Partial Fractions | 部分分式

📚 Partial Fractions | 部分分式

In A-Level Pure Mathematics, partial fractions is a technique used to express a single rational function as a sum of simpler fractions. This simplifies integration, binomial expansion, and solving differential equations.

在A-Level纯数学中，部分分式是一种将单个有理函数表示成几个简单分式之和的方法，可以简化积分、二项式展开以及微分方程的求解。

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

A rational function is a fraction where both numerator and denominator are polynomials. Partial fractions decomposition rewrites one complicated fraction as a sum of two or more ‘partial’ fractions with simpler denominators.

有理函数是分子和分母都是多项式的分式。部分分式分解将一个复杂的分式改写为两个或更多分母更简单的“部分”分式的和。

For example, (2x+1)/[(x+1)(x-3)] can be expressed as A/(x+1) + B/(x-3) where A and B are constants to be found.

例如，(2x+1)/[(x+1)(x-3)] 可以表示为 A/(x+1) + B/(x-3)，其中 A 和 B 是待确定的常数。

2. The Main Idea: Decomposition | 核心思想：分解

The denominators in the sum are the factors of the original denominator. We assume unknown numerators (constants for linear factors, linear expressions for quadratic factors) and then determine them by clearing denominators and comparing coefficients or substituting suitable values of x.

求和中的分母是原分母的因式。我们假设未知的分子（对线性因式是常数，对二次因式是一次表达式），然后通过去分母并比较系数或代入合适的 x 值来确定它们。

We only decompose proper fractions (degree of numerator less than denominator). If improper, we must first perform polynomial long division.

我们只分解真分式（分子次数低于分母）。如果是假分式，必须先进行多项式长除法。

3. Linear Factors (Distinct) | 不同线性因式

If the denominator factorises into distinct linear factors (ax+b), each factor gives a term of the form A/(ax+b).

如果分母可以分解为不同的一次因式 (ax+b)，每个因式都会产生一项 A/(ax+b)。

General form: P(x)/[(a₁x+b₁)(a₂x+b₂)…] = A₁/(a₁x+b₁) + A₂/(a₂x+b₂) + …

一般形式：P(x)/[(a₁x+b₁)(a₂x+b₂)…] = A₁/(a₁x+b₁) + A₂/(a₂x+b₂) + …

Example decomposition: 5/[(x-1)(x+2)] = A/(x-1) + B/(x+2)

分解示例：5/[(x-1)(x+2)] = A/(x-1) + B/(x+2)

4. Worked Example: Distinct Linear Factors | 例题：不同线性因式

Decompose (7x-3)/[(x-2)(x+1)] into partial fractions.

将 (7x-3)/[(x-2)(x+1)] 分解为部分分式。

Step 1: Write (7x-3)/[(x-2)(x+1)] = A/(x-2) + B/(x+1). Multiply through by (x-2)(x+1): 7x-3 = A(x+1) + B(x-2).

步骤1：写出 (7x-3)/[(x-2)(x+1)] = A/(x-2) + B/(x+1)。整式乘以 (x-2)(x+1)：7x-3 = A(x+1) + B(x-2)。

Step 2: Substitute convenient x-values. Let x=2: 14-3 = A(3) ⇒ 11 = 3A ⇒ A = 11/3. Let x=-1: -7-3 = B(-3) ⇒ -10 = -3B ⇒ B = 10/3.

步骤2：代入方便的 x 值。令 x=2：14-3 = A(3) ⇒ 11 = 3A ⇒ A = 11/3。令 x=-1：-7-3 = B(-3) ⇒ -10 = -3B ⇒ B = 10/3。

Thus, (7x-3)/[(x-2)(x+1)] = (11/3)/(x-2) + (10/3)/(x+1).

因此，(7x-3)/[(x-2)(x+1)] = (11/3)/(x-2) + (10/3)/(x+1)。

Check: combine the right side over common denominator to verify it gives back (7x-3)/[(x-2)(x+1)].

检验：将右式通分合并，验证是否回到 (7x-3)/[(x-2)(x+1)]。

5. Repeated Linear Factors | 重复线性因式

When the denominator contains a repeated factor like (ax+b)ⁿ, we include terms for each power from 1 to n: A₁/(ax+b) + A₂/(ax+b)² + … + Aₙ/(ax+b)ⁿ.

当分母包含重复因式如 (ax+b)ⁿ 时，需要包含从1次幂到n次幂的每一项：A₁/(ax+b) + A₂/(ax+b)² + … + Aₙ/(ax+b)ⁿ。

This ensures all possible degrees of denominator are represented during decomposition, which is crucial for correct integration.

