📚 Partial Fractions: Complete A-Level Guide | 部分分式：A-Level 完整指南
Partial fractions are an algebraic technique used to break a complicated rational expression into a sum of simpler fractions. This makes integration, binomial expansion and the analysis of rational functions much easier. In Edexcel A Level Mathematics, partial fractions appear regularly in Pure Mathematics, especially within integration and series work, and they also support later topics such as differential equations and proof by decomposition.
部分分式是一种代数技巧，用于把一个复杂的有理式拆分成若干简单分式之和。这使得积分、二项展开以及有理函数的分析变得容易得多。在 Edexcel A Level 数学中，部分分式经常在纯数学中出现，尤其是在积分与级数部分，同时也为微分方程、分式恒等式证明等后续内容打基础。
1. What Are Partial Fractions? | 什么是部分分式
Partial fractions are the reverse of adding rational expressions. Given a single algebraic fraction with a factorisable denominator, we split it into a sum of simpler fractions called partial fractions. This is especially useful in integration, binomial expansion and solving differential equations, because the simpler pieces follow standard results that you already know.
部分分式是代数分式相加的逆运算。给定一个分母可因式分解的单一代数分式，我们将其拆分为若干更简单的分式之和，这些简单分式称为部分分式。这在积分、二项展开和微分方程求解中特别有用，因为拆开后的每一块都能对应到你已经掌握的标准结果。
For example, 1/[(x+1)(x+2)] can be written as 1/(x+1) − 1/(x+2), which is much easier to integrate term by term.
例如，1/[(x+1)(x+2)] 可以写成 1/(x+1) − 1/(x+2)，这样逐项积分要容易得多。
The symbol ≡ must hold for every value of x for which both sides are defined. Examiners frequently expect you to write an identity statement first and then solve for the unknown constants.
当等式必须对所有使两边都有意义的 x 都成立时，通常用恒等号 ≡ 而不是 =。考官常常要求你先写出恒等式，再求未知常数。
2. The Key Idea: Denominator Factors | 核心思想：分母因式
Before splitting, factorise the denominator completely into linear factors and, where necessary, irreducible quadratic factors. The form of the partial fractions depends on the types of factors present: distinct linear factors, repeated linear factors, and irreducible quadratic factors.
拆分之前，先将分母完全因式分解为线性因式，必要时还包括不可约二次因式。部分分式的形式取决于因式类型：不同线性因式、重复线性因式以及不可约二次因式。
Each factor in the denominator generates exactly one term on the right-hand side, and a repeated factor of power n generates n separate terms with increasing powers. This 'one term per factor' rule is the single most useful guideline in the whole topic.
分母中每一个因式恰好对应右端一项，而 n 次重复因式要产生 n 个次数递增的分项。这个"每个因式对应一项"的规则是本章最有用的判断准则。
Always check the degree of the numerator. If the degree of the numerator is greater than or equal to the degree of the denominator, the fraction is improper and must be divided first.
始终检查分子的次数。如果分子的次数大于或等于分母的次数，则该分式为假分式，必须先进行多项式除法。
3. Proper and Improper Rational Fractions | 真分式与假分式
A rational expression is proper when the degree of the numerator is strictly less than the degree of the denominator. If the degree of the numerator is greater than or equal to that of the denominator, the expression is improper and you must carry out algebraic long division first, writing the result as a polynomial plus a proper remainder.
当分子的次数严格小于分母的次数时，该有理式为真分式。如果分子的次数大于或等于分母的次数，则为假分式，必须先做多项式长除法，把结果写成一个多项式加上一个真分式余项。
For instance, (x² + 3x + 5)/(x + 1) divides to give x + 2 + 3/(x + 1), and only the final proper fraction needs to be decomposed. Skipping this step is one of the most common ways to lose marks.
例如，(x² + 3x + 5)/(x + 1) 相除后得到 x + 2 + 3/(x + 1)，只有最后这个真分式需要继续拆分。跳过这一步是丢分最常见的原因之一。
4. Distinct Linear Factors | 不同线性因式
Suppose the denominator factorises into distinct linear factors, for example (x + 1)(x − 2). We write an identity with one constant numerator above each factor.
假设分母可因式分解为不同线性因式，例如 (x + 1)(x − 2)。我们设一个恒等式，在每个因式上方放一个常数分子。
f(x)/[(x + 1)(x − 2)] ≡ A/(x + 1) + B/(x − 2)
Multiplying through by the whole denominator gives f(x) ≡ s B, and substituting x = −1 isolates A, because the other bracket becomes zero each time.
两边同乘整个分母，得到 f(x) ≡ A(x − 2) + B(x + 1)。代入 x = 2 可以单独解出 B，代入 x = −1 可以单独解出 A，因为每次另一个括号都变成零。
efficients of x and of the constant term gives the same result and acts as a useful check.
