📚 Integration by Parts: Formula, LIATE Rule and Worked Examples | 分部积分法：公式、LIATE 法则与典型例题
Integration by parts is one of the most powerful techniques in the Edexcel A-Level Pure Mathematics specification. It is used to integrate products such as x sin x and x eˣ, and also single functions whose derivatives are simpler, such as ln x and arctan x. The technique reverses the product rule for differentiation, so mastering it allows you to handle a wide range of integrals that cannot be solved by direct reverse differentiation.
分部积分法是 Edexcel A-Level 纯数学考纲中最有力的技巧之一。它用于求 x sin x、x eˣ 等乘积的积分，也用于求导后变得更简单的单个函数，如 ln x 和 arctan x 的积分。该技巧逆转了乘法求导法则，因此掌握它可以帮助你解决大量无法直接反求导的积分。
This article rebuilds the topic from first principles to exam-level fluency: the derivation from the product rule, the LIATE choice rule, five fully worked integrals, the cyclic case, definite integrals with limits, and a checklist of the errors that most often cost marks.
本文从基本原理一路讲到考试级熟练度：由乘积法则出发的推导、LIATE 选择法则、五道完整例题、循环型积分、带上下限的定积分，以及一张最常丢分的错误清单。
1. What Integration by Parts Is and When to Use It | 分部积分法是什么、什么时候用
Integration by parts (often abbreviated IBP) is a technique for evaluating an integral by rewriting it as a product minus a second, easier integral. It does not produce an answer by itself; it trades a hard integral for a friendlier one that you already know how to finish.
分部积分法（常缩写为 IBP）是一种通过把原积分改写为"一个乘积减去另一个更简单的积分"来求值的技巧。它本身不会直接给出答案，而是用一个更容易完成的积分去交换原来那个难处理的积分。
You should reach for it whenever the integrand is a product of two different types of function, or when the integrand is a single function that you cannot integrate directly but can differentiate into something simpler.
当被积函数是两类不同函数的乘积，或者被积函数是单个函数、你无法直接积分但求导后会变得更简单时，就应该想到它。
- Products of a polynomial and a trigonometric function, e.g. ∫ x cos x dx —— 多项式与三角函数之积，例如 ∫ x cos x dx
- Products of a polynomial and an exponential, e.g. ∫ x eˣ dx —— 多项式与指数函数之积，例如 ∫ x eˣ dx
- Products involving a logarithm with no obvious substitution, e.g. ∫ x² ln x dx —— 含对数且无明显换元的乘积，例如 ∫ x² ln x dx
- Single inverse trigonometric functions, e.g. ∫ arctan x dx —— 单个反三角函数，例如 ∫ arctan x dx
- Single logarithmic functions, e.g. ∫ ln x dx —— 单个对数函数，例如 ∫ ln x dx
- Cyclic products such as ∫ eˣ sin x dx, where the same integral reappears —— 循环型乘积，如 ∫ eˣ sin x dx，原积分会再次出现
+ 3)⁵ dx are almost always faster by substitution, because the derivative of the inner function is already sitting in the integrand. Choosing the wrong method is one of the most common sources of lost time in the exam.
相反，像 ∫ x(x² + 3)⁵ dx 这样的积分，用换元法几乎总是更快，因为内层函数的导数已经出现在被积函数中。选错方法，是考试中最常见的时间浪费来源之一。
2. The Formula and Its Derivation from the Product Rule | 公式及其由乘积法则的推导
The integration by parts formula is usually written as:
分部积分公式通常写作：
∫ u (dv/dx) dx = uv − ∫ v (du/dx) dx
Here u and v are functions of x. In words: the integral of u times dv/dx equals u times v minus the integral of v times du/dx. A compact version, used when the dx is understood, is:
其中 u 和 v 都是 x 的函数。用文字表述：u 乘以 dv/dx 的积分等于 u 乘以 v 减去 v 乘以 du/dx 的积分。当 dx 被默认省略时，也可使用紧凑写法：
∫ u dv = uv − ∫ v du
The formula comes from the product rule for differentiation:
该公式来自乘法求导法则：
d/dx (uv) = u (dv/dx) + v (du/dx)
v) − v (du/dx), and integrating both sides with respect to x produces the formula above, because the integral of d/dx (uv) is simply uv.
移项得 u (dv/dx) = d/dx (uv) − v (du/dx)，两边对 x 积分即可得到上述公式，因为 d/dx (uv) 的积分就是 uv。
to integrate. Notice that the second integral on the right is ∫ v (du/dx) dx, not ∫ v u dx — a single forgotten bracket here will change your answer completely.
其目的是通过选择合适的 u 使它在求导后变简单，并选择容易积分的 dv/dx，从而用一个更简单的积分替代原来的积分。注意右边第二个积分是 ∫ v (du/dx) dx，而不是 ∫ v u dx——这里少写一个括号，答案就会完全不同。
3. Choosing u and dv/dx: The LIATE Priority List | 选取 u 与 dv/dx：LIATE 优先级表
The hardest part is often deciding which factor should be u. A common priority list is LIATE, read from left to right: the factor that appears earliest in the list becomes u, and everything else (with the dx) becomes dv/dx.
