Integration by Parts Techniques | 分部积分法解题技巧

📚 Integration by Parts Techniques | 分部积分法解题技巧

Integration by parts is one of the most powerful techniques in A-Level Calculus. It transforms a difficult integral into a simpler one by reversing the product rule. This article provides a systematic breakdown of the method, with worked examples and exam-style advice.

分部积分法是 A-Level 微积分中最强大的工具之一。它通过逆用乘积法则，将一个较难的积分转化为一个更简单的积分。本文系统地讲解这一方法，并附有典型例题和考试技巧。

1. Formula and Derivation | 公式与推导

The integration by parts formula is derived directly from the product rule. If u = u(x) and v = v(x), then the product rule states that (uv)’ = u’v + uv’. Integrating both sides with respect to x gives the integration by parts formula.

分部积分公式直接由乘积法则推导而来。若 u = u(x)，v = v(x)，则乘积法则给出 (uv)’ = u’v + uv’。两边对 x 积分，即得分部积分公式。

∫u dv = uv – ∫v du

Equivalently, in the notation used in A-Level examinations:

等价地，在 A-Level 考试常用的记号中可写成：

∫u(dv/dx)dx = uv – ∫v(du/dx)dx

The key idea is to choose u and dv wisely. The term u is differentiated to obtain du, while dv is integrated to obtain v. The resulting integral ∫v du should be simpler than the original.

关键在于明智地选择 u 和 dv。对 u 求导得到 du，对 dv 积分得到 v。化简后的积分 ∫v du 应当比原积分更简单。

2. The LIATE Rule | LIATE 选择规则

A common and reliable guideline for choosing u is the LIATE rule. The earlier a function type appears in the acronym, the more suitable it is as u. This is because these functions generally become simpler when differentiated.

选择 u 的一个常用且可靠的指南是 LIATE 规则。该缩略词中排位越靠前的函数类型，越适合作为 u。这是因为这些函数求导后通常会变得更简单。

L	Logarithmic functions	对数函数 ln x
I	Inverse trigonometric functions	反三角函数 arctan x, arcsin x
A	Algebraic functions	代数函数 x, x², 2x + 3
T	Trigonometric functions	三角函数 sin x, cos x
E	Exponential functions	指数函数 eˣ

For example, in ∫x ln x dx, ln x is a logarithmic function and x is algebraic. Since L comes before A, we choose u = ln x and dv = x dx.

例如，在 ∫x ln x dx 中，ln x 是对数函数，x 是代数函数。由于 L 排在 A 之前，我们选择 u = ln x，dv = x dx。

3. Polynomial × Exponential | 多项式乘指数函数

When the integrand is a product of a polynomial and an exponential function, choose the polynomial as u and the exponential as dv. The degree of the polynomial is reduced by one after each application of integration by parts.

当被积函数是多项式与指数函数的乘积时，选择多项式作为 u，指数函数作为 dv。每应用一次分部积分，多项式的次数就降低一次。

Example: Evaluate ∫x eˣ dx.

例题：求 ∫x eˣ dx。

Let u = x and dv = eˣ dx. Then du = dx and v = eˣ. Applying the formula:

令 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

The new integral ∫eˣ dx is elementary, so the problem is solved in one step. For ∫x² eˣ dx, apply integration by parts twice.

新积分 ∫eˣ dx 是基本积分，因此一步即解。对于 ∫x² eˣ dx，则需要连续应用两次分部积分。

4. Polynomial × Trigonometric | 多项式乘三角函数

This type follows the same strategy: differentiate the polynomial and integrate the trigonometric function. One application reduces the polynomial degree by one.

此类型采用相同策略：对多项式求导，对三角函数积分。每应用一次即可将多项式次数降低一次。

Example: Evaluate ∫x sin x dx.

例题：求 ∫x sin x dx。

Choose u = x and dv = sin x dx. Then du = dx and v = -cos x. Thus:

取 u = x，dv = sin x dx。则 du = dx，v = -cos x。于是：

∫x sin x dx = -x cos x – ∫(-cos x)dx = -x cos x + sin x + C

Notice the careful handling of the negative sign. A common error is to write -x cos x – ∫cos x dx; always verify the sign after substituting v.

注意负号的处理。常见错误是写成 -x cos x – ∫cos x dx；代入 v 后务必检查符号。

5. Logarithmic and Inverse Trigonometric | 对数和反三角函数

For integrals involving ln x or arctan x, these functions are chosen as u even when they appear alone. In this case, set dv = dx.

对于含 ln x 或 arctan x 的积分，即使它们单独出现，也应作为 u 选取。此时令 dv = dx。

Example: Evaluate ∫ln x dx.

例题：求 ∫ln x dx。

Let u = ln x and dv = dx. Then du = (1/x)dx and v = x. Applying the formula:

令 u = ln x，dv = dx。则 du = (1/x)dx，v = x。代入公式：

∫ln x dx = x ln x – ∫x·(1/x)dx = x ln x – x + C

Similarly, for ∫arctan x dx, choose u = arctan x and dv = dx. Then du = 1/(1 + x²)dx, leading to a standard result.

类似地，对于 ∫arctan x dx，取 u = arctan x，dv = dx。则 du = 1/(1 + x²)dx，从而得到标准结果。

∫arctan x dx = x arctan x – ½ ln(1 + x²) + C

6. Exponential × Trigonometric | 指数乘三角函数（循环法）

When the integrand is eˣ times sin x or cos x, neither function becomes simpler after differentiation. In this case, apply integration by parts twice and solve for the original integral algebraically. This is called the cyclic method.

