Partial Fractions | 部分分式

📚 Partial Fractions | 部分分式

In A-Level Edexcel Mathematics, partial fractions are a powerful algebraic tool used to rewrite a single rational function as a sum of simpler fractions. This technique is essential for integration, series expansion, and solving differential equations. Mastering partial fractions will allow you to break down complicated algebraic expressions, making them much easier to integrate or analyse. In Pure Mathematics 3, you are expected to handle denominators with distinct linear factors, repeated linear factors, and irreducible quadratic factors, as well as cases where the rational function is improper.

在 A-Level 爱德思考数学中，部分分式是一种强大的代数工具，可将一个单独的有理函数改写为几个较简单分式之和。这一技巧对于积分、级数展开和求解微分方程至关重要。掌握部分分式后，你就能分解复杂的代数表达式，使其更易于积分或分析。在纯数 3 中，你需要处理分母为不同线性因式、重复线性因式以及不可约的二次因式的情形，还要处理有理函数为假分式的情况。

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1. What are Partial Fractions? | 什么是部分分式？

A rational function is a quotient of two polynomials, such as (3x+5)/((x+1)(x-2)). The idea behind partial fractions is to express this quotient as a sum of fractions like A/(x+1) + B/(x-2), where A and B are constants to be found. This decomposition is valid only if the degree of the numerator is less than the degree of the denominator; otherwise, you must first perform polynomial division.

有理函数是两个多项式的商，例如 (3x+5)/((x+1)(x-2))。部分分式的核心思想是将这个商表示为如 A/(x+1) + B/(x-2) 的分式之和，其中 A 和 B 是待定常数。只有当分子的次数低于分母的次数时，这种分解才成立；否则必须先进行多项式除法。

Once the rational function is expressed in partial fractions, it becomes straightforward to integrate each simple term using the natural logarithm or to expand them as binomial series. This technique is a cornerstone of many exam problems, so understanding its logic and methodology is vital.

一旦有理函数用部分分式表示，就可以直接利用自然对数或二项展开对每一项进行积分或展开。这一技巧是许多试题的基石，因此理解其逻辑和方法至关重要。

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2. The General Form for Denominators with Linear Factors | 分母为线性因式的一般形式

When the denominator factorises into distinct linear factors, say (ax+b)(cx+d), the partial fractions will take the form A/(ax+b) + B/(cx+d). If there are three distinct linear factors, you add a third constant, and so on. It is important to assign an unknown constant to each linear factor in the denominator.

当分母可分解为不同的线性因式，如 (ax+b)(cx+d)，部分分式的形式将是 A/(ax+b) + B/(cx+d)。如果有三个不同的线性因式，就需要加上第三个常数，依此类推。为分母中的每一个线性因式分配一个未知常数是非常重要的。

(px+q)/((ax+b)(cx+d)) = A/(ax+b) + B/(cx+d)

Once the form is set up, you multiply through by the original denominator to obtain an identity involving A and B, which can then be solved using substitution or by equating coefficients.

建立好形式后，将等式两边同时乘以原分母，得到一个关于 A 和 B 的恒等式，然后可以用代入法或比较系数法求解。

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3. Decomposition with Distinct Linear Factors | 不同线性因式的分解

1)(2x+1)) = A/(x-1) + B/(2x+1). Multiplying by (x-1)(2x+1) gives: 7x-3 = A(2x+1) + B(x-1).

考虑表达式 (7x-3)/((x-1)(2x+1))。分母由两个不同的线性因式组成，因此我们写成：(7x-3)/((x-1)(2x+1)) = A/(x-1) + B/(2x+1)。两边乘以 (x-1)(2x+1) 得到：7x-3 = A(2x+1) + B(x-1)。

½-1) → -6.5 = -1.5B → B = 13/3. Thus the partial fractions are (4/3)/(x-1) + (13/3)/(2x+1).

为求出 A，代入 x = 1（抵消 B）：7(1)-3 = A(2(1)+1) + 0 → 4 = 3A → A = 4/3。为求出 B，代入 x = -½（抵消 A）：7(-½)-3 = 0 + B(-½-1) → -6.5 = -1.5B → B = 13/3。因此部分分式为 (4/3)/(x-1) + (13/3)/(2x+1)。

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4. Decomposition with Repeated Linear Factors | 重复线性因式的分解

If a factor appears more than once, for example (x+1)², the partial fractions must include terms for each power of that factor up to the highest power: A/(x+1) + B/(x+1)². More generally, for a factor (ax+b)ⁿ, you write a series A₁/(ax+b) + A₂/(ax+b)² + … + Aₙ/(ax+b)ⁿ.

