📚 Partial Fractions in IGCSE Mathematics | IGCSE 数学：偏微分 考点精讲
In IGCSE Mathematics, the topic often referred to in some translated materials as ‘partial differentiation’ is actually the algebraic technique of partial fractions. This method allows us to break down a complicated rational expression into a sum of simpler fractions, which is essential for integration, solving equations, and further calculus topics. It is important not to confuse this with partial differentiation in multivariable calculus, which is not part of the IGCSE syllabus. This article provides a comprehensive revision guide to mastering partial fractions for your IGCSE exam.
在IGCSE数学课程中，一些教学材料中提到的“偏微分”实为部分分式（Partial Fractions）这一代数技巧。该方法将复杂的有理分式拆分成几个简单分式之和，这对后续的积分运算、方程求解以及更高级的微积分学习都至关重要。请注意，此处并非多元微积分中的偏导数，IGCSE并不涉及那些内容。本文将系统梳理部分分式的所有核心考点，助你高效备考。
1. Introduction to Partial Fractions | 部分分式简介
Partial fraction decomposition is the reverse process of adding algebraic fractions. Given a single rational function, we express it as a sum of two or more simpler fractions whose denominators are factors of the original denominator. For example, a fraction like (2x+3)/(x(x+1)) can be split into A/x + B/(x+1). This technique is heavily tested in IGCSE Additional Mathematics and appears in some extended syllabi.
部分分式分解是将一个复杂代数分式拆分为若干个简单分式之和的过程，本质上是分式加法的逆运算。例如，形如 (2x+3)/(x(x+1)) 的分式可拆分为 A/x + B/(x+1) 的形式。此考点在IGCSE附加数学中尤为常见，部分扩展课程也会涉及。
2. Proper and Improper Fractions | 真分式与假分式
Before starting any decomposition, we must check whether the given rational expression is proper or improper. A proper fraction has a numerator with a lower degree than the denominator. If the degree of the numerator is greater than or equal to that of the denominator, the fraction is improper. For improper fractions, we must first perform polynomial long division to rewrite it as a polynomial plus a proper fraction. Only the proper part is then decomposed into partial fractions.
分解前务必先判断所给有理式是真分式还是假分式。当分子次数低于分母次数时，称为真分式；若分子次数大于或等于分母次数，则为假分式。对于假分式，须先通过多项式长除法，将它转化为一个多项式与一个真分式之和，再对真分式部分进行部分分式分解。
3. Distinct Linear Factors | 不同的线性因式
When the denominator can be factorised into distinct linear factors, such as (x – a)(x – b), the partial fraction form will be A/(x – a) + B/(x – b). Here A and B are constants to be determined. This is the most fundamental case and the starting point for all partial fraction problems.
当分母可分解为不同的一次因式（如 (x – a)(x – b)）时，部分分式的形式为 A/(x – a) + B/(x – b)，其中 A 与 B 为待定常数。这是最基本的情形，也是所有部分分式问题的起点。
P(x) / [(x – a)(x – b)] = A/(x – a) + B/(x – b)
P(x) / [(x – a)(x – b)] = A/(x – a) + B/(x – b)
4. Worked Example: Distinct Linear Factors | 例题：不同线性因式
Decompose (3x + 5) / [x(x + 2)] into partial fractions.
将 (3x + 5) / [x(x + 2)] 分解为部分分式。
Step 1: Write the form as A/x + B/(x + 2).
第1步：设部分分式形式为 A/x + B/(x + 2)。
Step 2: Multiply through by the denominator x(x + 2):
第2步：两边同乘分母 x(x + 2)：
3x + 5 = A(x + 2) + Bx
3x + 5 = A(x + 2) + Bx
Step 3: Choose convenient values for x. Let x = 0: 5 = A(2) => A = 5/2.
第3步：选取方便的 x 值。令 x = 0：5 = A(2)，故 A = 5/2。
Step 4: Let x = -2: 3(-2) + 5 = B(-2) => -1 = -2B => B = 1/2.
第4步：令 x = -2：3(-2) + 5 = B(-2)，得 -1 = -2B，故 B = 1/2。
Step 5: Write the final decomposition: (5/2)/x + (1/2)/(x + 2), or (5/(2x)) + (1/(2(x + 2))).
