Partial Fractions | IGCSE 数学：部分分式考点精讲

📚 Partial Fractions | IGCSE 数学：部分分式考点精讲

Partial fractions is a key skill in IGCSE Additional Mathematics, allowing you to break a complex algebraic fraction into a sum of simpler fractions. This technique is essential for integration, series expansion, and solving equations. In this article, we systematically review every essential type of partial fraction decomposition you need to know for the CIE exam, complete with worked examples and examiner tips.

部分分式是 IGCSE 附加数学的核心技能之一，它让你把一个复杂的有理分式分解为几个简单分式的和。这一技巧对于积分、级数展开和方程求解至关重要。本篇文章将系统梳理 CIE 考试中你必须掌握的各类部分分式分解方法，并配有例题和考官提分指南。

1. Understanding Partial Fractions | 认识部分分式

A rational expression P(x)/Q(x) where degree of P(x) is less than Q(x) can often be expressed as a sum of simpler fractions. For example, (3x+1)/(x²-1) can be written as A/(x-1) + B/(x+1). The process of finding constants A and B is called decomposition into partial fractions.

当有理分式 P(x)/Q(x) 中分子次数低于分母时，往往可以写成几个较简单分式的和。例如 (3x+1)/(x²-1) 可写成 A/(x-1) + B/(x+1) 的形式。求常数 A 和 B 的过程就称为部分分式分解。

2. When to Use Partial Fractions | 何时使用部分分式

You will need partial fractions when integrating rational functions, expanding expressions as power series, or simplifying algebraic expressions. In IGCSE 0606, questions usually ask: ‘Express in partial fractions’ or ‘Find constants A, B, C such that…’. It is also tested in the context of binomial expansion of rational functions.

你会需要在积分有理函数、将表达式展开成幂级数或化简代数式时使用部分分式。在 IGCSE 0606 考试中，题目通常会要求“表示成部分分式”或“求常数 A, B, C 使得……”。在有理函数的二项式展开中也常会涉及此考点。

3. Denominator with Distinct Linear Factors | 分母为不同的一次因式

If the denominator factors into distinct linear factors, write: f(x)/[(x-a)(x-b)] = A/(x-a) + B/(x-b). Multiply through by the denominator and either equate coefficients or substitute convenient x-values to find A and B. For example: (5x-1)/[(x-2)(x+3)] = A/(x-2) + B/(x+3). Substituting x=2 gives A=9/5, x=-3 gives B=16/5.

如果分母可分解为不同的一次因式，设待分解式为：f(x)/[(x-a)(x-b)] = A/(x-a) + B/(x-b)。两边同乘分母后，可通过比较系数或代入特定 x 值求出 A 和 B。例如：(5x-1)/[(x-2)(x+3)] = A/(x-2) + B/(x+3)。代入 x=2 求得 A=9/5，代入 x=-3 求得 B=16/5。

4. Denominator with Repeated Linear Factors | 分母含有重复一次因式

If a linear factor (ax+b) is repeated n times, you must include a series of terms: A₁/(ax+b) + A₂/(ax+b)² + … + Aₙ/(ax+b)ⁿ. For example, (3x+2)/(x-1)² decomposes as A/(x-1) + B/(x-1)². Multiply both sides by (x-1)² and solve.

如果一次因式 (ax+b) 出现 n 次重复，就必须加入一系列分式：A₁/(ax+b) + A₂/(ax+b)² + … + Aₙ/(ax+b)ⁿ。例如 (3x+2)/(x-1)² 应分解为 A/(x-1) + B/(x-1)²。两边同乘 (x-1)² 即可求解。

5. Denominator with Irreducible Quadratic Factor | 分母中含有不可分解的二次因式

When the denominator contains a quadratic factor that cannot be factorised (e.g., x²+1), the corresponding partial fraction has the form (Ax+B)/(x²+1). For more complex expressions, ensure the numerator is linear. Example: (2x)/(x²+4)(x-1) = (Ax+B)/(x²+4) + C/(x-1).

当分母含有无法再分解的二次因式（如 x²+1），对应的部分分式应设为 (Ax+B)/(x²+1)。如果表达式更复杂，确保分子的次数为一次。例如：(2x)/(x²+4)(x-1) = (Ax+B)/(x²+4) + C/(x-1)。

6. Improper Fractions: When Degree of Numerator ≥ Denominator | 假分式：分子次数 ≥ 分母次数

If the degree of the numerator is equal to or greater than the denominator, first perform polynomial long division. The result will be a polynomial plus a proper fraction, which can then be decomposed further. Example: (x³+2x)/(x²-1) = x + (3x)/(x²-1), then decompose (3x)/(x²-1) into partial fractions.

若分子次数大于或等于分母次数，应先进行多项式长除法。结果将是一个多项式加上一个真分式，再将真分式分解成部分分式。例如：(x³+2x)/(x²-1) = x + (3x)/(x²-1)，然后分解 (3x)/(x²-1) 为部分分式。

7. Methods for Finding Constants: Substitution vs Equating Coefficients | 求待定常数的方法：代入法与比较系数法

Substitution method: Choose x-values that make a factor zero to directly find one constant. Useful for distinct linear factors. Equating coefficients: Expand the right-hand side and match coefficients of powers of x with the left numerator. This method works in all cases, especially when substitution becomes messy (quadratic denominators). Often a combination of both is quickest.

