ (−e⁻ˣ) dx = −a e⁻ᵃ + 0 + ∫₀ᵃ e⁻ˣ dx = −a e⁻ᵃ + [−e⁻ˣ]₀ᵃ = −a e⁻ᵃ − e⁻ᵃ + 1 = 1 − e⁻ᵃ(a + 1). As a → ∞, W → 1, meaning the total work done over an infinite displacement is finite, which is a physically interesting result.

分部积分法在 Edexcel A-Level 力学中不可或缺。考虑一个粒子在变力 F(x) = x e⁻ˣ 作用下沿 x 轴运动。当粒子从 x = 0 移动到 x = a 时，该力所做的功为 W = ∫₀ᵃ F(x) dx = ∫₀ᵃ x e⁻ˣ dx。令 u = x，dv/dx = e⁻ˣ，所以 du = dx，v = −e⁻ˣ。那么 W = [−x e⁻ˣ]₀ᵃ − ∫₀ᵃ (−e⁻ˣ) dx = −a e⁻ᵃ + 0 + ∫₀ᵃ e⁻ˣ dx = −a e⁻ᵃ + [−e⁻ˣ]₀ᵃ = −a e⁻ᵃ − e⁻ᵃ + 1 = 1 − e⁻ᵃ(a + 1)。当 a → ∞ 时，W → 1，意味着在无限位移上做的总功是有限的，这是一个有趣的物理结果。

Edexcel Exam Technique and Mark Schemes — Edexcel 考试技巧与评分标准

Edexcel examiners award marks for specific steps in integration by parts questions. The mark scheme typically allocates one mark for correctly identifying u and dv/dx, one mark for finding du/dx and v, one mark for correctly substituting into the formula, one or two marks for evaluating the resulting integral, and a final mark for the correct simplified answer including the constant of integration where required. Always show your working explicitly. Write “Let u = …” and “dv/dx = …” on separate lines. For definite integrals, show the evaluation of [uv] at the limits as a separate step. If the question asks for an exact answer, leave your answer in terms of e or π rather than giving a decimal approximation. Common examiner comments note that students lose marks by omitting brackets around negative signs and by failing to simplify their final answer fully.

Edexcel 考官对分部积分题目中的特定步骤给分。评分标准通常为：正确识别 u 和 dv/dx 得一分，求出 du/dx 和 v 得一分，正确代入公式得一分，计算所得积分得一到两分，最后正确简化答案（包括所需的积分常数）得一分。务必明确展示你的解题过程。在单独的行上写”令 u = …”和”dv/dx = …”。对于定积分，将 [uv] 在上下限处的求值作为单独的步骤展示。如果题目要求精确答案，请以 e 或 π 的形式给出答案，而不是给出小数近似值。考官的常见评语指出，学生因省略负号周围的括号以及未能完全简化最终答案而失分。

Choosing Between Substitution and Integration by Parts — 在代换法和分部积分法之间选择

One of the key skills tested in Edexcel A-Level is recognising which integration technique to apply. As a general rule, if the integrand is a product of two different types of function (for example, algebraic and exponential, or logarithmic and trigonometric), integration by parts is likely the correct approach. If the integrand involves a composite function where the derivative of the inner function appears as a factor, substitution is more appropriate. For instance, ∫ x e^(x²) dx should be tackled by substitution (let u = x²) rather than integration by parts, because the derivative of x², namely 2x, appears as a factor. Meanwhile, ∫ x eˣ dx requires integration by parts because x and eˣ are unrelated function types with no derivative link. Developing the instinct to distinguish these cases comes from extensive practice with past paper questions.

Edexcel A-Level 考查的关键技能之一是识别应使用哪种积分技巧。作为一般规则，如果被积函数是两种不同类型函数的乘积（例如代数函数和指数函数，或对数函数和三角函数），分部积分法很可能是正确的方法。如果被积函数涉及复合函数，其中内部函数的导数作为一个因式出现，那么代换法更合适。例如，∫ x e^(x²) dx 应通过代换法（令 u = x²）来解决，而不是分部积分法，因为 x² 的导数 2x 作为因式出现。同时，∫ x eˣ dx 需要分部积分法，因为 x 和 eˣ 是不相关的函数类型，没有导数联系。培养区分这些情况的直觉来自于对历年真题的大量练习。