如果某一因式出现多次，例如 (x+1)²，部分分式必须包含从该因式的一次幂到最高次幂的各个项：A/(x+1) + B/(x+1)²。更一般地，对于因式 (ax+b)ⁿ，你需要写出 A₁/(ax+b) + A₂/(ax+b)² + … + Aₙ/(ax+b)ⁿ。

1). You can find C quickly by substituting x = -2, and then solve for A and B using other straightforward substitutions or equating coefficients.

以 (5x+4)/((x+2)²(x-1)) 为例，正确的形式是 A/(x-1) + B/(x+2) + C/(x+2)²。去掉分母后得到：5x+4 = A(x+2)² + B(x-1)(x+2) + C(x-1)。你可通过代入 x = -2 快速求出 C，然后用其他简单的代入值或比较系数法求出 A 和 B。

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5. Decomposition with Irreducible Quadratic Factors | 不可约二次因式的分解

Sometimes the denominator contains a quadratic factor that cannot be factorised over the real numbers, such as x²+1 or x²+x+1. In such cases, the corresponding partial fraction has a linear numerator: (Ax+B)/(quadratic factor). The constants A and B are then determined along with any others from the decomposition.

有时分母包含一个在实数范围内无法再分解的二次因式，例如 x²+1 或 x²+x+1。这种情况下，对应的部分分式需要以一个线性表达式作为分子：(Ax+B)/(二次因式)。接着将 A、B 与其他常数一同求出。

can then expand the right-hand side, equate coefficients, or substitute convenient values to obtain A, B, and C.

例如，要分解 (2x²+3x+5)/((x²+1)(x-2))，我们设定：(Ax+B)/(x²+1) + C/(x-2)。两边乘以分母得 2x²+3x+5 = (Ax+B)(x-2) + C(x²+1)。然后你可以将右边展开、比较系数，或代入合适的值求出 A、B 和 C。

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6. Handling Improper Rational Functions | 处理假分式

When the degree of the numerator is greater than or equal to the degree of the denominator, the rational function is said to be improper. You cannot directly apply the partial fraction technique; you must first divide the numerator by the denominator using polynomial long division (or synthetic division if applicable) to express it as a polynomial plus a proper rational part.

当分子的次数大于或等于分母的次数时，这个有理函数被称为假分式。此时不能直接应用部分分式法；你必须先用多项式长除法（或适用情况下的综合除法）将分子除以分母，将其表示为一个多项式加上一个真分式的形式。

oper fraction (8x+1)/((x+1)(x-2)) can then be split into partial fractions in the usual manner.

对于 (x³+2x²-5)/(x²-x-2)，进行除法得到：x³+2x²-5 = (x+3)(x²-x-2) + (8x+1)。因此 (x³+2x²-5)/(x²-x-2) = x+3 + (8x+1)/(x²-x-2)。之后便可将真分式 (8x+1)/((x+1)(x-2)) 按常规方法分解为部分分式。

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7. Method 1: Substitution Method | 方法一：代入法

The substitution method is often the quickest way to find unknown constants. After multiplying by the denominator, you obtain an identity that holds for all x values. By choosing x values that make some brackets zero, you can directly solve for individual constants one at a time.

代入法通常是求未知常数最快的方法。在乘以分母之后，你会得到一个对所有 x 值都成立的恒等式。通过选择使某些括号为零的 x 值，你可以逐一直接解出各个常数。

e to use the substitution method first for as many constants as possible, then use coefficient comparison for the remaining ones.

例如，给定 A(x-2) + B(x+1) = 5x+3。令 x = 2：B(3) = 13 → B = 13/3。令 x = -1：A(-3) = -2 → A = 2/3。建议先用代入法求出尽可能多的常数，再用比较系数法处理剩余的常数。

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8. Method 2: Equating Coefficients | 方法二：比较系数法

When substitution cannot easily isolate all constants, especially with repeated or quadratic factors, equating coefficients is a reliable alternative. You expand both sides of the identity and then equate the coefficients of like powers of x, obtaining a system of simultaneous equations.

当代入法无法轻易分离出所有常数时，尤其是在涉及重复因式或二次因式时，比较系数是一种可靠的方法。你需要将恒等式两边展开，然后令 x 的同次幂系数相等，从而得到一个关于常数的方程组。

. Equate coefficients of x², x, and constant term to form three equations. Solve simultaneously for A, B, and C. This method is systematic but can be longer; it is often used to check results obtained via substitution.

