📚 Partial Fractions | 部分分式
In A-Level Edexcel Pure Mathematics, partial fractions are a powerful algebraic tool used to decompose rational expressions into simpler fractions. This technique is essential for integration, solving differential equations, and simplifying complex algebraic functions. Mastery of partial fractions will allow you to break down intimidating rational functions into manageable parts, especially when linear, repeated, or quadratic denominators appear.
在Edexcel A-Level纯数学中，部分分式是一种强大的代数工具，用于将有理表达式分解为更简单的分式。这项技巧对于积分、解微分方程以及化简复杂代数函数至关重要。掌握部分分式能让你将令人望而生畏的有理函数拆分为易于处理的小部分，特别是在分母出现一次因式、重复因式或二次因式的时候。
1. What Are Partial Fractions? | 什么是部分分式？
Partial fractions express a single algebraic fraction as the sum of two or more simpler fractions. The numerator and denominator are polynomials, and the degree of the numerator must be strictly less than the degree of the denominator. If not, polynomial division must be applied first. The method relies on factorising the denominator completely over the real numbers, which then dictates the form of the partial fraction decomposition.
部分分式将一个单一的代数分式表示为两个或更多简单分式的和。分子与分母均为多项式，且分子的次数必须严格小于分母的次数。若不满足，需先进行多项式除法。该方法的基础是将分母在实数范围内完全分解因式，然后根据因式类型决定分式拆分的结构。
2. When to Use Partial Fractions in the Edexcel Syllabus | 在 Edexcel 考纲中何时使用部分分式
You will encounter partial fractions primarily in integration problems (Pure Year 2), but also in binomial expansions and sometimes in functions. The exam may ask you to express a given rational function in partial fractions and then integrate it, or to use the expansion for series approximations. Recognising the denominator structure quickly helps you apply the appropriate decomposition.
你在考试中主要会在积分问题（纯数第二年）中遇到部分分式，也会出现在二项展开式中，偶尔在函数中使用。考题可能要求你把一个有理函数写成部分分式然后积分，或利用展开式进行级数逼近。快速识别分母的结构能帮助你采用正确的分解方式。
3. Linear Factors: The Simplest Case | 一次因式：最简单的情况
When the denominator factorises into distinct linear factors, e.g., (x+2)(x-1), we write the fraction as A/(x+2) + B/(x-1), where A and B are constants to be determined. This is the most frequent form seen in exam questions. The constants can be found by either substituting convenient values of x that make each bracket zero, or by equating coefficients.
当分母分解为互异的一次因式，如 (x+2)(x-1)，我们将分式写为 A/(x+2) + B/(x-1)，其中 A 和 B 是待定常数。这是考试中最常见的形式。常数可以通过代入使各个括号为零的 x 值，或通过比较系数来求出。
4. Repeated Linear Factors | 重复一次因式
If a linear factor appears squared or cubed, such as (2x-1)², the decomposition must include all powers from 1 up to that repeated term. For example, for a denominator containing (x-3)², the form is A/(x-3) + B/(x-3)². For (x+1)³, include A/(x+1) + B/(x+1)² + C/(x+1)³. Missing any lower power is a common mistake that leads to inconsistent equations.
如果某个一次因式以平方或立方出现，例如 (2x-1)²，分解时必须包含从一次到该重复次数的所有项。例如，分母包含 (x-3)²，形式为 A/(x-3) + B/(x-3)²。若是 (x+1)³，则需包含 A/(x+1) + B/(x+1)² + C/(x+1)³。漏掉任何一个低次幂都是常见错误，会导致方程矛盾。
5. Irreducible Quadratic Factors | 不可约二次因式
When the factor is a quadratic that does not factorise further (e.g., x²+1 or x²+x+1), the numerator in the partial fraction must be of the form Ax+B, not just a constant. For a denominator with (x²+4), the term is (Ax+B)/(x²+4). This ensures the decomposition is valid for all x. The constants are then found by the usual algebraic methods.
当因式为不能再分解的二次式（如 x²+1 或 x²+x+1），部分分式中的分子必须为 Ax+B 的一次式，而非单一常数。若分母含有 (x²+4)，对应项为 (Ax+B)/(x²+4)。这样才能保证分解对所有 x 成立。随后通过常规代数方法求出所有常数。
6. Combining Linear and Quadratic Factors | 一次与二次因式混合
Denominators such as (x-1)(x²+2) require a combination of techniques: A/(x-1) for the linear part, and (Bx+C)/(x²+2) for the irreducible quadratic. Always check the degree of the numerator: if it is already less than the denominator, proceed directly; if not, perform polynomial long division first. The resulting proper fraction is then decomposed.
像 (x-1)(x²+2) 这样的分母需要结合技巧：一次因式对应 A/(x-1)，不可约二次因式对应 (Bx+C)/(x²+2)。务必检查分子次数：若已小于分母，直接分解；若否，先进行多项式长除法。得到的真分式再继续分解。
7. The Substitution Method to Find Constants | 代入法求常数
factor zero. For instance, let x=-1 to isolate A, and x=2 to isolate B. This method is fast and typically gives the constants immediately, but it works cleanly only for linear factors. For repeated or quadratic factors, you still need to equate coefficients after substituting a few easy values.
