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Integration by Parts: A-Level Calculus Part 5 | 分部积分法：A-Level 微积分第5部分

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📚 Integration by Parts: A-Level Calculus Part 5 | 分部积分法：A-Level 微积分第5部分

Integration by parts is one of the most powerful tools in A-level calculus. It transforms a product of functions into a simpler integral by using the product rule for differentiation in reverse. In this part of the revision series, we focus on when and how to apply the method, with clear examples and exam tips.

分部积分法是 A-level 微积分中最强大的工具之一。它通过反向使用微分的乘积法则，将函数的乘积转化为更简单的积分。在本期复习系列中，我们将重点讲解何时以及如何应用该方法，并给出清晰的例题和考试提示。

1. What Integration by Parts Is and When to Use It | 什么是分部积分法以及何时使用它

When an integral involves a product of two different types of functions, such as x eˣ or x sin x, ordinary substitution often fails. Integration by parts allows us to transfer the derivative from one factor to another, making the new integral easier to evaluate.

当积分涉及两种不同类型函数的乘积时，例如 x eˣ 或 x sin x，普通的换元法通常会失效。分部积分法允许我们将导数从一个因子转移到另一个因子上，从而使新的积分更容易计算。

This technique is especially useful when one factor becomes simpler after differentiation, such as a polynomial xⁿ, while the other factor is easy to integrate, such as eˣ, sin x, or cos x. The method is standard in A-level pure mathematics and frequently appears in exam papers.

当其中一个因子在求导后变得更简单时，例如多项式 xⁿ，而另一个因子容易积分，例如 eˣ、sin x 或 cos x，这种方法尤其有用。该方法是 A-level 纯数学中的标准内容，在考试卷中经常出现。

A quick recognition checklist helps you decide whether to reach for this method. If your integrand is a product, and if the two factors belong to different families of functions, then integration by parts is almost certainly the intended route. Typical triggers include a polynomial multiplied by an exponential, a polynomial multiplied by a trigonometric function, a logarithm multiplied by any polynomial, and an inverse trigonometric function multiplied by a polynomial.

一个快速识别的清单能帮你判断是否该使用这种方法。如果被积函数是一个乘积，并且两个因子属于不同的函数族，那么分部积分几乎肯定是命题人期望的解法。典型的触发信号包括：多项式乘以指数函数、多项式乘以三角函数、对数乘以任意多项式，以及反三角函数乘以多项式。

2. The Formula and Its Derivation | 公式及其推导

The formula comes directly from the product rule. Differentiate the product uv with respect to x and you obtain the familiar expression below.

该公式直接来自乘积法则。对乘积 uv 关于 x 求导，就得到下面这个熟悉的表达式。

d/dx (u v) = u · dv/dx + v · du/dx

Now integrate both sides with respect to x. The integral of a derivative simply returns the original function, so the left-hand side becomes uv, while the right-hand side splits into two separate integrals.

现在对两边关于 x 积分。导数的积分会直接还原为原函数，因此左边变成 uv，而右边则拆分成两个独立的积分。

u v = ∫ u dv + ∫ v du

are usually more convenient to handle. You must always identify u and dv before applying the rule, and then compute du and v carefully.

重新整理即可得到下面所有例题中使用的标准公式。在微分形式中，du = u′ dx 与 dv = v′ dx 通常更便于操作。在使用该法则之前，你必须先确定 u 和 dv，然后仔细求出 du 和 v。

∫ u dv = u v − ∫ v du

The same rule is often written with explicit derivatives, which is the form you will see in most textbooks and mark schemes. Both versions are identical; choose whichever you can apply without making sign errors under exam pressure.

同一条法则也常写成带显式导数的形式，这是大多数教材和评分标准中出现的写法。两种写法完全等价，选择考试压力下最不易出符号错误的那一种即可。

∫ u (dv/dx) dx = u v − ∫ v (du/dx) dx

3. Choosing u and dv: The LIATE Rule | 选择 u 和 dv：LIATE 规则

The LIATE rule is a practical guide for choosing u. The order from highest priority to lowest is: L for logarithmic functions, I for inverse trigonometric functions, A for algebraic (polynomial) functions, T for trigonometric functions, and E for exponential functions. Whatever comes first in this list should normally be your u, and the remaining factor becomes dv.