取分子 5x + 1，则 A = (5(−1) + 1)/(−1 − 2) = 4/3，B = (5(2) + 1)/(2 + 1) = 11/3。比较 x 的系数与常数项也能得到同样结果，可用作验算。
5. Repeated Linear Factors | 重复线性因式
If a linear factor is repeated, every power up to the full power must appear. A denominator containing (x + 2)² therefore needs two terms, with denominators (x + 2) and (x + 2)².
若某个线性因式是重复的，则从一次幂直到最高次幂都必须出现。因此分母含 (x + 2)² 时需要两个分项，分母分别为 (x + 2) 与 (x + 2)²。
(4x + 1)/(x + 2)² ≡ A/(x + 2) + B/(x + 2)²
Multiplying by (x + 2)² gives 4x + 1 ≡ omparing the coefficient of x gives A = 4, so the decomposition is 4/(x + 2) − 7/(x + 2)².
两边同乘 (x + 2)²，得 4x + 1 ≡ A(x + 2) + B。代入 x = −2 得 B = −7，比较 x 的系数得 A = 4，所以拆分为 4/(x + 2) − 7/(x + 2)²。
For a triple factor such as (x − 1)³, you must include (x − 1), (x − 1)² and (x − 1)³. Missing the middle power is a classic error that produces an unsolvable system of equations.
对于 (x − 1)³ 这样的三重因式，必须同时写出 (x − 1)、(x − 1)² 与 (x − 1)³。漏掉中间那一次幂是典型错误，会导致方程组无解。
6. Irreducible Quadratic Factors | 不可约二次因式
An irreducible quadratic factor cannot be factorised into real linear factors, for example x² + 1, x² + 4 or x² + x + 1. For such a factor you must place a general linear numerator Bx + C above it, not a single constant.
不可约二次因式无法分解为实系数线性因式，例如 x² + 1、x² + 4 或 x² + x + 1。对这类因式，分子必须设成一般的一次式 Bx + C，而不能只设一个常数。
(2x + 3)/[(x + 1)(x² + 1)] ≡ A/(x + 1) + (Bx + C)/(x² + 1)
Multiplying out gives 2x + 3 ≡ nstants yields C = 5/2. The decomposition is therefore 1/[2(x + 1)] + (5 − x)/[2(x² + 1)].
展开得 2x + 3 ≡ A(x² + 1) + (Bx + C)(x + 1)。代入 x = −1 得 A = ½，比较 x² 系数得 B = −½，比较常数项得 C = 5/2。因此拆分为 1/[2(x + 1)] + (5 − x)/[2(x² + 1)]。
7. Summary Table of Decomposition Forms | 拆分形式速查表
The table below summarises which numerator to place above each type of denominator factor. Learn this table, because writing the correct form is worth method marks even before any constants are found.
下表总结了每种分母因式应当配什么样的分子。务必记住这张表，因为即使还没求出常数，写对形式本身就能拿到方法分。
| Denominator factor | Matching term | 中文说明 |
| (ax + b) | A/(ax + b) | 不同线性因式，分子为常数 |
| (ax + b)² | A/(ax + b) + B/(ax + b)² | 重复线性因式，逐次幂各一项 |
| (ax + b)³ | A/(ax + b) + B/(ax + b)² + C/(ax + b)³ | 三次重复，三项齐全 |
| (ax² + bx + c), irreducible | (Bx + C)/(ax² + bx + c) | 不可约二次因式，分子设一次式 |
| Improper fraction | Polynomial + proper remainder | 先做长除法再拆分 |
8. How to Find the Constants: Substitution | 求常数：代入法
Once the identity is written, multiply both sides by the original denominator. The result is a polynomial identity, and there are two standard routes to obtain A, B, C and so on: substituting convenient values of x, or comparing coefficients.
写出恒等式后，两边同乘原分母。结果是一个多项式恒等式，此后有两条标准路径求 A、B、C 等：代入合适的 x 值，或比较各项系数。
Substitution is fastest when the factors are linear and distinct, because choosing the root of each factor eliminates every other unknown at once. Comparing coefficients is more reliable when quadratic factors are present, or when only one root is available.
当因式都是不同的线性因式时，代入法最快，因为取每个因式的根可以一次性消掉其余所有未知数。当出现二次因式，或只能取到一个根时，比较系数更稳妥。
The best exam strategy is usually a hybrid: use substitution to pin down the easy constants, then compare coefficients of the highest and lowest powers to finish the job. Always verify by checking one extra value of x.
考试中最佳策略通常是混合使用：先用代入法锁定容易求的常数，再比较最高次与最低次项系数完成计算。最后务必多代一个 x 值进行验算。
9. The Cover-Up Method | 遮盖法
er the factor (x − a) in the original fraction and substitute x = a into whatever remains.
遮盖法是处理不同线性因式的快捷技巧。要求 (x − a) 上方的常数，只需在原分式中用手遮住因式 (x − a)，再把 x = a 代入剩下的部分。
A = value of f(x)/[remaining factors] at x = a
It works because the identity, evaluated at the root, makes every other term vanish. Cover-up fails for repeated linear factors and for irreducible quadratic factors, so fall back on substitution and coefficient comparison in those cases.
