Rational Functions and Partial Fractions | 有理函数与部分分式分解

📚 Rational Functions and Partial Fractions | 有理函数与部分分式分解

A rational function is a ratio of two polynomials. Understanding how to decompose a rational function into simpler partial fractions is a key algebraic skill that appears in calculus, integration, differential equations, and many examination problems. In this topic guide, we will explore the definition of rational functions, the conditions under which partial fraction decomposition is possible, the different cases involving linear and quadratic factors, and how to apply these techniques efficiently.

有理函数是两个多项式的比值。理解如何将一个有理函数分解为更简单的部分分式，是微积分、积分、微分方程以及众多考试问题中的关键代数技能。在本专题中，我们将探讨有理函数的定义、部分分式分解成立的条件、涉及线性因子和二次因子的不同情形，以及如何高效运用这些技巧。

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1. Definition of Rational Functions | 有理函数的定义

A rational function is expressed in the form f(x) = P(x)Q(x) , where P(x) and Q(x) are polynomials and Q(x) ≠ 0 . The domain of a rational function consists of all real numbers except those that make the denominator zero. The degree of a polynomial is the highest power of the variable that appears with a nonzero coefficient.

有理函数可表示为 f(x) = P(x)Q(x) 的形式，其中 P(x) 和 Q(x) 是多项式，且 Q(x) ≠ 0 。有理函数的定义域由所有使得分母不为零的实数组成。多项式的次数是变量出现的最高幂次，且该幂次对应的系数不为零。

Proper and improper rational functions: If the degree of P(x) is less than the degree of Q(x) , the rational function is called proper. If the degree of P(x) is greater than or equal to the degree of Q(x) , it is called improper. Before applying partial fraction decomposition, an improper rational function must first be simplified by polynomial long division.

真分式与假分式：如果 P(x) 的次数小于 Q(x) 的次数，则称该有理函数为真分式。如果 P(x) 的次数大于或等于 Q(x) 的次数，则称其为假分式。在应用部分分式分解之前，必须先通过多项式长除法对假分式进行化简。

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2. Why Partial Fractions Matter | 部分分式为何重要

Partial fraction decomposition rewrites a complicated rational expression as a sum of simpler rational expressions. This technique is especially useful when integrating rational functions, because each simpler fraction can often be integrated directly using basic formulas. It is also used in solving differential equations, in Laplace transforms, and in certain series expansions.

部分分式分解将一个复杂的有理表达式改写为若干个更简单的有理表达式之和。该技巧在积分有理函数时特别有用，因为每个更简单的分式通常可以直接使用基本公式积分。它还被用于求解微分方程、拉普拉斯变换以及某些级数展开中。

For example, the expression ^3x+5⁄_x²+x-2 can be decomposed as Ax+2 + Bx-1 . Once the unknown constants A and B are determined, integrating the original function becomes much more straightforward.

例如，表达式 ^3x+5⁄_x²+x-2 可以分解为 Ax+2 + Bx-1 。一旦确定了未知常数 A 和 B ，对原函数进行积分就会变得简单得多。

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3. Step 1: Ensure the Fraction Is Proper | 第一步：确保分式为真分式

If the degree of the numerator is greater than or equal to the degree of the denominator, perform polynomial long division first. The result will be a polynomial plus a proper rational function. Only the proper rational part needs to be decomposed into partial fractions.

如果分子的次数大于或等于分母的次数，首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。只有真分式部分才需要分解为部分分式。

Example: Simplify ^x3+2x⁄_x²-1 using long division.

示例：使用长除法化简 ^x3+2x⁄_x²-1 。

Dividing x³+2x by x²-1 , we get quotient x and remainder 3x . Therefore:

用 x²-1 除以 x³+2x ，得到商 x 和余数 3x 。因此：

^x3+2x⁄_x²-1 = x + ^3x⁄_x²-1

The remaining fraction is proper because the degree of 3x is 1, which is less than the degree of x²-1 , which is 2. We then decompose ^3x⁄_x²-1 further.

剩余的分式是真分式，因为 3x 的次数为 1，小于 x²-1 的次数 2。我们进一步分解 ^3x⁄_x²-1 。

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4. Step 2: Factor the Denominator Completely | 第二步：将分母完全因式分解

Partial fraction decomposition relies on the factors of the denominator. Factor the denominator into linear factors of the form (ax+b) and irreducible quadratic factors of the form (ax²+bx+c) . A quadratic factor is irreducible if it cannot be factored into real linear factors, meaning its discriminant b²-4ac < 0 .