代入法：选取能让某个因式为零的 x 值，直接求出某个常数。对不同的一次因式很有效。比较系数法：将右侧展开后，令各次幂系数与左侧分子对应相等。适用于所有情况，尤其当分母含二次式时代入法不便时。实际解题中两者结合往往最快。

8. Worked Example 1: Two Distinct Linear Factors | 例题 1：两个不同的一次因式

Express (7x-3)/(x²-5x+6) in partial fractions. Factorise denominator: (x-2)(x-3). Let (7x-3)/[(x-2)(x-3)] = A/(x-2) + B/(x-3). Multiply by (x-2)(x-3): 7x-3 = A(x-3) + B(x-2). Substitute x=2: 14-3 = A(-1) => A = -11. Substitute x=3: 21-3 = B(1) => B = 18. Hence, (7x-3)/(x²-5x+6) = -11/(x-2) + 18/(x-3).

将 (7x-3)/(x²-5x+6) 表示成部分分式。分母因式分解为 (x-2)(x-3)。设 (7x-3)/[(x-2)(x-3)] = A/(x-2) + B/(x-3)。两边同乘 (x-2)(x-3)：7x-3 = A(x-3) + B(x-2)。代入 x=2：14-3 = A(-1) => A = -11。代入 x=3：21-3 = B(1) => B = 18。所以 (7x-3)/(x²-5x+6) = -11/(x-2) + 18/(x-3)。

9. Worked Example 2: Repeated Linear Factor | 例题 2：重复一次因式

Decompose (4x+1)/(x+2)². Write as A/(x+2) + B/(x+2)². Multiply: 4x+1 = A(x+2) + B. Expand: 4x+1 = Ax + 2A + B. Equate coefficients of x: A = 4. Constant term: 2A+B = 1 => 8+B=1 => B=-7. Therefore, (4x+1)/(x+2)² = 4/(x+2) – 7/(x+2)².

分解 (4x+1)/(x+2)²。设为 A/(x+2) + B/(x+2)²。两边同乘分母：4x+1 = A(x+2) + B。展开：4x+1 = Ax + 2A + B。比较 x 系数：A=4。常数项：2A+B=1 => 8+B=1 => B=-7。故 (4x+1)/(x+2)² = 4/(x+2) – 7/(x+2)²。

10. Worked Example 3: Quadratic Denominator | 例题 3：二次分母

Given (3x²+2x+1)/[(x²+1)(x-1)], write as (Ax+B)/(x²+1) + C/(x-1). Multiply through: 3x²+2x+1 = (Ax+B)(x-1) + C(x²+1). Expand RHS: Ax² – Ax + Bx – B + Cx² + C = (A+C)x² + (-A+B)x + (-B+C). Equate coefficients: x²: A+C=3; x: -A+B=2; constant: -B+C=1. Solve: from second eq B=A+2; from first C=3-A. Substitute into third: -(A+2) + (3-A) = 1 => -A-2+3-A=1 => -2A+1=1 => A=0. Then B=2, C=3. So result is 2/(x²+1) + 3/(x-1).

将 (3x²+2x+1)/[(x²+1)(x-1)] 分解为 (Ax+B)/(x²+1) + C/(x-1)。两边同乘分母：3x²+2x+1 = (Ax+B)(x-1) + C(x²+1)。右侧展开：Ax² – Ax + Bx – B + Cx² + C = (A+C)x² + (-A+B)x + (-B+C)。比较系数：x²：A+C=3；x：-A+B=2；常数：-B+C=1。解方程组：由第二式 B=A+2；由第一式 C=3-A；代入第三式：-(A+2)+(3-A)=1 => -2A+1=1 => A=0。则 B=2，C=3。结果为 2/(x²+1) + 3/(x-1)。

11. Common Mistakes and Examiner Advice | 常见错误与考官建议

Many students forget to include all terms for repeated factors, or use an incorrect numerator for quadratic factors. Always check the degree of the numerator versus denominator before starting. Misplacing a sign during substitution is a frequent careless error. Use brackets carefully and double-check your expansions. In CIE exams, marks are awarded for correct setup even if solving is flawed, so always write the general form clearly.

很多学生忘记为重复因式设置完整的分式串，或在二次因式上使用了错误的分子形式。开始前务必检查分子分母的次数。代入时弄错正负号是常见的粗心错误。小心使用括号，并复查你的展开过程。在 CIE 考试中，即使计算有误，只要设式正确也能得分，因此始终清晰地写出通用的分解形式。

12. Connecting to Binomial Expansion and Integration | 链接二项式展开与积分

Once in partial fractions, expressions like A/(x+a) can be rewritten as A(a+x)⁻¹ and expanded using the binomial theorem for |x|<|a|. Similarly, partial fractions allow term-by-term integration, turning difficult integrals into simple ln|ax+b| or arctan forms. This is why the topic is so heavily tested.

表示成部分分式后，像 A/(x+a) 这样的表达式可以改写为 A(a+x)⁻¹，并利用二项式定理在 |x|<|a| 条件下展开。同样地，部分分式使逐项积分成为可能，把困难的积分化为简单的 ln|ax+b| 或 arctan 形式。这是此考点如此频繁考查的原因。

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Partial Fractions | IGCSE 数学：部分分式考点精讲