A-Level数学 三大积分法 换元 分部 分式

A-Level数学 三大积分法 换元 分部 分式

Introduction to Integration Methods 积分方法简介

Integration is one of the two foundational operations of calculus, alongside differentiation. While differentiation follows a clear set of rules (power rule, product rule, chain rule), integration is more of an art than a science: there is no single algorithm that guarantees a solution for every integrable function. A-Level Mathematics tests your ability to recognise the structure of an integral and choose the appropriate technique. The three most powerful tools in your integration toolkit are substitution, integration by parts, and partial fractions. This article covers each method in depth, with worked examples and common pitfalls.
积分是微积分两大基础运算之一，与微分并列。微分遵循一套清晰的规则（幂法则、乘积法则、链式法则），而积分更像一门艺术而非科学：没有一种单一算法能保证对每个可积函数都有解。A-Level数学考察的是你识别积分结构并选择合适的技巧的能力。你的积分工具箱中最强大的三个工具是换元法、分部积分法和部分分式法。本文深入介绍每种方法，配以详细例题和常见陷阱。

Integration by Substitution 换元积分法

Substitution is the inverse of the chain rule. When you spot a composite function ： one function nested inside another ： substitution often turns a complicated integral into a manageable one. The key insight is to let u equal the inner function, then express everything in terms of u and du. For A-Level exams, you will be given the substitution unless it is an obvious reverse chain rule pattern, but knowing how to choose your own substitution is essential for harder questions.
换元法是链式法则的逆运算。当你发现一个复合函数（一个函数嵌套在另一个内部）时，换元法往往能将复杂的积分转化为可处理的积分。关键思路是令u等于内层函数，然后将所有内容用u和du表示。在A-Level考试中，除非是明显的反向链式法则模式，通常会给出换元公式，但知道如何自行选择换元对于更难的题目至关重要。

Consider the integral: ∫ 2x·(x² + 3)⁴ dx. Here the derivative of the inner function x² + 3 is 2x, which appears as a factor. Let u = x² + 3, then du/dx = 2x, so du = 2x dx. The integral becomes ∫ u⁴ du = u⁵/5 + C. Substituting back gives (x² + 3)⁵/5 + C. Notice how the 2x dx was absorbed perfectly into du ： this is the hallmark of a well-chosen substitution.
考虑积分：∫ 2x·(x² + 3)⁴ dx。这里内层函数x² + 3的导数是2x，正好作为因式出现。令u = x² + 3，则du/dx = 2x，因此du = 2x dx。积分变为∫ u⁴ du = u⁵/5 + C。代回得(x² + 3)⁵/5 + C。注意2x dx完美地被du吸收，这是选对换元的标志。

For definite integrals, substitution requires an additional step: you must change the limits of integration. If x goes from a to b, then u goes from g(a) to g(b) where u = g(x). This is often faster than finding the antiderivative in terms of u, converting back to x, and then applying the original limits. Many students lose marks by forgetting to change the limits and mistakenly applying x-limits to a u-expression.
对于定积分，换元法需要一个额外步骤：必须更改积分上下限。如果x从a到b，那么u从g(a)到g(b)，其中u = g(x)。这通常比用u求出原函数、再转回x、然后代入原始上下限更快。许多学生因忘记更改上下限，错误地将x的上下限应用于u的表达式而丢分。

Trigonometric substitutions are particularly common in A-Level exams. When you see √(a² − x²) inside an integral, substituting x = a sin θ transforms the radical into a cos θ, exploiting the identity sin²θ + cos²θ = 1. For √(a² + x²), the substitution x = a tan θ turns the expression into a sec θ using 1 + tan²θ = sec²θ. These patterns are worth memorising because they transform algebraic integrands into trigonometric ones that are often much simpler to integrate. After integration, remember to convert your answer back to x using a right-triangle diagram to express sin θ, cos θ, or tan θ in terms of x.
三角换元在A-Level考试中特别常见。当你在积分中看到√(a² − x²)时，令x = a sin θ可将根式化为cos θ，利用sin²θ + cos²θ = 1。对于√(a² + x²)，令x = a tan θ将表达式变为sec θ，利用1 + tan²θ = sec²θ。这些模式值得记住，因为它们将代数被积函数转化为三角被积函数，通常更容易积分。积分后，记得用直角三角形图将sin θ、cos θ或tan θ用x表示，把答案转回x。

