Partial Fractions | 分部积分法

📚 Partial Fractions | 分部积分法

Partial fractions is a powerful algebraic technique used to decompose rational functions into simpler fractions, making integration, differentiation and series expansion far more manageable. In the Edexcel A-Level Mathematics syllabus, this method appears in both Pure Mathematics and Further Mathematics, underpinning many calculus and algebraic manipulation problems.

分部积分法是一种将有理函数拆分为更简单分式的代数技巧，能极大简化积分、微分和级数展开的计算。在 Edexcel A-Level 数学大纲中，该方法贯穿纯数和进阶数学，是许多微积分与代数变形题目的基础。

This article explores the core principles, step-by-step procedures and typical exam-style applications of partial fractions, ensuring you can handle linear, repeated and quadratic factors with confidence. Worked examples will reinforce the theory, while common pitfalls are addressed to sharpen your problem-solving technique.

本文将深入探讨分部积分法的核心原理、分解步骤以及典型的考试应用，帮助你自信应对线性因子、重复因子和二次因子。通过例题巩固理论，同时指出常见错误，提升你的解题技巧。

1. What Are Partial Fractions? | 什么是分部积分法？

Partial fractions decompose a single algebraic fraction with a polynomial denominator into a sum of two or more simpler fractions. For example, (3x+5)/(x²+3x+2) can be expressed as 1/(x+1) + 2/(x+2). This decomposition is only possible when the degree of the numerator is strictly less than the degree of the denominator; otherwise, algebraic long division must be performed first to obtain a proper rational function.

分部积分法将一个分母为多项式的代数分式拆分成两个或多个更简单的分式之和。例如，(3x+5)/(x²+3x+2) 可表示为 1/(x+1) + 2/(x+2)。这种拆分只有在分子的次数严格小于分母的次数时才可直接进行；否则需先通过多项式长除法转化为真分式。

The method relies on the fact that every polynomial with real coefficients can be factorised into linear and irreducible quadratic factors. By assigning unknown constants to each factor, we set up an identity that holds for all values of the variable, enabling us to solve for the constants systematically.

该方法基于这样一个事实：每个实系数多项式都可以分解为线性因子和不可约二次因子。通过为每个因子分配未知常数，我们建立起一个对所有变量值都成立的恒等式，从而系统地解出常数。

2. The Three Core Denominator Types | 三种核心分母类型

Edexcel A-Level focuses on three denominator structures: distinct linear factors, repeated linear factors and irreducible quadratic factors. A distinct linear factor such as (x+1) produces a term A/(x+1). A repeated linear factor like (x+2)² gives A/(x+2) + B/(x+2)². An irreducible quadratic factor such as (x²+1) requires the form (Ax+B)/(x²+1).

Edexcel A-Level 重点考察三种分母结构：不同的线性因子、重复线性因子和不可约二次因子。对于不同的线性因子如 (x+1)，分项形式为 A/(x+1)；重复线性因子例如 (x+2)²，则需拆分为 A/(x+2) + B/(x+2)²；不可约二次因子如 (x²+1)，需要形式 (Ax+B)/(x²+1)。

Understanding these templates is essential because the form of the decomposition directly affects how the constants are determined. Using the wrong template will lead to identities that cannot be satisfied for all x, wasting valuable exam time.

理解这些模板至关重要，因为拆分的正确形式直接影响常数的求解。使用错误的模板将导致无法对所有 x 成立的恒等式，浪费宝贵的考试时间。

Denominator factor	Denominator Factor	Partial fraction form
Distinct linear (ax+b)	不同的线性因子	A/(ax+b)
Repeated linear (ax+b)ⁿ	重复线性因子	A₁/(ax+b) + A₂/(ax+b)² + ... + Aₙ/(ax+b)ⁿ
Irreducible quadratic (ax²+bx+c)	不可约二次因子	(Ax+B)/(ax²+bx+c)

3. Step-by-Step Method: Distinct Linear Factors | 逐步分解法：不同的线性因子

rite (2x+3)/((x-1)(x+2)) = A/(x-1) + B/(x+2). Multiply through by the common denominator to clear fractions: 2x+3 = A(x+2) + B(x-1).

以真分式 (2x+3)/(x²+x-2) 为例。首先分母因式分解：x²+x-2 = (x-1)(x+2)。两者为不同的线性因子，故设 (2x+3)/((x-1)(x+2)) = A/(x-1) + B/(x+2)。两边同乘公分母去分母：2x+3 = A(x+2) + B(x-1)。

d B, substitute key values of x that make each bracket zero. Let x=1: 2(1)+3 = A(3) + B(0) ⇒ 5 = 3A ⇒ A = 5/3. Let x=-2: 2(-2)+3 = A(0) + B(-3) ⇒ -1 = -3B ⇒ B = 1/3. Thus the partial fractions are 5/(3(x-1)) + 1/(3(x+2)).