示例：由 A(x-1)(x+2) + B(x+2) + C = x²+5x+4，展开并合并同类项。令 x²、x 和常数项的系数分别相等，得到三个方程。联立求解即可得到 A、B 和 C。此方法系统性强，但可能较费时；它常被用于验证代入法所得的结果。

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9. Worked Example: Distinct Linear Factors | 示例：不同线性因式

Let us fully decompose (4x²+15x+7)/((x+1)(x+2)(x+3)). Since the denominator is already in factorised form with three distinct linear factors and the numerator is of degree 2 (less than 3), we write:
(4x²+15x+7)/((x+1)(x+2)(x+3)) = A/(x+1) + B/(x+2) + C/(x+3).

让我们完整分解 (4x²+15x+7)/((x+1)(x+2)(x+3))。由于分母已因式分解为三个不同的线性因式，且分子的次数为 2（小于 3），我们写成：
(4x²+15x+7)/((x+1)(x+2)(x+3)) = A/(x+1) + B/(x+2) + C/(x+3)。

Multiply both sides by (x+1)(x+2)(x+3) to give:
4x²+15x+7 = A(x+2)(x+3) + B(x+1)(x+3) + C(x+1)(x+2).
)(1) = 16 -30 +7 → -B = -7 → B = 7. Let x = -3 → C(-2)(-1) = 36 -45 +7 → 2C = -2 → C = -1.
Hence the decomposition is -2/(x+1) + 7/(x+2) - 1/(x+3).

两边乘以 (x+1)(x+2)(x+3) 得到：
4x²+15x+7 = A(x+2)(x+3) + B(x+1)(x+3) + C(x+1)(x+2)。
现在用代入法：令 x = -1 → A(1)(2) = 4 -15 +7 → 2A = -4 → A = -2。令 x = -2 → B(-1)(1) = 16 -30 +7 → -B = -7 → B = 7。令 x = -3 → C(-2)(-1) = 36 -45 +7 → 2C = -2 → C = -1。
因此分解结果为 -2/(x+1) + 7/(x+2) - 1/(x+3)。

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10. Worked Example: Repeated Linear Factors | 示例：重复线性因式

Decompose (3x²-5x+2)/((x-1)(x-2)²). The denominator has a distinct factor (x-1) and a repeated factor (x-2)², so we set up:
(3x²-5x+2)/((x-1)(x-2)²) = A/(x-1) + B/(x-2) + C/(x-2)².
Multiplying through gives: 3x²-5x+2 = A(x-2)² + B(x-1)(x-2) + C(x-1).

分解 (3x²-5x+2)/((x-1)(x-2)²)。分母含有一个不同因式 (x-1) 和一个重复因式 (x-2)²，因此我们设定：
(3x²-5x+2)/((x-1)(x-2)²) = A/(x-1) + B/(x-2) + C/(x-2)²。
两边相乘得：3x²-5x+2 = A(x-2)² + B(x-1)(x-2) + C(x-1)。

onvenient value. For instance, expand the identity with A=0 and C=4: 3x²-5x+2 = B(x-1)(x-2) + 4(x-1). Substitute x = 0: 2 = B(2) + (-4) → 6 = 2B → B = 3. The result is 0/(x-1) + 3/(x-2) + 4/(x-2)², which simplifies to 3/(x-2) + 4/(x-2)².

代入 x = 2 求 C：3(4)-10+2 = C(1) → 4 = C。接着用 x = 1 求 A：3-5+2 = A(1) → 0 = A。然后通过比较系数或使用另一个方便的值求 B。例如，将 A=0 和 C=4 代入恒等式展开：3x²-5x+2 = B(x-1)(x-2) + 4(x-1)。令 x = 0：2 = B(2) + (-4) → 6 = 2B → B = 3。结果为 0/(x-1) + 3/(x-2) + 4/(x-2)²，可简化为 3/(x-2) + 4/(x-2)²。

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11. Applications: Integration using Partial Fractions | 应用：利用部分分式进行积分

One of the main reasons for learning partial fractions at A-Level is to integrate rational functions. After splitting a fraction, you typically obtain terms like A/(ax+b), which integrate to (A/a) ln|ax+b|. Terms like B/(ax+b)ⁿ integrate using the reverse chain rule or by rewriting as B(ax+b)⁻ⁿ.

A-Level 学习部分分式的主要原因之一就是为了对有理函数进行积分。将分式拆分后，通常会得到如 A/(ax+b) 的项，其积分为 (A/a) ln|ax+b|。而对于形如 B/(ax+b)ⁿ 的项，可以利用反向链式法则或改写为 B(ax+b)⁻ⁿ 进行积分。

(x+3)) dx. Using our previous decomposition, this becomes ∫ [-2/(x+1) + 7/(x+2) - 1/(x+3)]

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