LIATE 规则是选择 u 的实用指南。优先级从高到低依次为：L 代表对数函数，I 代表反三角函数，A 代表代数（多项式）函数，T 代表三角函数，E 代表指数函数。通常应把列表中排在最前的那一类作为 u，剩下的因子则作为 dv。

Letter Function family Typical choice for u
L Logarithmic ln x, log x
I Inverse trigonometric arcsin x, arctan x
A Algebraic (polynomial) x, x², 3x + 1
T Trigonometric sin x, cos x
E Exponential eˣ, aˣ

Two common-sense rules back up the mnemonic. First, whatever you choose as dv must be something you can actually integrate. Second, whatever you choose as u should become simpler when you differentiate it. If both conditions hold, the new integral will be easier than the original one.

有两条常识性规则为该口诀提供支撑。第一，你选作 dv 的部分必须是你确实能够积分的。第二，你选作 u 的部分在求导后应当变得更简单。如果两个条件都满足，那么新的积分就会比原来的更容易。

take u = ln x and dv = dx, so that v = x. This trick turns an apparently non-product integral into a standard application of the rule.

当被积函数是单个函数时会出现一种特殊情况，例如 ln x 或 arctan x。此时你把它写成乘积 ln x × 1，取 u = ln x、dv = dx，于是 v = x。这一技巧把一个看似非乘积形式的积分转化为该法则的标准应用。

Some textbooks use ILATE instead of LIATE, swapping the first two positions. For A-level purposes the difference rarely matters, because questions mixing a logarithm with an inverse trigonometric function are unusual. Follow whichever version your teacher or textbook uses, and always justify your choice briefly in written solutions.

有些教材使用 ILATE 而不是 LIATE，把前两位调换。就 A-level 而言，这一差别很少产生影响，因为同时含对数与反三角函数的题目并不常见。按你老师或教材使用的版本即可，并在书面解答中简要说明你的选择理由。

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

This is the classic first example. The integrand is a product of an algebraic function and an exponential function, so by LIATE the algebraic part takes priority and becomes u.

这是最经典的第一道例题。被积函数是代数函数与指数函数的乘积，因此按 LIATE，代数部分优先，作为 u。

Substituting into the formula gives the following line of working. Notice that the remaining integral is now a basic exponential integral that can be evaluated immediately.

代入公式后得到下面这一步推导。注意剩下的积分现在是一个可以直接求出的基本指数积分。

∫ x eˣ dx = x eˣ − ∫ eˣ · 1 dx = x eˣ − eˣ + C

The answer can be written in factorised form as eˣ(x − 1) + C, which is often what mark schemes show. Always include the constant of integration for indefinite integrals; a missing + C can cost a mark in a short question.

答案也可以写成因式分解形式 eˣ(x − 1) + C，这通常是评分标准给出的形式。不定积分一定要写上积分常数；在小题中漏写 + C 可能会被扣分。

You can verify the result by differentiating eˣ(x − 1). The product rule gives eˣ(x − 1) + eˣ · 1 = x eˣ, which is exactly the original integrand, so the answer is correct.

你可以通过对 eˣ(x − 1) 求导来验证结果。乘积法则给出 eˣ(x − 1) + eˣ · 1 = x eˣ，正好是原来的被积函数，因此答案正确。

5. Worked Example 2: Polynomials with Trigonometric Functions | 例题2：多项式与三角函数

Polynomial-trigonometric products follow exactly the same pattern. The polynomial is differentiated and disappears, while the trigonometric function is integrated twice in a cycle of sin and cos, so signs must be tracked very carefully.

多项式与三角函数的乘积遵循完全相同的模式。多项式被求导后消失，而三角函数在 sin 与 cos 之间循环积分，因此必须非常小心地跟踪符号。

irst, evaluate ∫ x sin x dx. Take u = x and dv = sin x dx, so that du = dx and v = −cos x.

首先计算 ∫ x sin x dx。取 u = x，dv = sin x dx，于是 du = dx，v = −cos x。

∫ x sin x dx = −x cos x − ∫ (−cos x) dx = −x cos x + sin x + C

n x. Here no negative sign appears at the front, which makes the working slightly shorter.