求 A 和 B 时，代入使各括号为零的特殊 x 值。令 x=1：2(1)+3 = A(3) + B(0) ⇒ 5 = 3A ⇒ A = 5/3。令 x=-2：2(-2)+3 = A(0) + B(-3) ⇒ -1 = -3B ⇒ B = 1/3。因此分部积分结果为 5/(3(x-1)) + 1/(3(x+2))。

stem A+B=2 and 2A-B=3, solving to the same constants. Choosing values is typically faster.

也可用比较系数法：将右侧展开为 (A+B)x + (2A-B)，与 2x+3 比较，得方程组 A+B=2 和 2A-B=3，解出相同常数。通常代入特殊值更快捷。

4. Repeated Linear Factors | 重复线性因子

ling to include the lower power term is a common exam mistake. For (3x+1)/(x+1)², write (3x+1)/(x+1)² = A/(x+1) + B/(x+1)². Multiply through: 3x+1 = A(x+1) + B.

当分母含有重复线性因子时，例如 (x+1)²，分解式中必须包含每个幂次对应的分项：A/(x+1) + B/(x+1)²。漏掉低次项是考试中的常见错误。对于 (3x+1)/(x+1)²，设 (3x+1)/(x+1)² = A/(x+1) + B/(x+1)²。去分母：3x+1 = A(x+1) + B。

Here, substituting x=-1 eliminates A: 3(-1)+1 = A(0)+B ⇒ -2 = B. To find A, use any other value, say x=0: 1 = A(1) -2 ⇒ 2/(x+1)². Always check by combining the result to ensure it matches the original fraction.

此时代入 x=-1 可消去 A：3(-1)+1 = A(0)+B ⇒ B=-2。求 A 可用任意其他值，如 x=0：1 = A(1) -2 ⇒ A=3。故分解为 3/(x+1) - 2/(x+1)²。务必通分验证结果是否与原分式一致。

For higher multiplicities, e.g. (x-2)³, the decomposition is A/(x-2) + B/(x-2)² + C/(x-2)³. The same substitution and equating approach works, but solving for all constants may require comparing coefficients as the system grows.

对于更高次重复，如 (x-2)³，分解形式为 A/(x-2) + B/(x-2)² + C/(x-2)³。同样的代入和比较系数方法仍然有效，但随着方程组增大，可能需要结合使用系数比较法。

5. Irreducible Quadratic Factors | 不可约二次因子

x+B)/(x²+1). For example, decompose (5x²+2x+3)/((x²+1)(x-1)). Write = (Ax+B)/(x²+1) + C/(x-1). Multiply through by the denominator: 5x²+2x+3 = (Ax+B)(x-1) + C(x²+1).

不可约二次因子如 x²+1 不能分解为实线性因子。分部积分时必须包含线性分子：(Ax+B)/(x²+1)。例如分解 (5x²+2x+3)/((x²+1)(x-1))。设 = (Ax+B)/(x²+1) + C/(x-1)。两边同乘分母：5x²+2x+3 = (Ax+B)(x-1) + C(x²+1)。

ents with the left side: A+C=5, -A+B=2, -B+C=3. Solving simultaneously yields A=2, B=4, C=3. So the result is (2x+4)/(x²+1) + 3/(x-1).

展开并合并同类项：(Ax+B)(x-1) = Ax² - Ax + Bx - B = Ax² + (-A+B)x - B。右侧变为 (A+C)x² + (-A+B)x + (-B+C)。比较左侧系数得方程组 A+C=5, -A+B=2, -B+C=3。联立解得 A=2, B=4, C=3。结果为 (2x+4)/(x²+1) + 3/(x-1)。

Substituting x=1 simplifies finding C directly: 5+2+3 = (A+B)×0 + C(2) ⇒ 10=2C ⇒ C=5. Wait, that gives C=5, but previous equating gave C=3? Let's check: (5+2+3)=10, denominator (x²+1)(x-1) at x=1: (2)(0)=0? identity is 5x²+2x+3 = (Ax+B)(x-1) + C(x²+1). Set x=1: LHS=5+2+3=10, RHS=(A+B)(0)+C(2)=2C ⇒ C=5. Our earlier equating must have an error. Let' uating coefficients: (A+C)x² + (-A+B)x + (-B+C) = 5x²+2x+3. So -B+C=3. If C=5, then -B+5=3 ⇒ B=2. Then -A+B=2 ⇒ -A+2=2 ⇒ bstitution method caught the mistake; always verify with a quick substitution if possible.