Integration by Parts 分部积分法

Integration by parts is derived from the product rule for differentiation. The formula is ∫ u dv = uv − ∫ v du, or in the A-Level notation: ∫ u (dv/dx) dx = uv − ∫ v (du/dx) dx. This technique is ideal when the integrand is a product of two functions where one simplifies upon differentiation and the other is easy to integrate. Classic candidates include x·eˣ, x·sin x, ln x, and x·ln x.
分部积分法源自微分的乘积法则。公式为∫ u dv = uv − ∫ v du，在A-Level符号中写为：∫ u (dv/dx) dx = uv − ∫ v (du/dx) dx。当被积函数是两个函数的乘积，其中一个微分后化简、另一个容易积分时，该技巧最为适用。经典适用对象包括x·eˣ、x·sin x、ln x和x·ln x。

The choice of u and dv is crucial. A useful mnemonic is LIATE (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential): choose u from the earlier category. For ∫ x·eˣ dx, u = x (Algebraic) and dv = eˣ dx (Exponential). Then du = dx and v = eˣ. Applying the formula: ∫ x·eˣ dx = x·eˣ − ∫ eˣ dx = x·eˣ − eˣ + C = eˣ(x − 1) + C. If you had chosen u = eˣ and dv = x dx instead, the resulting integral would become MORE complicated, not less ： a classic beginner mistake.
u和dv的选择至关重要。一个有用的助记法是LIATE（对数、反三角、代数、三角、指数）：从较前类别中选择u。对于∫ x·eˣ dx，u = x（代数）而dv = eˣ dx（指数）。则du = dx，v = eˣ。代入公式：∫ x·eˣ dx = x·eˣ − ∫ eˣ dx = x·eˣ − eˣ + C = eˣ(x − 1) + C。如果当初选了u = eˣ和dv = x dx，结果积分会变得更复杂而不是更简单：这是一个经典的初学者错误。

For ∫ ln x dx, you might think there is no product, but it can be written as ∫ 1·ln x dx. Here choose u = ln x and dv = 1·dx. Then du = (1/x) dx and v = x. The formula yields: ∫ ln x dx = x·ln x − ∫ x·(1/x) dx = x·ln x − ∫ 1 dx = x·ln x − x + C. This is a common A-Level question and worth memorising the result.
对于∫ ln x dx，你可能会觉得没有乘积，但它可以写成∫ 1·ln x dx。这里选择u = ln x和dv = 1·dx。则du = (1/x) dx，v = x。公式得出：∫ ln x dx = x·ln x − ∫ x·(1/x) dx = x·ln x − ∫ 1 dx = x·ln x − x + C。这是A-Level常见题，结果值得记下来。

Sometimes integration by parts must be applied twice. This happens when the integral cycles back to its original form, as with ∫ eˣ·sin x dx. Let u = sin x, dv = eˣ dx. After the first application you get a new integral involving cos x. Apply parts again with u = cos x, and the original integral reappears. Bring it to one side and solve algebraically to find the answer.
有时分部积分法需要应用两次。当积分循环回到原始形式时就会发生这种情况，比如∫ eˣ·sin x dx。令u = sin x，dv = eˣ dx。第一次应用后得到一个含cos x的新积分。再次用分部积分令u = cos x，原始积分重新出现。将其移到一边代数求解即可得到答案。

Another common pattern is ∫ x²·eˣ dx, where the algebraic factor requires two rounds of integration by parts. First apply parts with u = x², dv = eˣ dx, reducing the power of x by one. The resulting integral ∫ 2x·eˣ dx still contains x·eˣ, so apply parts a second time with u = x, dv = eˣ dx. This two-step reduction is typical when an algebraic term has a power greater than one paired with an exponential or trigonometric function. A helpful check: after each application of parts, the algebraic power should decrease by one; if it does not, you have chosen u and dv the wrong way around.
另一种常见模式是∫ x²·eˣ dx，其中代数因式需要两轮分部积分。先用u = x²、dv = eˣ dx进行分部，将x的幂次降低一次。结果积分∫ 2x·eˣ dx仍包含x·eˣ，因此用u = x、dv = eˣ dx再次分部。这种两步降幂法在代数项幂次大于1且与指数或三角函数配对时很典型。一个有用的检验：每次分部后代数幂次应降低一次；如果没有降低，说明你把u和dv选反了。