其次计算 ∫ x cos x dx。此时取 u = x，dv = cos x dx，于是 du = dx，v = sin x。此处前面不出现负号，因此推导略短一些。

∫ x cos x dx = x sin x − ∫ sin x dx = x sin x + cos x + C

A useful check is to remember the sign pattern. When the trigonometric factor is sin x, you get a minus sign in front of the uv term and a plus sign before the final sin x. When it is cos x, both signs turn positive. Writing the bracket around the substituted expression prevents most sign slips.

一个有用的检查方法是记住符号规律。当三角因子是 sin x 时，uv 项前是负号，最后的 sin x 前是正号；当它是 cos x 时，两个符号都为正。把代入的表达式用括号括起来，可以避免大多数符号失误。

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6. Worked Example 3: Logarithmic Functions | 例题3：对数函数

hen the integrand is simply ln x, write it as ln x × 1 so that the formula can be applied.

对数在 A-level 中无法直接积分，因此它们总是扮演 u 的角色。当被积函数就是 ln x 时，把它写成 ln x × 1，以便应用公式。

For ∫ ln x dx, take u = ln x with du = (1/x) dx, and dv = dx with v = x.

对于 ∫ ln x dx，取 u = ln x，du = (1/x) dx；取 dv = dx，v = x。

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

is the standard result that every A-level student should recognise instantly, and it is often quoted directly in mark schemes.

剩下的积分因为 x × (1/x) = 1 而完美简化。这是每位 A-level 学生都应立刻认出的标准结果，评分标准中也常直接引用。

n x dx. Here LIATE selects the logarithm as u, so u = ln x and dv = x dx, giving v = x²/2.

现在考虑 ∫ x ln x dx。此时 LIATE 选择对数为 u，故 u = ln x，dv = x dx，得 v = x²/2。

∫ x ln x dx = (x²/2) ln x − ∫ (x²/2)(1/x) dx = (x²/2) ln x − x²/4 + C

x instead, the new integral would have contained ln x again and would have been no easier.

注意第二个积分中 x 的次数减少了一，这正是该方法奏效的原因。如果你改选 u = x，新的积分中会再次出现 ln x，难度丝毫没有降低。

7. Applying the Rule Twice: ∫ x² eˣ dx | 多次应用：∫ x² eˣ dx

When the polynomial factor has degree two or higher, a single application of integration by parts leaves another polynomial-exponential product behind, so the method must be applied again. Each round reduces the degree of the polynomial by one until only the exponential remains.

当多项式因子的次数为二次或更高时，应用一次分部积分后仍会留下另一个多项式与指数函数的乘积，因此必须再次使用该方法。每一轮都会把多项式的次数降低一，直到只剩下指数函数。

First round: take u = x² and dv = eˣ dx, so du = 2x dx and v = eˣ.

第一轮：取 u = x²，dv = eˣ dx，于是 du = 2x dx，v = eˣ。

∫ x² eˣ dx = x² eˣ − 2 ∫ x eˣ dx

Second round: the integral ∫ x eˣ dx is exactly the result from Example 1, namely x eˣ − eˣ. Substituting this back and collecting terms gives the final answer.

第二轮：积分 ∫ x eˣ dx 正是例题1的结果，即 x eˣ − eˣ。把它代回并合并同类项，即得最终答案。

∫ x² eˣ dx = x² eˣ − 2(x eˣ − eˣ) + C = eˣ(x² − 2x + 2) + C

For higher powers you can use the tabular method, sometimes called the DI method, which organises repeated integration by parts into two columns. In the D column you repeatedly differentiate the polynomial until it reaches zero, and in the I column you repeatedly integrate the exponential or trigonometric factor. Then you multiply diagonally with alternating signs, starting with plus.

对于更高次幂，你可以使用表格法，有时也称为 DI 法，它把反复的分部积分整理成两列。在 D 列中反复对多项式求导直到得到零，在 I 列中反复对指数或三角函数积分。然后沿对角线相乘，符号交替变化，以正号开始。

8. The Cyclic Case: ∫ eˣ sin x dx | 循环情形：∫ eˣ sin x dx

When both factors belong to the exponential or trigonometric families, differentiating never simplifies either one. Instead, after two applications the original integral reappears, and you solve for it algebraically. This is the cyclic case.