最难的部分通常是决定哪个因式应该作为 u。一个常用的优先级列表是 LIATE，从左到右阅读：出现得越靠前的因式就当作 u，剩下的部分（连同 dx）就是 dv/dx。
| Letter 字母 | Type 类型 | Examples 例子 | Why 理由 |
| L | Logarithmic 对数 | ln x, log₂ x | No elementary antiderivative; but 1/x is simple 无初等原函数，但导数 1/x 很简单 |
| I | Inverse trigonometric 反三角 | arcsin x, arctan x | Derivative is algebraic 导数是有理式 |
| A | Algebraic 代数（多项式） | x, x², 3x + 1 | Differentiating lowers the power 求导后降幂 |
| T | Trigonometric 三角 | sin x, cos x, tan x | Easy to integrate 容易积分 |
| E | Exponential 指数 | eˣ, e²ˣ, 3ˣ | Self-similar under integration 积分后形式不变 |
n x dx the logarithmic part ln x comes first in LIATE, so u = ln x and dv/dx = x. In ∫ x eˣ dx the algebraic factor x beats the exponential, so u = x and dv/dx = eˣ. In ∫ x sin x dx, algebraic beats trigonometric, so u = x.
例如，在 ∫ x ln x dx 中，对数部分 ln x 在 LIATE 中排在前面，所以 u = ln x、dv/dx = x；在 ∫ x eˣ dx 中，代数因式 x 优先于指数函数，所以 u = x、dv/dx = eˣ；在 ∫ x sin x dx 中，代数优先于三角，所以 u = x。
LIATE is a heuristic, not a law. If a choice makes the new integral worse rather than better, swap the roles and try again. The real test is always: does the resulting integral look easier than the one you started with?
LIATE 只是经验法则，不是定律。如果某个选择让新积分变得更糟而不是更好，就交换角色再试一次。真正的判断标准永远是：得到的积分看起来是否比原来的更容易？
4. Worked Example 1: ∫ x sin x dx | 例题一：∫ x sin x dx
Step 1 — Write down the four ingredients explicitly, because examiners award method marks for them.
第一步——选择。按 LIATE，代数因式 x 应取作 u，因此 dv/dx = sin x。把这四个要素明确写出来，因为考官会为此给出方法分。
u = x ⇒ du/dx = 1    dv/dx = sin x ⇒ v = −cos x
Step 2 — stitute into the formula. Take care with the double negative that appears when v = −cos x.
第二步——代入公式。注意当 v = −cos x 时出现的双负号。
∫ x sin x dx = x(−cos x) − ∫ (−cos x)(1) dx
Step 3 — simplify and finish.
第三步——化简并完成。
= −x cos x + ∫ cos x dx = −x cos x + sin x + C
n x) = −cos x + x sin x + cos x = x sin x. The check works, and it costs about ten seconds — far less than redoing the question.
通过求导检验：d/dx (−x cos x + sin x) = −cos x + x sin x + cos x = x sin x。检验成立，而且只需十秒左右，远比重做一遍划算。
5. Worked Example 2: ∫ x eˣ dx and Repeated Application | 例题二：∫ x eˣ dx 与重复使用
Here u = x and dv/dx = eˣ, so du/dx = 1 and v = eˣ. The formula gives:
这里取 u = x、dv/dx = eˣ，于是 du/dx = 1、v = eˣ。公式给出：
∫ x eˣ dx = x eˣ − ∫ eˣ dx = x eˣ − eˣ + C = eˣ(x − 1) + C
factor has a higher power, apply the method as many times as the power demands. For ∫ x² eˣ dx the first application with u = x² gives:
如果代数因式的次数更高，就要按次数重复使用该方法。对于 ∫ x² eˣ dx，第一次取 u = x² 得到：
∫ x² eˣ dx = x² eˣ − 2 ∫ x eˣ dx
The remaining integral is exactly the one just found, so substitute the known result rather than integrating from scratch:
剩下的积分正是刚才求出的那一个，所以直接代入已知结果，无需从头再积：
= x² eˣ − 2 eˣ(x − 1) + C = eˣ(x² − 2x + 2) + C
Notice the pattern: each round of parts reduces the power of x by one. That is why LIATE tells you to differentiate the polynomial — it is the factor that eventually disappears.
注意这个模式：每做一轮分部积分，x 的次数就降低一次。这正是 LIATE 要求你对多项式求导的原因——它是最终会消失的那个因式。
6. Worked Example 3: ∫ x² ln x dx and the Power-Reducing Pattern | 例题三：∫ x² ln x dx 与降幂模式
is the part you can integrate. Take u = ln x and dv/dx = x², so du/dx = 1/x and v = x³/3.
这道题没有选择余地：只有 ln x 能扮演 u 的角色，因为 x² 是可以积分的那部分。取 u = ln x、dv/dx = x²，于是 du/dx = 1/x、v = x³/3。
∫ x² ln x dx = (x³/3) ln x − ∫ (x³/3)(1/x) dx
The second integrand simplifies beautifully — the 1/x cancels one power of x:
第二个被积函数化简得非常漂亮——1/x 消去了 x 的一次幂：
= (x³/3) ln x − (1/3) ∫ x² dx = (x³/3) ln x − x³/9 + C
A cleaner form combines the two terms: x³(3 ln x − 1)/9 + C. Both forms are acceptable, but factorised answers are easier to check and less likely to hide an arithmetic slip.
更整洁的形式是把两项合并：x³(3 ln x − 1)/9 + C。两种写法都可以接受，但提取公因式后的答案更容易检验，也更不容易掩盖算术失误。
ⁿ ln x dx: the logarithm is differentiated (turning into a power of x), and the power is integrated (raising the degree), after which the whole thing collapses to an ordinary polynomial integral.
这就是所有 ∫ xⁿ ln x dx 的标准形态：对数被求导（变成 x 的幂），幂函数被积分（次数升高），之后整个式子就塌缩成一个普通的代数积分。
7. Integrating a Single Function: ln x and arctan x | 单一函数的积分：ln x 与 arctan x
A function that is not a product at all can still be handled by parts, thanks to the trick of writing it as the product of
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