部分分式分解依赖于分母的因子。将分母分解为形如 (ax+b) 的线性因子和形如 (ax²+bx+c) 的不可约二次因子。如果一个二次因子不能分解为实线性因子，即其判别式 b²-4ac < 0 ，则该二次因子是不可约的。

Example: Factor the denominator of ^2x+1⁄_x³-x .

示例：因式分解 ^2x+1⁄_x³-x 的分母。

We have x³-x = x(x²-1) = x(x-1)(x+1) . This gives three distinct linear factors. Each distinct linear factor will contribute one partial fraction term in the decomposition.

我们有 x³-x = x(x²-1) = x(x-1)(x+1) 。这给出了三个不同的线性因子。每个不同的线性因子将在分解中贡献一个部分分式项。

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5. Case 1: Distinct Linear Factors | 情形一：不同的线性因子

If the denominator factors into n distinct linear factors (a₁x+b₁)(a₂x+b₂)⋯(a_nx+b_n) , then the partial fraction decomposition has the form:

如果分母因式分解为 n 个不同的线性因子 (a₁x+b₁)(a₂x+b₂)⋯(a_nx+b_n) ，则部分分式分解的形式为：

P(x)Q(x) = ^A₁⁄_a1x+b1 + ^A₂⁄_a2x+b2 + ⋯ + ^A_n⁄_anx+bn

Each constant A_i can be found by multiplying both sides by the original denominator and then substituting convenient values of x , typically the roots of each linear factor.

每个常数 A_i 可以通过将等式两边乘以原分母，然后代入方便的 x 值来求出，通常代入每个线性因子的根。

Worked example: Decompose ^3x+5⁄_x²+x-2 .

例题：分解 ^3x+5⁄_x²+x-2 。

Factor the denominator: x²+x-2 = (x+2)(x-1) . Write:

将分母因式分解： x²+x-2 = (x+2)(x-1) 。写出：

3x+5(x+2)(x-1) = Ax+2 + Bx-1

Multiply through by (x+2)(x-1) :

两边乘以 (x+2)(x-1) ：

3x+5 = A(x-1) + B(x+2)

Let x = 1 : 3(1)+5 = B(1+2) ⟹ 8 = 3B ⟹ B = 83 .

令 x = 1 ： 3(1)+5 = B(1+2) ⟹ 8 = 3B ⟹ B = 83 。

Let x = -2 : 3(-2)+5 = A(-2-1) ⟹ -1 = -3A ⟹ A = 13 .

令 x = -2 ： 3(-2)+5 = A(-2-1) ⟹ -1 = -3A ⟹ A = 13 。

Thus:

因此：

^3x+5⁄_x²+x-2 = 1/3x+2 + 8/3x-1

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6. Case 2: Repeated Linear Factors | 情形二：重复的线性因子

If a linear factor (ax+b) appears with multiplicity k , then the partial fraction decomposition must include k terms for that factor:

如果线性因子 (ax+b) 以重数 k 出现，则部分分式分解必须包含该因子的 k 个项：

^P(x)⁄_(ax+b)^k = ^A₁⁄_ax+b + ^A₂⁄_(ax+b)² + ⋯ + ^A_k⁄_(ax+b)^k

Each term has a different constant numerator. This pattern is essential because the highest power term alone is not sufficient to represent all possible numerators when the denominator has a repeated factor.

每一项都有不同的常数分子。这种模式是必要的，因为当分母有重复因子时，仅最高次幂项不足以表示所有可能的分子。

Worked example: Decompose ^2x+3⁄_x²(x+1) .

例题：分解 ^2x+3⁄_x²(x+1) 。

The denominator has a repeated linear factor x² and a distinct factor (x+1) . Write:

分母有一个重复线性因子 x² 和一个不同因子 (x+1) 。写出：

^2x+3⁄_x²(x+1) = Ax + ^B⁄_x² + Cx+1

Multiply through by x²(x+1) :

两边乘以 x²(x+1) ：

2x+3 = A x(x+1) + B(x+1) + C x²

Let x = 0 : 3 = B(0+1) ⟹ B = 3 .

令 x = 0 ： 3 = B(0+1) ⟹ B = 3 。

Let x = -1 : 2(-1)+3 = C(1) ⟹ 1 = C .

令 x = -1 ： 2(-1)+3 = C(1) ⟹ 1 = C 。

To find A , compare coefficients of x² : on the left, the coefficient is 0; on the right, the coefficient is A + C . Thus 0 = A + 1 ⟹ A = -1 .

