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Partial Fraction Decomposition: A Complete Method Guide | 部分分式分解方法精讲

2026-08-30 18:51 · 本文部分内容由AI辅助创作

📚 Partial Fraction Decomposition: A Complete Method Guide | 部分分式分解方法精讲

Partial fraction decomposition is a technique that rewrites a single complicated rational expression as a sum of simpler rational expressions. It is an essential tool in A-Level Mathematics, particularly for integration, binomial expansion and solving differential equations.

部分分式分解是一种将一个复杂的有理分式改写为若干个更简单的有理分式之和的方法。它是A-Level数学中的重要工具,尤其在积分、二项式展开和求解微分方程中经常使用。

1. What Is Partial Fraction Decomposition? | 什么是部分分式分解

A rational expression is of the form P(x)/Q(x), where P(x) and Q(x) are polynomials. Partial fractions reverse the process of adding fractions: instead of combining terms over a common denominator, we split the denominator into factors and assign numerators.

有理分式具有 P(x)/Q(x) 的形式,其中 P(x) 和 Q(x) 都是多项式。部分分式分解是做加法的逆过程:我们不是合并分母,而是将分母分解成因式,并为每个因式设置对应的分子。

For example, the expression 1/(x-1) - 1/(x+2) combines to 3/((x-1)(x+2)). Partial fractions decompose the combined form back into the two simpler pieces.

例如,1/(x-1) - 1/(x+2) 合并后得到 3/((x-1)(x+2))。部分分式分解就是把这个合并后的形式还原成两个更简单的分式。


2. Proper and Improper Rational Expressions | 真分式与假分式

A rational expression is proper when the degree of the numerator is less than the degree of the denominator. For example, (3x+1)/(x²-1) is proper because the numerator has degree 1 and the denominator has degree 2.

当分子的次数小于分母的次数时,这个有理分式称为真分式。例如,(3x+1)/(x²-1) 是真分式,因为分子次数为1,小于分母次数2。

If the degree of the numerator is greater than or equal to the degree of the denominator, the expression is improper. We must first divide the numerator by the denominator before applying the standard partial fraction process.

如果分子的次数大于或等于分母的次数,则该分式称为假分式。在使用标准的部分分式分解步骤之前,我们必须先用分子除以分母,即做多项式除法。


3. Distinct Linear Factors | 不同线性因子

When the denominator factorises into different linear factors, each factor produces one partial fraction with a constant numerator.

当分母分解为互不相同的线性因子时,每一个因子对应一个分子为常数的部分分式。

For Q(x) = (x-a)(x-b), set:

若 Q(x) = (x-a)(x-b),设:

P(x)/((x-a)(x-b)) = A/(x-a) + B/(x-b)

ator to obtain P(x) = A(x-b) + B(x-a). Then substitute x = a and x = b to isolate A and B.

其中 A 和 B 为待定常数。两边同时乘以分母,得到 P(x) = A(x-b) + B(x-a)。然后分别代入 x = a 和 x = b,就可以求出 A 和 B。

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

gives 5x-1 = A(x+1) + B(x-2). Let x = 2: 9 = 3A, so A = 3. Let x = -1: -6 = -3B, so B = 2.

例子:(5x-1)/((x-2)(x+1)) = A/(x-2) + B/(x+1)。两边乘分母得 5x-1 = A(x+1) + B(x-2)。令 x = 2:9 = 3A,所以 A = 3;令 x = -1:-6 = -3B,所以 B = 2。


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4. Repeated Linear Factors | 重线性因子

If the denominator contains a repeated factor such as (x-a)², the decomposition must include terms for both (x-a) and (x-a)².

如果分母含有重复因子,例如 (x-a)²,则分解式中必须同时包含 (x-a) 和 (x-a)² 对应的两项。

For Q(x) = (x-a)²(x-b), write:

若 Q(x) = (x-a)²(x-b),设:

P(x)/((x-a)²(x-b)) = A/(x-a) + B/(x-a)² + C/(x-b)

The first two terms handle the repeated factor; the third handles the distinct factor. Multiply by (x-a)²(x-b) and use substitution or compare coefficients.

前两项处理重复因子,第三项处理非重复因子。两边乘以 (x-a)²(x-b),再用代入法或比较系数法求解。

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

. Substituting x = 1 gives C = 5. Comparing coefficients of x² gives 0 = A + B, so B = -2.

两边乘分母后,3x+2 = A(x-1)² + Bx(x-1) + Cx。代入 x = 0 得 A = 2;代入 x = 1 得 C = 5;比较 x² 的系数得 0 = A + B,因此 B = -2。


Published by TutorHao | Mathematics Revision Series | aleveler.com

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