令 x=1 可直接求 C：5+2+3 = (A+B)×0 + C(2) ⇒ 10=2C ⇒ C=5。等等，这与前面比较系数得到的 C=3 矛盾。检查：代入 x=1 时，5x²+2x+3 = 5+2+3=10，右侧 (Ax+B)(0) + C(2)=2C ⇒ C=5。之前的系数比较有误。重新计算：右侧展开为 (A+C)x² + (-A+B)x + (-B+C)，与左侧 5,2,3 对应，则 -B+C=3，若 C=5，则 -B+5=3 ⇒ B=2；-A+B=2 ⇒ -A+2=2 ⇒ A=0；A+C=5 成立。正确常数为 A=0, B=2, C=5。所以分解为 2/(x²+1) + 5/(x-1)。代入特殊值法及时发现了错误；应尽量验证。

6. Improper Fractions and Long Division | 假分式与长除法

us a proper fraction. For example, (x³+2x²+x+1)/(x²+x). Divide: x³/x² = x, multiply divisor by x: x³+x², subtract from numerator: (x³+2x²) - (x³+x²) = x², bring down +x+1. Next, x²/x² = 1, multiply: x²+x, subtract: (x²+x) - (x²+x)=0, remainder +1. So quotient is x+1, remainder 1, giving x+1 + 1/(x²+x). Then factorise denominator x(x+1) and decompose 1/(x(x+1)) into partial fractions: 1/x - 1/(x+1). Final answer: x+1 + 1/x - 1/(x+1).

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若分子的次数大于或等于分母的次数，必须先进行多项式长除法，得到一个多项式加一个真分式。例如 (x³+2x²+x+1)/(x²+x)。长除：x³/x² = x，除数乘 x 得 x³+x²，被除数减去得 x²，移下 +x+1。接着 x²/x² = 1，乘除数得 x²+x，减去后余数 +1。商式为 x+1，余数为 1，因此原式 = x+1 + 1/(x²+x)。分母因式分解为 x(x+1)，再对 1/(x(x+1)) 进行分部积分：1/x - 1/(x+1)。最终结果为 x+1 + 1/x - 1/(x+1)。

Neglecting long division is a classic error that leads to an invalid decomposition setup. Always check degrees before starting.

忽略长除法是一个典型错误，会导致无法设立正确的分解形式。分解前务必核对次数。

7. Integration Using Partial Fractions | 利用分部积分法求积分

tly, typically yielding natural logarithms or arctangent functions. For instance, ∫(5x+3)/(x²-1) dx. Factor denominator: (x-1)(x+1), decompose into A/(x-1)+B/(x+1). Solving gives A=4, B=1. Then ∫[4/(x-1)+1/(x+1)]dx = 4ln|x-1|+ln|x+1|+C.

Edexcel A-Level 中最常见的应用之一就是有理函数的积分。分解后每一项均可直接积分，通常得到自然对数或反正切函数。例如 ∫(5x+3)/(x²-1) dx。分母因式分解为 (x-1)(x+1)，拆分为 A/(x-1)+B/(x+1)。解得 A=4, B=1。于是 ∫[4/(x-1)+1/(x+1)]dx = 4ln|x-1|+ln|x+1|+C。

to a combination of ln and arctan. For example, ∫(2x+1)/(x²+2x+5) dx. Here the denominator' +1)²+4) dx - ∫1/((x+1)²+4) dx. The first part is a ln form: ln((x+1)²+4). The second part uses formula ∫1/(a²+u²)du = (1/a) arctan(u/a), giving (1/2) arctan((x+1)/2). But if we had decomposed a proper fraction with irreducible quadratic factor, integration would proceed similarly, often using substitution after completing square. However, partial fractions for such integrals are usually reserved for cases where the quadratic factor cannot be split further, and integration then requires the linear numerator to be integrated as ln|quadratic| plus an arctan term.

当出现二次因子时，积分会得到 ln 和 arctan 的组合。例如 ∫(2x+1)/(x²+2x+5) dx。分母判别式为负，先配方：x²+2x+5 = (x+1)²+4。分子重写为 2x+1 = 2(x+1)-1。拆分积分：∫2(x+1)/((x+1)²+4) dx - ∫1/((x+1)²+4) dx。第一部分为 ln 形式：ln((x+1)²+4)。第二部分用公式 ∫1/(a²+u²)du = (1/a) arctan(u/a)，得 (1/2) arctan((x+1)/2)。但如果是对含有不可约二次因子的真分式进行分部积分，积分过程类似，通常在配方后通过换元得到 ln 和 arctan 项。

Thus proficiency in partial fractions directly boosts integration skills, particularly for exam questions on the trapezium rule's analytical counterpart and differential equations.