为求 A ，比较 x² 的系数：左边系数为 0；右边系数为 A + C 。因此 0 = A + 1 ⟹ A = -1 。

Therefore:

因此：

^2x+3⁄_x²(x+1) = -1x + ³⁄_x² + 1x+1

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7. Case 3: Irreducible Quadratic Factors | 情形三：不可约二次因子

If the denominator contains an irreducible quadratic factor ax²+bx+c , then the corresponding partial fraction term has a linear numerator:

如果分母包含不可约二次因子 ax²+bx+c ，则对应的部分分式项具有线性分子：

^P(x)⁄_(ax²+bx+c) = ^Ax+B⁄_ax²+bx+c

This is because a constant numerator cannot account for all possible expressions over a quadratic denominator; a linear numerator provides the necessary flexibility.

这是因为常数分子无法涵盖二次分母上的所有可能表达式；线性分子提供了必要的灵活性。

Worked example: Decompose ^x+1⁄_x(x²+1) .

例题：分解 ^x+1⁄_x(x²+1) 。

Here x is a linear factor and x²+1 is an irreducible quadratic. Write:

这里 x 是线性因子， x²+1 是不可约二次因子。写出：

^x+1⁄_x(x²+1) = Ax + ^Bx+C⁄_x²+1

Multiply through by x(x²+1) :

两边乘以 x(x²+1) ：

x+1 = A(x²+1) + (Bx+C)x

Expanding: x+1 = Ax² + A + Bx² + Cx . Group terms:

展开： x+1 = Ax² + A + Bx² + Cx 。合并同类项：

x+1 = (A+B)x² + Cx + A

Compare coefficients:

比较系数：

• x² coefficient: 0 = A+B
• x coefficient: 1 = C
• Constant term: 1 = A

Thus A = 1 , B = -1 , C = 1 . Therefore:

因此 A = 1 ， B = -1 ， C = 1 。因此：

^x+1⁄_x(x²+1) = 1x + ^-x+1⁄_x²+1

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8. Case 4: Repeated Quadratic Factors | 情形四：重复的二次因子

If an irreducible quadratic factor (ax²+bx+c) appears with multiplicity k , then the decomposition includes k terms:

如果不可约二次因子 (ax²+bx+c) 以重数 k 出现，则分解包括 k 个项：

^P(x)⁄_{(ax²+bx+c)^k} = ^A₁x+B₁⁄_ax²+bx+c + ^A₂x+B₂⁄_{(ax²+bx+c)²} + ⋯ + ^A_kx+B_k⁄_{(ax²+bx+c)^k}

Each term has a linear numerator with distinct constants. This is the most general case and can be combined with any of the previous cases when the denominator has multiple factor types.

每一项都有带有不同常数的线性分子。这是最一般的情形，当分母有多种因子类型时，可以与前面任何一种情形结合使用。

Worked example: Decompose ^2x⁄_(x²+1)² .

例题：分解 ^2x⁄_(x²+1)² 。

Write the decomposition form:

写出分解形式：

^2x⁄_(x²+1)² = ^Ax+B⁄_x²+1 + ^Cx+D⁄_(x²+1)²

Multiply through by (x²+1)² :

两边乘以 (x²+1)² ：

2x = (Ax+B)(x²+1) + Cx + D

Expand: 2x = Ax³ + Ax + Bx² + B + Cx + D . Group:

展开： 2x = Ax³ + Ax + Bx² + B + Cx + D 。合并：

2x = Ax³ + Bx² + (A+C)x + (B+D)

Compare coefficients:

比较系数：

• x³ : 0 = A
• x² : 0 = B
• x : 2 = A+C ⟹ C=2
• constant: 0 = B+D ⟹ D=0

Thus ^2x⁄_(x²+1)² = ^2x⁄_(x²+1)² , which shows that sometimes the original form is already a valid partial fraction. This example illustrates that the general form always works, even if some terms have zero coefficients.

因此 ^2x⁄_(x²+1)² = ^2x⁄_(x²+1)² ，这表明有时原始形式已经是有效的部分分式。这个例子说明一般形式总是有效的，即使某些项的系数为零。

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9. The Cover-Up Method | 遮盖法

The cover-up method is a fast way to find the constant corresponding to a single distinct linear factor. To find A_i for the factor (a_ix+b_i) , cover up that factor in the original denominator and evaluate the remaining expression at the root of that factor.