这确保了分解过程中包含了分母的所有可能次数，这对正确积分至关重要。

Example: (x+3)/(x-1)² = A/(x-1) + B/(x-1)²

示例：(x+3)/(x-1)² = A/(x-1) + B/(x-1)²

6. Worked Example: Repeated Linear Factors | 例题：重复线性因式

Decompose (4x+1)/(x+2)² into partial fractions.

将 (4x+1)/(x+2)² 分解为部分分式。

Write (4x+1)/(x+2)² = A/(x+2) + B/(x+2)². Multiply through by (x+2)²: 4x+1 = A(x+2) + B.

写出 (4x+1)/(x+2)² = A/(x+2) + B/(x+2)²。整式乘以 (x+2)²：4x+1 = A(x+2) + B。

Expand: 4x+1 = Ax + 2A + B. Compare coefficients: for x: 4 = A; for constants: 1 = 2A + B. Substitute A=4: 1 = 8 + B ⇒ B = -7.

展开：4x+1 = Ax + 2A + B。比较系数：x项：4 = A；常数项：1 = 2A + B。代入 A=4：1 = 8 + B ⇒ B = -7。

Therefore, (4x+1)/(x+2)² = 4/(x+2) – 7/(x+2)².

因此，(4x+1)/(x+2)² = 4/(x+2) – 7/(x+2)²。

7. Quadratic Factors (Irreducible) | 不可约二次因式

If the denominator includes an irreducible quadratic factor ax²+bx+c (discriminant Δ < 0), its partial fraction has a linear numerator: (Ax+B)/(ax²+bx+c).

如果分母包含一个不可约的二次因式 ax²+bx+c（判别式 Δ < 0），其部分分式的分子为一次式：(Ax+B)/(ax²+bx+c)。

This accounts for the fact that the numerator must have degree one less than the factor. We solve for A and B using the same methods.

这是因为分子次数必须比因式次数低一次。我们使用相同的方法求解 A 和 B。

Example: (2x+1)/[(x²+1)(x-2)] = (Ax+B)/(x²+1) + C/(x-2)

示例：(2x+1)/[(x²+1)(x-2)] = (Ax+B)/(x²+1) + C/(x-2)

8. Worked Example: Quadratic Factor | 例题：二次因式

Express (3x²-2x+5)/[(x²+1)(x-3)] in partial fractions.

将 (3x²-2x+5)/[(x²+1)(x-3)] 表示为部分分式。

Set up: (3x²-2x+5)/[(x²+1)(x-3)] = (Ax+B)/(x²+1) + C/(x-3). Multiply by denominator: 3x²-2x+5 = (Ax+B)(x-3) + C(x²+1).

设：(3x²-2x+5)/[(x²+1)(x-3)] = (Ax+B)/(x²+1) + C/(x-3)。乘以分母：3x²-2x+5 = (Ax+B)(x-3) + C(x²+1)。

Let x=3: 27 – 6 + 5 = C(9+1) ⇒ 26 = 10C ⇒ C = 2.6 or 13/5.

令 x=3：27 – 6 + 5 = C(9+1) ⇒ 26 = 10C ⇒ C = 2.6 即 13/5。

Expand remaining: (Ax+B)(x-3) = Ax² – 3Ax + Bx – 3B. Total: (A+C)x² + (-3A+B)x + (-3B+C) = 3x² -2x +5. Compare coefficients: A+C = 3 ⇒ A=3 – 13/5 = 2/5. -3A+B = -2 ⇒ -3(2/5)+B = -2 ⇒ B = -2 + 6/5 = -4/5. Constant check: -3(-4/5)+13/5 = 12/5+13/5=25/5=5, matches.

展开剩余部分：(Ax+B)(x-3) = Ax² – 3Ax + Bx – 3B。总计：(A+C)x² + (-3A+B)x + (-3B+C) = 3x² -2x +5。比较系数：A+C = 3 ⇒ A=3 – 13/5 = 2/5。-3A+B = -2 ⇒ -3(2/5)+B = -2 ⇒ B = -2 + 6/5 = -4/5。常数项验证：-3(-4/5)+13/5 = 12/5+13/5=25/5=5，一致。

Thus: (2x/5 – 4/5)/(x²+1) + (13/5)/(x-3).