因此，熟练运用分部积分法直接提升积分能力，尤其有助于解决涉及梯形法则的解析对照和微分方程的考题。

8. Binomial Expansion and Series | 二项展开与级数

orm A/(ax+b)ⁿ or (Ax+B)/(quadratic), each can be expressed as a power series. For example, expand (3x+5)/((1-2x)(1+x)) in ascending powers up to x³. First, decompose: (3x+5)/((1-2x)(1+x)) = 4/(1-2x) + 1/(1+x). Now rewrite as 4(1-2x)⁻¹ + (1+x)⁻¹. Expand: (1-2x)⁻¹ = 1 + 2x + 4x² + 8x³ + ..., so 4 times that yields 4 + 8x + 16x² + 32x³ + ... . (1+x)⁻¹ = 1 - x + x² - x³ + ... . Sum: 5 + 7x + 17x² + 31x³ + ... . Always state the range of validity: |2x|<1 ⇒ |x|<1/2 for the first, |x|<1 for the second, so overall |x|<1/2.

分部积分法还可配合二项式定理对有理函数进行级数展开。将分式分解为 A/(ax+b)ⁿ 或 (Ax+B)/(quadratic) 的形式后，每一项均可展为幂级数。例如将 (3x+5)/((1-2x)(1+x)) 展开到 x³ 项。先分解：设 (3x+5)/((1-2x)(1+x)) = 4/(1-2x) + 1/(1+x)。重写为 4(1-2x)⁻¹ + (1+x)⁻¹。展开：(1-2x)⁻¹ = 1 + 2x + 4x² + 8x³ + ...，乘以 4 得 4 + 8x + 16x² + 32x³ + ... 。(1+x)⁻¹ = 1 - x + x² - x³ + ...。求和得 5 + 7x + 17x² + 31x³ + ...。必须注明有效性范围：第一个级数 |2x|<1 ⇒ |x|<1/2，第二个 |x|<1，故整体 |x|<1/2。

This technique is heavily tested in Edexcel FP1 and FP2, where combining series expansions with partial fractions forms a routine 6-8 mark question.

这一技巧在 Edexcel FP1 和 FP2 中考查频繁，将分部积分与级数展开结合的题目通常占 6-8 分。

9. Solving Differential Equations | 求解微分方程

te: dy/(y²+1) = dx/(x²-1). The right side decomposes: 1/(x²-1) = 1/2[1/(x-1) - 1/(x+1)]. Integrate both sides: arctan(y) = (1/2)ln|(x-1)/(x+1)| + C. This elegant solution is typical of A2 Maths modelling contexts.

许多可分离变量微分方程会产生需要分部积分的有理被积函数。例如 dy/dx = (y²+1)/(x²-1)。分离变量：dy/(y²+1) = dx/(x²-1)。右侧分解：1/(x²-1) = 1/2[1/(x-1) - 1/(x+1)]。两边积分：arctan(y) = (1/2)ln|(x-1)/(x+1)| + C。这类巧妙解法常见于 A2 数学的建模题目中。

Equilibrium solutions and logistic growth often involve partial fractions after separation, so mastering this algebraic tool streamlines the calculus.

平衡解和 Logistic 增长模型在分离变量后常涉及分部积分，因此熟练掌握这一代数工具可以简化微积分过程。

10. Common Mistakes and Exam Tips | 常见错误与应试技巧

Common pitfalls include: forgetting to check the degree condition, misidentifying the form for repeated or quadratic factors, sign errors when substituting, and failing to simplify the final partial fractions. Also, many students forget to write the absolute value inside logarithms during integration or neglect the integration constant.

常见错误包括：忘记检查次数条件、重复因子或二次因子的拆分形式写错、代入时出现符号错误、以及未将最终分部积分结果化为最简。此外，许多学生在积分时忘记在 ln 内加绝对值，或遗漏积分常数。

Exam technique: after decomposition, quickly test with a convenient x-value to verify identity. In integration problems, simplify coefficients before integrating. For series expansions, clearly state the validity range; marks are often allocated for this. When faced with quadratic factors, complete the square early.

应试技巧：分解后快速用方便的 x 值验证恒等式；在积分题中，积分前先化简系数；对于级数展开，清晰注明有效性范围，往往有分值；遇到二次因子时，尽早配方。

Also, in the Edexcel formula booklet, the standard integrals for 1/(a²+x²) and 1/√a²-x² are provided; learn to recognise when to use them post-decomposition.