其原理是：在根处求值时，其余各项都变为零。遮盖法对重复线性因式与不可约二次因式不适用，此时应改用代入法与比较系数法。
The method is a legitimate time-saver in timed exams but should always be verified with a quick substitution, since a single arithmetic slip in the denominator is easy to miss.
在限时考试中该方法确实能节省时间，但一定要用一次快速代入验算，因为分母上的算术失误很容易被忽略。
10. Partial Fractions in Integration | 部分分式在积分中的应用
The main reason partial fractions appear on the specification is integration. Each linear partial fraction integrates to a natural logarithm, while each repeated one integrates to a negative power.
大纲中安排部分分式的主要目的就是积分。每个线性分项积分后得到自然对数，重复分项则积分成负次幂。
∫ A/(ax + b) dx = (A/a) ln|ax + b| + C
∫ B/(ax + b)² dx = −B/[a(ax + b)] + C
For example, ∫ 1/[(x + 1)(x + 2)] dx = ln|x + 1| − ln|x + 2| + C, which may also be written as ln|(x + 1)/(x + 2)| + C. Definite integrals are often easier when the two logarithms are combined first.
例如，∫ 1/[(x + 1)(x + 2)] dx = ln|x + 1| − ln|x + 2| + C，也可以写成 ln|(x + 1)/(x + 2)| + C。对于定积分，先把两个对数合并往往更易计算。
When an irreducible quadratic appears, the numerator is split into a multiple of the derivative of the quadratic plus a constant, producing a logarithm plus an arctangent term.
当出现不可约二次因式时，把分子拆成"二次因式的导数倍数 + 常数"，从而分别得到对数项与反正切项。
11. Partial Fractions in Binomial Expansion | 部分分式在二项展开中的应用
Partial fractions also convert a rational function with a factorised denominator into pieces that can be expanded with the binomial theorem for negative or fractional powers.
部分分式还能把分母已因式分解的有理函数转化为可用负指数或分数指数二项式定理展开的形式。
(1 + x)⁻¹ = 1 − x + x² − x³ + … for |x| < 1
(1 − 2x)⁻¹ = 1 + 2x + 4x² + 8x³ + … for |x| < ½
For instance, 1/[(1 + x)(1 − 2x)] decomposes to (1/3)/(1 + x) + (2/3)/(1 − 2x). Expanding and collecting terms gives 1 + x + 3x² + 5x³ + …, valid when |x| < ½, the tighter of the two conditions.
例如，1/[(1 + x)(1 − 2x)] 拆分为 (1/3)/(1 + x) + (2/3)/(1 − 2x)。展开并合并同类项得 1 + x + 3x² + 5x³ + …，在 |x| < ½ 时成立，取两者中更严格的条件。
Examiners often ask for the range of validity, so state the modulus inequality explicitly. When the two expansions have different radii of convergence, the answer is always the smaller one.
考官常要求写出成立范围，所以要明确写出模不等式。当两个展开式的收敛半径不同时，答案永远取较小的那个。
12. Common Exam Mistakes and Final Checklist | 常见考试错误与检查清单
Most lost marks in this topic come from a small number of repeated slips rather than from genuine conceptual difficulty. Reading through the checklist below before an exam is a highly efficient use of revision time.
本章丢分大多来自少数反复出现的疏漏，而不是真正的理解困难。考前过一遍下面的检查清单，是效率很高的复习方式。
- Forgetting to divide first when the fraction is improper | 假分式忘记先做多项式除法
- Writing a constant numerator above a repeated or quadratic factor | 重复因式或二次因式上方误写常数分子
- Omitting one of the powers of a repeated factor | 漏写重复因式的某个次幂
- Sign errors when substituting negative roots | 代入负根时符号出错
- Not stating the range of validity in binomial expansion | 二项展开未写出成立范围
- Dropping the constant C in indefinite integration | 不定积分漏写常数 C
- Forgetting the modulus signs in logarithmic answers | 对数答案漏写绝对值符号
- Not verifying the identity with one extra value of x | 未用额外一个 x 值验算恒等式
n and your decomposition. If the two values agree, the arithmetic is almost certainly sound, and any remaining error is likely to be a missing factor rather than a numerical one.
一个可靠的最后检查是把 x = 0 分别代入原分式与拆分结果。如果两边数值一致，说明算术几乎无误；若仍有问题，多半是漏掉了某个因式，而不是数值算错。
Mastery of partial fractions pays off far beyond a single question: the same manipulation underpins the integration of rational functions, the summing of series, the solution of certain differential equations, and the standard results that appear throughout the A Level Pure Mathematics papers.
掌握部分分式的收益远不止一道题：同样的变形支撑着有理函数的积分、级数求和、某些微分方程的求解，以及贯穿 A Level 纯数学各卷的诸多标准结果。
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