当两个因子都属于指数或三角函数族时，求导无法使任何一个变简单。相反，在使用两次之后原来的积分会重新出现，这时需要用代数方法解出它。这就是循环情形。

Let I = ∫ eˣ sin x dx. Take u = sin x and dv = eˣ dx, so du = cos x dx and v = eˣ.

令 I = ∫ eˣ sin x dx。取 u = sin x，dv = eˣ dx，于是 du = cos x dx，v = eˣ。

I = eˣ sin x − ∫ eˣ cos x dx

Apply the rule again to ∫ eˣ cos x dx, this time with u = cos x, du = −sin x dx, dv = eˣ dx and v = eˣ. Watch the negative sign carefully.

对 ∫ eˣ cos x dx 再次应用该法则，这次取 u = cos x，du = −sin x dx，dv = eˣ dx，v = eˣ。请仔细处理负号。

I = eˣ sin x − (eˣ cos x + ∫ eˣ sin x dx) = eˣ sin x − eˣ cos x − I

ntegral I has reappeared with a minus sign, move it to the left-hand side to obtain 2I = eˣ refore the answer below. The constant of integration must be added at the final step only.

由于原积分 I 带着负号重新出现，把它移到左边得到 2I = eˣ(sin x − cos x)，于是得到下面的答案。积分常数只需在最后一步添加。

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

An entirely analogous calculation with cos x in place of sin x gives ½ eˣ completely, you have either chosen u and dv inconsistently between the two rounds or made a sign error.

把 sin x 换成 cos x 进行完全类似的计算，可得 ½ eˣ (sin x + cos x) + C。如果你得到 0 = 0 或积分完全抵消，那么你要么在两轮中 u 和 dv 的选择不一致，要么出现了符号错误。

9. Definite Integrals and the Limits Notation | 定积分与上下限记号

For a definite integral, integrate by parts as usual but apply the limits to the uv term immediately. The remaining integral keeps the same limits, which means you never have to write the constant of integration. The formula becomes the expression below.

对于定积分，照常进行分部积分，但要立刻把上下限代入 uv 项。剩下的积分保持相同的上下限，这意味着你完全不需要写积分常数。公式变为下面的表达式。

∫ₐᵇ u (dv/dx) dx = [u v]ₐᵇ − ∫ₐᵇ v (du/dx) dx

Example: evaluate ∫₀¹ x eˣ dx. Using u = x, dv = eˣ dx, v = eˣ aluated between the limits is (1 · e¹) − (0 · e⁰) = e, and the remaining integral is ∫₀¹ eˣ dx = e − 1.

例：计算 ∫₀¹ x eˣ dx。取 u = x，dv = eˣ dx，v = eˣ，uv 项在上下限之间的值为 (1 · e¹) − (0 · e⁰) = e，而剩余积分为 ∫₀¹ eˣ dx = e − 1。

∫₀¹ x eˣ dx = e − (e − 1) = 1

· 1 − 1 · 0 = e, while the remaining integral is ∫₁ᵉ 1 dx = e − 1. The final answer is 1.

例：计算 ∫₁ᵉ ln x dx。此处 u = ln x，dv = dx，v = x，uv 项给出 [x ln x]₁ᵉ = e · 1 − 1 · 0 = e，而剩余积分为 ∫₁ᵉ 1 dx = e − 1。最终答案为 1。

maining integral is ∫₀^π/2 sin x dx = [−cos x]₀^π/2 = 0 + 1 = 1, so the answer is π/2 − 1.

第三个有用的例子是 ∫₀^π/2 x cos x dx。取 u = x，dv = cos x dx，v = sin x，uv 项为 [x sin x]₀^π/2 = (π/2)(1) − 0 = π/2，剩余积分为 ∫₀^π/2 sin x dx = [−cos x]₀^π/2 = 0 + 1 = 1，所以答案为 π/2 − 1。

10. Common Mistakes and Exam Technique | 常见错误与考试技巧

Most lost marks in this topic come from sign errors, from choosing u and dv badly, and from omitting intermediate steps. Showing your choice of u and dv explicitly in the first line of working is a simple habit that earns method marks even when the final answer goes wrong.

这一主题上失分大多来自符号错误、u 与 dv 选择不当以及省略中间步骤。在解答的第一行明确写出你对 u 和 dv 的选择，是一个简单的好习惯，即使最终答案出错也能拿到方法分。