第5步：写出最终分解结果： (5/2)/x + (1/2)/(x + 2)，即 5/(2x) + 1/(2(x + 2))。
5. Repeated Linear Factors | 重复线性因式
If the denominator contains a repeated linear factor, such as (x – a)², the decomposition includes terms for each power up to the multiplicity. For a squared factor, the form is A/(x – a) + B/(x – a)². If the factor is cubed, you would include C/(x – a)³, but IGCSE typically sticks to second powers. The constant B can be found directly by the cover-up method, while A requires substitution or coefficient comparison.
若分母含有重复线性因式（如 (x – a)²），则分解式中需包含从一次到最高次幂的每一项。对于平方因式，形式为 A/(x – a) + B/(x – a)²。若是三次方，还需加上 C/(x – a)³，但IGCSE通常只要求到二次方。常数 B 可用覆盖法直接求得，A 则需通过代入或比较系数确定。
6. Worked Example: Repeated Linear Factors | 例题：重复线性因式
Express (2x + 3) / (x – 1)² in partial fractions.
将 (2x + 3) / (x – 1)² 表示为部分分式。
Step 1: Write the form A/(x – 1) + B/(x – 1)².
第1步：设形式为 A/(x – 1) + B/(x – 1)²。
Step 2: Multiply both sides by (x – 1)²:
第2步：两边同乘 (x – 1)²：
2x + 3 = A(x – 1) + B
2x + 3 = A(x – 1) + B
Step 3: Let x = 1: 2(1) + 3 = B => B = 5.
第3步：令 x = 1：2(1) + 3 = B，得 B = 5。
Step 4: Let x = 0: 3 = A(-1) + 5 => -A = -2 => A = 2.
第4步：令 x = 0：3 = A(-1) + 5，得 -A = -2，故 A = 2。
Step 5: Write the answer: 2/(x – 1) + 5/(x – 1)².
第5步：写出结果：2/(x – 1) + 5/(x – 1)²。
7. Irreducible Quadratic Factor | 不可约二次因式
Sometimes the denominator contains a quadratic factor that cannot be factorised into real linear factors, such as x² + 1 or x² + x + 1. In such cases, the numerator of the partial fraction corresponding to this factor must be a linear expression Ax + B, not just a constant. The general form for a denominator (x + a)(x² + bx + c) is A/(x + a) + (Bx + C)/(x² + bx + c).
当分母中含有不可分解的二次因式（如实数范围内不能分解的 x² + 1 或 x² + x + 1）时，对应部分分式的分子必须设为一次式 Ax + B，而不能仅为常数。对于形如 (x + a)(x² + bx + c) 的分母，部分分式形式为 A/(x + a) + (Bx + C)/(x² + bx + c)。
8. Worked Example: Quadratic Factor | 例题：二次因式
Decompose (x² + 2x + 3) / [(x + 1)(x² + 1)] into partial fractions.
将 (x² + 2x + 3) / [(x + 1)(x² + 1)] 分解为部分分式。
Step 1: The form is A/(x + 1) + (Bx + C)/(x² + 1).
第1步：设形式为 A/(x + 1) + (Bx + C)/(x² + 1)。
Step 2: Multiply through by (x + 1)(x² + 1):
第2步：两边同乘 (x + 1)(x² + 1)：
x² + 2x + 3 = A(x² + 1) + (Bx + C)(x + 1)
x² + 2x + 3 = A(x² + 1) + (Bx + C)(x + 1)
Step 3: Let x = -1: 1 – 2 + 3 = A(2) => 2 = 2A => A = 1.
第3步：令 x = -1：1 – 2 + 3 = A(2)，得 2 = 2A，故 A = 1。
Step 4: Expand and substitute A = 1: x² + 2x + 3 = (x² + 1) + (Bx + C)(x + 1). Simplify to 2x + 2 = (Bx + C)(x + 1).
第4步：展开并代入 A = 1：x² + 2x + 3 = (x² + 1) + (Bx + C)(x + 1)。化简得 2x + 2 = (Bx + C)(x + 1)。
Step 5: Divide both sides by (x + 1) (valid for all x ≠ -1): 2 = Bx + C. Therefore B = 0, C = 2.
第5步：两边同除以 (x + 1)（当 x ≠ -1 时成立），得 2 = Bx + C。因而 B = 0，C = 2。
Step 6: The final decomposition is 1/(x + 1) + 2/(x² + 1).
第6步：最终分解结果为 1/(x + 1) + 2/(x² + 1)。
9. Methods for Finding Constants | 求常数的方法
There are two primary strategies for determining the unknown constants A, B, C, etc. The Published by TutorHao | IGCSE Mathematics Revision Series | aleveler.com
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