Integration by Partial Fractions 部分分式积分法

When the integrand is a rational function (a ratio of two polynomials), partial fractions decompose it into a sum of simpler fractions that can be integrated directly. The denominator must be factorised first. For a proper fraction where the numerator’s degree is less than the denominator’s, decompose into partial fractions with unknown constants, solve for the constants, then integrate term by term. This technique is particularly powerful because the resulting integrals are typically natural logarithms.
当被积函数是有理函数（两个多项式的比值）时，部分分式将之分解为可直接积分的简单分式之和。分母必须先因式分解。对于真分数（分子次数低于分母），将其分解为含未知常数的部分分式，解出常数，然后逐项积分。这个技巧特别强大，因为积分结果通常是对数函数。

Consider ∫ (3x + 5)/(x² − 1) dx. Factor the denominator: x² − 1 = (x − 1)(x + 1). Write: (3x + 5)/(x² − 1) = A/(x − 1) + B/(x + 1). Multiply both sides by (x − 1)(x + 1): 3x + 5 = A(x + 1) + B(x − 1). Expand: 3x + 5 = (A + B)x + (A − B). Equate coefficients: A + B = 3 and A − B = 5. Solving gives A = 4, B = −1. The integral becomes ∫ 4/(x−1) dx − ∫ 1/(x+1) dx = 4ln|x−1| − ln|x+1| + C.
考虑∫ (3x + 5)/(x² − 1) dx。对分母因式分解：x² − 1 = (x − 1)(x + 1)。写出：(3x + 5)/(x² − 1) = A/(x − 1) + B/(x + 1)。两边同乘(x − 1)(x + 1)：3x + 5 = A(x + 1) + B(x − 1)。展开：3x + 5 = (A + B)x + (A − B)。比较系数：A + B = 3且A − B = 5。解得A = 4，B = −1。积分变为∫ 4/(x−1) dx − ∫ 1/(x+1) dx = 4ln|x−1| − ln|x+1| + C。

For denominators with repeated linear factors like (x + 2)², you need terms for each power: A/(x + 2) + B/(x + 2)². For irreducible quadratic factors like (x² + 4), the partial fraction takes the form (Ax + B)/(x² + 4). The integration of such terms may require completing the square and using substitution, or recognising arctan forms. A-Level examiners often combine partial fractions with substitution or arctan integration to test your ability to chain multiple techniques.
对于有重复线性因式的分母如(x + 2)²，需要每个幂次各设一项：A/(x + 2) + B/(x + 2)²。对于不可约二次因式如(x² + 4)，部分分式形式为(Ax + B)/(x² + 4)。积分这类项可能需要配方后用换元法，或识别出arctan形式。A-Level出题人常将部分分式与换元法或arctan积分结合，以考察你串联多种技巧的能力。

A common trap: before decomposing into partial fractions, check if the rational function is improper (numerator degree ≥ denominator degree). If so, perform polynomial long division first. The quotient becomes a polynomial that integrates directly, and the remainder ： now a proper fraction ： is where partial fractions apply. Skipping this step produces an incorrect decomposition and wastes valuable exam time.
一个常见陷阱：在分解部分分式之前，检查有理函数是否为假分数（分子次数不小于分母次数）。如果是，先进行多项式长除法。商成为可直接积分的多项式，余数：现在是一个真分数：才是部分分式的应用对象。跳过这一步会产生错误的分解并浪费宝贵的考试时间。

Summary and Exam Tips 总结与考试技巧

Each integration method has its own signature: substitution shines when you see a function and its derivative paired together; integration by parts handles products where one factor simplifies upon differentiation; and partial fractions decompose rational functions into log-integrable pieces. The hardest A-Level questions mix these techniques ： for example, a substitution that transforms a radical into a rational function, followed by partial fractions. Practise identifying which method to apply first: scan the integrand for composite structures (substitution), product patterns (by parts), or rational expressions (partial fractions). Always check your answer by differentiating, and remember that if one method leads to a dead end, simply start again with a different approach. With deliberate practice across all three methods, integration becomes a reliable source of marks rather than a source of anxiety.
每种积分法都有其特征信号：当看到一个函数与其导数成对出现时，换元法最为有效；分部积分法处理其中一个因式微分后化简的乘积；部分分式法将有理函数分解为可化为对数的片段。最难的A-Level题目融合这些技巧：例如，换元将根式转化为有理函数后再用部分分式。练习识别先应用哪种方法：扫描被积函数中是否有复合结构（换元）、乘积模式（分部）、或有理表达式（部分分式）。始终通过微分来验证答案，记住如果一种方法走进死胡同，只需用不同方法重新开始。通过对三种方法的有针对性练习，积分将成为可靠的得分来源而非焦虑来源。