确定 A/(x+1) + B/(x-2) 中的 A、B 时，将等式两边乘以原分母以消去分式，然后代入使每个因式为零的 x 值。例如，令 x=-1 求出 A，令 x=2 求出 B。此方法快捷，通常能立刻得到常数，但仅对一次因式最为干净。对于重复因式或二次因式，仍需在代入几个简单值后再比较系数。
8. Equating Coefficients: A Reliable Backup | 比较系数法：可靠的备选方案
Expand the right-hand side and collect like terms in x. Then equate the coefficients of corresponding powers of x on both sides to form simultaneous equations. This method works universally for any type of factor. Although more laborious, it provides a systematic check and is essential when substitution alone leaves unresolved constants, such as in repeated or quadratic factor cases.
展开等式右边并合并 x 的同次项。然后将两边对应 x 次幂的系数等同，形成联立方程组。此方法普适所有因式类型。尽管较为繁琐，但它提供了系统性的验证，且当单独使用代入法无法确定所有常数时（如重复或二次因式情况），比较系数法不可或缺。
9. Applying Partial Fractions to Integration | 部分分式在积分中的应用
Once decomposed, integrals like ∫(A/(x+1) + B/(x-2))dx become straightforward: A ln|x+1| + B ln|x-2| ter completing the square. Edexcel exam markers expect clear steps: factorise, write partial fractions, find constants, integrate term by term.
完成分解后，形如 ∫(A/(x+1) + B/(x-2))dx 的积分变得直接：A ln|x+1| + B ln|x-2| + C。对于二次分母，(Bx+C)/(x²+px+q) 的积分通常拆分为对数部分和反正切部分，经常需要先对分母配方。Edexcel 阅卷人期待清晰的步骤：因式分解、写出部分分式、求出常数、逐项积分。
10. Common Mistakes and How to Avoid Them | 常见错误与避坑指南
- Forgetting to reduce an improper fraction: always perform division if degree of numerator ≥ degree of denominator.
- 遗漏降次：若分子次数大于等于分母，务必先做除法。
- Missing a term in repeated factors: write all descending powers; do not skip the linear term in (Bx+C) over a quadratic.
- 重复因式漏项：写出所有递降幂次；对于二次因式，不要遗漏 (Bx+C) 中的一次项。
- Arithmetic errors when equating coefficients: expand brackets carefully, double-check sign errors.
- 比较系数时的算术失误：仔细展开括号，反复核对符号错误。
- Mishandling negative roots when substituting: if a factor is (3x+2), set x=-2/3, not 2/3.
- 代入时负根处理错误：若因式为 (3x+2)，令 x=-2/3 而非 2/3。
11. Worked Example: From Decomposition to Integration | 详细例题：从分解到积分
Problem: Express (5x+3)/((x+1)(x-2)) in partial fractions and hence find ∫ (5x+3)/((x+1)(x-2)) dx.
题目：将 (5x+3)/((x+1)(x-2)) 化为部分分式，并求 ∫ (5x+3)/((x+1)(x-2)) dx。
Solution: Write (5x+3)/((x+1)(x-2)) ≡ A/(x+1) + B/(x-2). Multiply by denominator: 5x+3 ≡ s are (2/3)/(x+1) + (13/3)/(x-2). Integration: ∫ [2/3·1/(x+1) + 13/3·1/(x-2)] dx = (2/3)ln|x+1| + (13/3)ln|x-2| + C.
解答：设 (5x+3)/((x+1)(x-2)) ≡ A/(x+1) + B/(x-2)。乘以分母：5x+3 ≡ A(x-2) + B(x+1)。代入 x=-1：-5+3 = A(-3) → A = 2/3。代入 x=2：10+3 = B(3) → B = 13/3。因此部分分式为 (2/3)/(x+1) + (13/3)/(x-2)。积分：∫ [2/3·1/(x+1) + 13/3·1/(x-2)] dx = (2/3)ln|x+1| + (13/3)ln|x-2| + C。
12. Exam Tips and Formula Summary | 考场贴士与公式总结
Memorise the standard partial fraction forms: A/(ax+b) for linear, A/(ax+b) + B/(ax+b)² for repeated linear, and (Ax+B)/(ax²+bx+c) for irreducible quadratic. Always write the identity symbol ≡ until constants are found. In integration, clearly state the logarithmic or inverse tangent results. Practice past papers, as partial fractions frequently combine with definite integration and trigonometric substitution.
熟记标准部分分式形式：一次因式对应 A/(ax+b)，重复一次因式对应 A/(ax+b) + B/(ax+b)²，不可约二次因式对应 (Ax+B)/(ax²+bx+c)。求出常数前始终使用恒等号 ≡。积分时清晰写出对数或反正切结果。多做历年真题，因为部分分式常与定积分和三角代换结合考查。
| Factor Type 因式类型 | Partial Fraction Form 部分分式形式 |
|---|---|
| Distinct linear factors e.g. (x+a)(x+b) | A/(x+a) + B/(x+b) |
| Repeated linear factor (ax+b)ⁿ | A₁/(ax+b) + A₂/(ax+b)² + ... + Aₙ/(ax+b)ⁿ |
| Irreducible quadratic ax²+bx+c | (Ax+B)/(ax²+bx+c) |
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