遮盖法是一种快速求出单个不同线性因子对应常数的方法。要求因子 (a_ix+b_i) 的 A_i ，在原分母中遮盖住该因子，然后在该因子的根处评估剩余的表达式。

Example: For 3x+5(x+2)(x-1) , to find A for x+2 , cover (x+2) and substitute x=-2 into 3x+5x-1 :

示例：对于 3x+5(x+2)(x-1) ，要求 x+2 的 A ，遮盖 (x+2) 并将 x=-2 代入 3x+5x-1 ：

A = 3(-2)+5-2-1 = -1-3 = 13

Similarly, to find B for x-1 , cover (x-1) and substitute x=1 :

类似地，要求 x-1 的 B ，遮盖 (x-1) 并代入 x=1 ：

B = 3(1)+51+2 = 83

This method works only for distinct linear factors. For repeated factors or quadratic factors, use coefficient comparison or substitution strategies.

该方法仅适用于不同的线性因子。对于重复因子或二次因子，需要使用系数比较或代入策略。

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10. Solving for Constants: Substitution and Coefficient Comparison | 求常数：代入法与系数比较法

Two main techniques are used to find the unknown constants after setting up the partial fraction form:

在建立部分分式形式后，求未知常数主要使用两种技巧：

• Substitution: Choose values of x that make each factor zero. This eliminates terms and directly solves for one constant at a time. This works best for distinct linear factors and often for simple cases.
• Coefficient comparison: Expand both sides and compare coefficients of the same powers of x . This creates a system of linear equations that can be solved simultaneously. This method is reliable for all cases, especially when substitution is difficult.

• 代入法：选择使每个因子为零的 x 值。这可以消去项并直接求解一个常数。该方法最适合不同的线性因子，通常也适用于简单情形。
• 系数比较法：展开等式两边，比较同次幂 x 的系数。这会形成线性方程组，可以联立求解。该方法对所有情形都可靠，尤其是在代入法较困难时。

In many questions, a combination of both methods is the most efficient strategy. For instance, use substitution for any distinct linear factors, then compare coefficients for the remaining constants.

在许多问题中，两种方法结合使用是最有效的策略。例如，对不同的线性因子使用代入法，然后对剩余常数使用系数比较法。

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11. Common Mistakes and How to Avoid Them | 常见错误及避免方法

Students often make the following errors when working with partial fractions:

学生在处理部分分式时常犯以下错误：

Mistake | 错误	Correction | 纠正
Forgetting long division on improper fractions	Always divide first if deg(P) ≥ deg(Q); then decompose the remainder fraction.
Forgetting the linear numerator Ax+B for quadratic factors	Use Ax+B , not just A , for irreducible quadratic factors.
Missing repeated factor terms	For (ax+b)k , include all terms from power 1 to power k.
Incorrect sign handling in substitution	Check signs carefully, especially when substituting negative roots.
Arithmetic errors when comparing coefficients	Expand carefully and align terms by power of x before comparing.

错误	纠正
忽略假分式的长除法	若 deg(P) ≥ deg(Q)，始终先除法，然后分解余数分式。
对二次因子忘记线性分子 Ax+B	对不可约二次因子使用 Ax+B ，而不仅仅是 A 。
遗漏重复因子项	对于 (ax+b)k ，包含从 1 次到 k 次的所有项。
代入时符号处理错误	代入负根时仔细检查符号。
比较系数时算术错误	先展开并按照 x 的幂次对齐项，再进行比较。

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12. Summary and Examination Tips | 总结与考试要点

Partial fraction decomposition follows a consistent process: check whether the fraction is proper, factor the denominator completely, write the correct decomposition form based on the factor types, and solve for the constants using substitution or coefficient comparison. The key to success is recognising which case applies and writing the correct form before solving.

部分分式分解遵循一致的过程：检查分式是否为真分式、将分母完全因式分解、根据因子类型写出正确的分解形式，再使用代入法或系数比较法求常数。成功的关键是识别适用的情形，并在求解前写出正确的形式。

In an exam setting, always verify your answer by combining the partial fractions back over a common denominator. This quick check can prevent careless errors from costing marks. Practising a variety of examples with mixed factor types is the best way to build speed and confidence.

在考试中，始终通过将部分分式重新合并到公分母上来验证答案。这个快速检查可以防止粗心错误丢分。练习包含混合因子类型的各种例题，是提高速度和信心的最佳方法。

Remember the golden rules:

记住黄金法则：

• Always long divide improper fractions first.
• Write one term for each distinct linear factor.
• Write k terms for each repeated factor of multiplicity k.
• Use a linear numerator for every irreducible quadratic factor.
• Check your result by summing the partial fractions.

• 假分式始终先进行长除法。
• 每个不同的线性因子写一个项。
• 每个重数为 k 的重复因子写 k 个项。
• 每个不可约二次因子使用线性分子。
• 通过将部分分式相加来检查结果。

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