另外，Edexcel 公式手册中提供了 1/(a²+x²) 和 1/√a²-x² 的标准积分公式；要懂得在分解后识别何时使用它们。

11. Worked Past Exam Question | 历年真题精解

Question: fractions. Hence find ∫(4x²+5x+3)/((x+1)²(x-2)) dx, giving your answer in simplified form.

题目：将 (4x²+5x+3)/((x+1)²(x-2)) 表示成分部积分的形式。并由此求 ∫(4x²+5x+3)/((x+1)²(x-2)) dx，给出最简形式。

1)(x-2) + B(x-2) + C(x+1)². Choose strategic x: x=-1: 4-5+3=2 on LHS, RHS: A(0)+B(-3)+C(0) ⇒ 2=-3B ⇒ B=-2/3. x=2: 16+10+3=29 on LHS, RHS: A(0)+B(0)+C(9) ⇒ 29=9C ⇒ 3)(-2) + (29/9)(1) = -2A + 4/3 + 29/9. Compute 4/3=12/9, so 12/9+29/9=41/9. So 3= -2A+41/9 ⇒ 27/9 = -2A+41/9 ⇒ -2A = -14/9 ⇒ A=7/9. Thus decomposition: 7/(9(x+1)) - 2/(3(x+1)²) + 29/(9(x-2)).

设 (4x²+5x+3)/((x+1)²(x-2)) = A/(x+1) + B/(x+1)² + C/(x-2)。乘开：4x²+5x+3 = A(x+1)(x-2) + B(x-2) + C(x+1)²。选择特征 x 值：令 x=-1：左侧 4-5+3=2，右侧 A(0)+B(-3)+C(0) ⇒ 2=-3B ⇒ B=-2/3。令 x=2：16+10+3=29，右侧 A(0)+B(0)+C(9) ⇒ 29=9C ⇒ C=29/9。为求 A，可展开比较 x² 系数或用其他 x 值。最简单：令 x=0：3 = A(1)(-2) + (-2/3)(-2) + (29/9)(1) = -2A + 4/3 + 29/9。将 4/3 化为 12/9，故 12/9+29/9=41/9。所以 3 = -2A+41/9 ⇒ 27/9 = -2A+41/9 ⇒ -2A = -14/9 ⇒ A=7/9。分解结果为：7/(9(x+1)) - 2/(3(x+1)²) + 29/(9(x-2))。

Integrate term by term: ∫7/(9(x+1))dx = (7/9)ln|x+1|; ∫-2/(3(x+1)²)dx = ∫(-2/3)(x+1)⁻² dx = (2/3)(x+1)⁻¹; ∫29/(9(x-2))dx = (29/9)ln|x-2|. So final answer: (7/9)ln|x+1| + (2/3(x+1)) + (29/9)ln|x-2| + C. Often simplified by factorising out 1/9: (1/9)[7ln|x+1| + 29ln|x-2|] + 2/(3(x+1)) + C. Ensure proper absolute values.

逐项积分：∫7/(9(x+1))dx = (7/9)ln|x+1|；∫-2/(3(x+1)²)dx = ∫(-2/3)(x+1)⁻² dx = (2/3)(x+1)⁻¹；∫29/(9(x-2))dx = (29/9)ln|x-2|。最终答案：(7/9)ln|x+1| + (2/3(x+1)) + (29/9)ln|x-2| + C。常提出 1/9 化简：(1/9)[7ln|x+1| + 29ln|x-2|] + 2/(3(x+1)) + C。务必使用绝对值。

12. Summary and Practice Advice | 总结与练习建议

Mastering partial fractions requires recognising denominator patterns swiftly, setting up the correct form, and accurately solving for constants. From distinct linear to repeated and irreducible quadratic factors, each case demands a methodical approach. Integration and series expansion applications reinforce the utility of this technique across the A-Level syllabus.

掌握分部积分法需要快速识别分母模式、设立正确分解形式并准确求解常数。从不同线性因子到重复及不可约二次因子，每种情形都需要有条不紊的方法。积分和级数展开的应用进一步突显该技巧在 A-Level 大纲中的广泛用途。

Practise with past papers, focusing on mixed-factor denominators and improper algebraic fractions. Aim for fluency in both coefficient comparison and strategic substitution. A solid command of partial fractions will significantly elevate your performance in Pure Mathematics, Further Calculus and Mechanics where integrals of rational functions appear frequently.

通过历年真题练习，重点攻克混合因子和假分式的题目。熟练掌握系数比较与特征值代入两种方法。扎实掌握分部积分法将显著提升你在纯数学、进阶微积分以及力学（常出现有理函数积分）中的表现。

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