因此：(2x/5 – 4/5)/(x²+1) + (13/5)/(x-3)。

9. Improper Fractions & Long Division | 假分式与长除法

A rational expression is improper if the degree of the numerator is greater than or equal to the degree of the denominator. In such cases, we first perform polynomial long division to obtain a polynomial plus a proper fraction.

如果分子的次数大于或等于分母的次数，该有理式就是假分式。在这种情况下，我们先使用多项式长除法得到一个多项式加上一个真分式。

The proper fractional part can then be decomposed using partial fractions, while the polynomial part remains outside.

然后真分式部分可以用部分分式分解，而多项式部分则保持在外。

Improper example: (x³+2x)/(x²-1) → divide to get x + (3x)/(x²-1)

假分式示例：(x³+2x)/(x²-1) → 长除后得到 x + (3x)/(x²-1)

10. Worked Example: Improper Fractions | 例题：假分式

Decompose (2x³+3x²-4x+1)/(x²-4) into polynomial + partial fractions.

将 (2x³+3x²-4x+1)/(x²-4) 分解为多项式加上部分分式。

Divide: (2x³+3x²-4x+1) ÷ (x²-4). Leading term: 2x³/x² = 2x. Multiply divisor by 2x: 2x³ – 8x. Subtract: (2x³+3x²-4x+1) – (2x³-8x) = 3x² +4x +1. Next term: 3x²/x² = 3. Multiply: 3x² -12. Subtract: (3x²+4x+1) – (3x²-12) = 4x+13. Quotient = 2x+3, remainder = 4x+13.

长除：(2x³+3x²-4x+1) ÷ (x²-4)。首项：2x³/x² = 2x。乘以除式：2x³ – 8x。相减：(2x³+3x²-4x+1) – (2x³-8x) = 3x² +4x +1。下一项：3x²/x² = 3。相乘：3x² -12。相减：(3x²+4x+1) – (3x²-12) = 4x+13。商 = 2x+3，余式 = 4x+13。

So (2x³+3x²-4x+1)/(x²-4) = 2x+3 + (4x+13)/(x²-4). Now decompose (4x+13)/[(x-2)(x+2)] = A/(x-2) + B/(x+2). X=2 ⇒ A=21/4; X=-2 ⇒ B=-5/4. Final answer: 2x+3 + (21/4)/(x-2) – (5/4)/(x+2).

因此 (2x³+3x²-4x+1)/(x²-4) = 2x+3 + (4x+13)/(x²-4)。现在分解 (4x+13)/[(x-2)(x+2)] = A/(x-2) + B/(x+2)。x=2 ⇒ A=21/4；x=-2 ⇒ B=-5/4。最终结果：2x+3 + (21/4)/(x-2) – (5/4)/(x+2)。

11. Tips for Solving and Checking | 求解与检验技巧

Always start by factorising the denominator completely. Decide the correct form based on factor types before assigning constants.

始终先将分母完全因式分解。基于因式类型确定正确形式，再设定常数。

Using the ‘substitution method’ with roots of linear factors often gives constants quickly; for repeated or quadratic factors, compare coefficients or combine substitution and coefficient matching.

使用代入法（代入线性因式的根）通常能快速求出常数；对于重复因式或二次因式，则比较系数或将代入法与系数法结合。

A useful check: recombine your partial fractions over a common denominator and verify the numerator matches the original. This practice helps catch sign errors and misassigned constants.

一个有用的检验方法：将你的部分分式通分合并，验证分子是否与原式一致。这种做法有助于发现符号错误和常数赋值错误。

For exams, remember that partial fractions lead directly to simpler integrations, especially with ln|x±a| and arctan forms.

对考试而言，记住部分分式能直接导向更简单的积分，尤其是 ln|x±a| 和 arctan 形式。

12. Summary & Exam Advice | 总结与考试建议

Partial fractions is a systematic method: factorise determinant form, set up decomposition, multiply through, solve for constants. Mastery comes from recognising patterns — distinct linear, repeated linear, and irreducible quadratic factors. Improper fractions require long division first.

部分分式是一种系统的方法：因式分解分母，建立分解式，整式相乘，求解常数。熟练的掌握源于识别模式——不同线性因式、重复线性因式和不可约二次因式。假分式需要先进行长除法。

Common exam tasks ask you to express a rational function in partial fractions and then integrate it. Practice with a variety of examples to ensure you can handle any combination of factors. Always present final answers in their simplest algebraic form.

常见的考试任务通常是要求将一个有理函数表示为部分分式，然后对其积分。通过大量例题练习，确保你能处理任意因式组合。最终答案始终要用最简代数形式呈现。

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Partial Fractions | 部分分式