当被积函数是 eˣ 与 sin x 或 cos x 的乘积时，两者求导后都不会变得更简单。此时连续应用两次分部积分，然后通过代数方程解出原积分。这种方法称为循环法。

Example: Evaluate I = ∫eˣ sin x dx.

例题：求 I = ∫eˣ sin x dx。

First application: let u = eˣ, dv = sin x dx. Then du = eˣ dx, v = -cos x. Hence I = -eˣ cos x + ∫eˣ cos x dx.

第一次应用：令 u = eˣ，dv = sin x dx。则 du = eˣ dx，v = -cos x。因此 I = -eˣ cos x + ∫eˣ cos x dx。

Second application: for ∫eˣ cos x dx, let u = eˣ, dv = cos x dx. Then du = eˣ dx, v = sin x. So ∫eˣ cos x dx = eˣ sin x – I.

第二次应用：对于 ∫eˣ cos x dx，令 u = eˣ，dv = cos x dx。则 du = eˣ dx，v = sin x。故 ∫eˣ cos x dx = eˣ sin x – I。

Substituting back: I = -eˣ cos x + eˣ sin x – I. Hence 2I = eˣ(sin x – cos x), giving the final answer.

代回原式：I = -eˣ cos x + eˣ sin x – I。于是 2I = eˣ(sin x – cos x)，得到最终答案。

∫eˣ sin x dx = ½ eˣ(sin x – cos x) + C

7. Reduction Formulas | 递推公式

For integrals involving powers such as ∫xⁿ eˣ dx or ∫sinⁿ x dx, repeated application of integration by parts leads to a reduction formula that connects Iₙ to Iₙ₋₁. This is particularly useful in A-Level Further Mathematics.

对于 ∫xⁿ eˣ dx 或 ∫sinⁿ x dx 等含幂次的积分，反复应用分部积分可以得到联系 Iₙ 与 Iₙ₋₁ 的递推公式。这在 A-Level 进阶数学中尤为有用。

Example: Let Iₙ = ∫xⁿ eˣ dx. Show that Iₙ = xⁿ eˣ – nIₙ₋₁.

例题：设 Iₙ = ∫xⁿ eˣ dx。证明 Iₙ = xⁿ eˣ – nIₙ₋₁。

Using integration by parts with u = xⁿ and dv = eˣ dx, we have du = nxⁿ⁻¹dx and v = eˣ. Therefore:

使用分部积分，取 u = xⁿ，dv = eˣ dx，则 du = nxⁿ⁻¹dx，v = eˣ。因此：

Iₙ = xⁿ eˣ – n∫xⁿ⁻¹ eˣ dx = xⁿ eˣ – nIₙ₋₁

This formula reduces the problem step by step until I₀ = ∫eˣ dx = eˣ + C is reached.

该公式逐步递推，直到最基础的 I₀ = ∫eˣ dx = eˣ + C 为止。

8. Definite Integrals | 定积分中的应用

For definite integrals, apply the same technique but evaluate each term at the limits of integration. The formula becomes:

对于定积分，方法相同，但需在积分上下限处对每一项求值。公式变为：

∫ₐᵇ u dv = [uv]ₐᵇ – ∫ₐᵇ v du

Example: Evaluate ∫₀¹ x eˣ dx.

例题：求 ∫₀¹ x eˣ dx。

Using u = x, dv = eˣ dx, we obtain:

取 u = x，dv = eˣ dx，可得：

∫₀¹ x eˣ dx = [x eˣ]₀¹ – ∫₀¹ eˣ dx = e – (e – 1) = 1

Always check whether the boundary term [uv]ₐᵇ can be simplified before proceeding to the remaining integral. This saves time and reduces errors.

计算前务必先化简边界项 [uv]ₐᵇ，这样可以节省时间并减少错误。

9. Common Pitfalls and Tips | 常见错误与技巧

Several errors appear frequently in examinations. Being aware of them will help you avoid losing marks unnecessarily.

考试中有几类高频错误。了解它们可以帮助你避免不必要的失分。

Wrong choice of u: Choosing u = eˣ in ∫x eˣ dx would lead to a harder integral. Always apply LIATE.
弱选择 u：在 ∫x eˣ dx 中选择 u = eˣ 会使积分变得更难。务必使用 LIATE 规则。
Missing the constant of integration: For indefinite integrals, always add + C in the final answer.
忘记常数 C：不定积分必须在最终答案中加入 + C。
Sign errors: When v is negative, such as v = -cos x, be careful with the minus signs in uv and in the integral.
符号错误：当 v 为负时，如 v = -cos x，要特别注意 uv 和积分号中的负号。
Stopping too early: After one application, check whether the new integral can be evaluated directly. If not, apply integration by parts again.
过早停止：应用一次后，检查新积分能否直接求出。若不能，就需要继续应用分部积分。

10. Summary and Practice | 总结与练习

Integration by parts is a systematic method that rewards careful selection of u and dv. Follow the LIATE rule, keep your working neat, and check signs at every stage. With regular practice, you will recognise patterns quickly and choose the correct substitution in seconds.

分部积分是一种系统化的方法，其核心在于谨慎选择 u 和 dv。遵循 LIATE 规则，保持步骤整洁，并在每一步仔细检查符号。通过规律练习，你很快就能识别题型模式，并在几秒内做出正确的选择。

For further practice, try the following integrals: ∫x² eˣ dx, ∫x cos 2x dx, ∫ln(x²) dx, and ∫e²ˣ cos 3x dx. Work through each one using the techniques outlined in this article.

为进一步练习，请尝试以下积分：∫x² eˣ dx、∫x cos 2x dx、∫ln(x²) dx 和 ∫e²ˣ cos 3x dx。运用本文介绍的方法逐一完成。

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Integration by Parts Techniques | 分部积分法解题技巧