Common Mistakes and How to Avoid Them — 常见错误及其避免方法

Several recurring mistakes cost students marks on integration by parts questions. First, incorrectly choosing u and dv is the most fundamental error. If after one round of integration by parts the new integral looks more complicated than the original, you have almost certainly chosen u incorrectly. Second, sign errors are pervasive. When v = −cos x and you substitute into the formula, remember that the term is uv − ∫ v du, so the subtraction sign interacts with the negative sign in v. Write − ∫ (−cos x) dx = + ∫ cos x dx explicitly to avoid confusion. Third, for definite integrals, do not forget to evaluate [uv] at both limits before subtracting the integral. Fourth, when using the tabular method, ensure the alternating signs start with positive for the first diagonal. Fifth, always include +C for indefinite integrals; this mark is almost always awarded explicitly in the mark scheme. Finally, check your answer by differentiation. If differentiating your result does not recover the original integrand, there is a mistake somewhere.

几个反复出现的错误让学生们在分部积分题目上失分。首先，错误选择 u 和 dv 是最根本的错误。如果经过一轮分部积分后，新积分看起来比原积分更复杂，你几乎肯定选错了 u。其次，符号错误普遍存在。当 v = −cos x 且代入公式时，记住该项是 uv − ∫ v du，因此减号与 v 中的负号相互作用。明确写出 − ∫ (−cos x) dx = + ∫ cos x dx 以避免混淆。第三，对于定积分，在减去积分之前不要忘记计算 [uv] 在两个上下限上的值。第四，使用表格法时，确保交替符号从第一条对角线的正号开始。第五，对于不定积分，务必加上 +C；评分标准中几乎总是明确给这个分数。最后，通过求导检查你的答案。如果对你的结果求导不能还原原始被积函数，说明某处有错误。

Practice Questions and Exam Strategy — 练习题与考试策略

To build fluency with integration by parts, practice with a systematic progression. Begin with straightforward polynomial-exponential products such as ∫ x e²ˣ dx and ∫ x² e³ˣ dx. Move on to polynomial-trigonometric combinations like ∫ x cos 2x dx and ∫ x² sin x dx. Then tackle logarithmic integrals including ∫ x ln x dx and ∫ (ln x)² dx. Finally, attempt the circular integral patterns: ∫ e²ˣ sin 3x dx and ∫ eˣ cos 2x dx. In the exam, allocate roughly one minute per mark. If a question is worth 7 marks, you should plan to spend about 7 minutes on it. If you become stuck, move on and return later. Integration by parts questions are often placed in the middle to later sections of the paper, alongside other challenging pure mathematics topics such as differential equations and parametric integration.

要熟练掌握分部积分法，请按系统性进阶进行练习。从简单的多项式指数函数乘积开始，如 ∫ x e²ˣ dx 和 ∫ x² e³ˣ dx。接着练习多项式三角函数组合，如 ∫ x cos 2x dx 和 ∫ x² sin x dx。然后攻克对数积分，包括 ∫ x ln x dx 和 ∫ (ln x)² dx。最后，尝试循环积分模式：∫ e²ˣ sin 3x dx 和 ∫ eˣ cos 2x dx。考试中，大约每分钟一分。如果一道题值 7 分，你应该计划花大约 7 分钟在这道题上。如果你卡住了，继续往下做，稍后再回来。分部积分法题目通常出现在试卷的中后段，与其他具有挑战性的纯数学话题如微分方程和参数积分一起出现。

Summary and Key Takeaways — 总结与要点

Integration by parts is a versatile and indispensable technique for Edexcel A-Level Mathematics. The LIATE rule provides a reliable framework for choosing u, but always verify that your choice simplifies the integral. Master the five standard patterns: polynomial times exponential, polynomial times trigonometric, logarithmic functions disguised as products with 1, repeated integration by parts for higher-degree polynomials, and the circular integral pattern for products of exponential and trigonometric functions. Remember the definite integral variant of the formula and always check your work by differentiation. With disciplined practice and careful attention to algebraic signs, integration by parts becomes a reliable tool rather than a source of anxiety on exam day.

分部积分法是 Edexcel A-Level 数学中一个多功能且不可或缺的技巧。LIATE 法则为选择 u 提供了可靠的框架，但务必验证你的选择是否简化了积分。掌握五种标准模式：多项式乘以指数函数、多项式乘以三角函数、伪装成与 1 乘积的对数函数、针对高次多项式的重复分部积分，以及针对指数函数和三角函数乘积的循环积分模式。记住公式的定积分变体，并始终通过求导检查你的答案。通过有纪律的练习和对代数符号的仔细关注，分部积分法将成为一个可靠的工具，而非考试当天的焦虑来源。