📚 Partial Fractions | 部分分式
Partial fractions form a key algebraic technique in A-level Mathematics, allowing a rational expression to be expressed as a sum of simpler fractions. Mastering this skill is vital, as it underpins integration of rational functions, facilitates binomial expansions of algebraic fractions, and appears frequently in both Pure and Applied mathematics modules across the Edexcel specification.
部分分式是A-level数学中一项关键的代数技巧，能将一个有理表达式表示为几个更简单分式的和。掌握这一技能至关重要，因为它为有理函数的积分奠定基础，有助于代数分式的二项式展开，并且在Edexcel考纲的纯数学与应用数学模块中频繁出现。
1. Understanding Rational Expressions and Partial Fractions | 理解有理表达式与部分分式
A rational expression is simply a fraction in which both the numerator and the denominator are polynomials. If the degree of the numerator is less than the degree of the denominator, the fraction is called proper; otherwise it is improper. The technique of partial fractions focuses on decomposing a proper rational expression into a sum of simpler component fractions, usually with linear or quadratic denominators that are easier to integrate or expand.
有理表达式就是分子和分母均为多项式的分式。如果分子的次数低于分母的次数，这个分式称为真分式；否则为假分式。部分分式技巧的核心是将一个真有理表达式分解成更简单的组成分式之和，这些分式的分母通常为线性或二次式，便于后续积分或展开。
For an expression like (px+q)/(ax+b)(cx+d), the goal is to write it in the form A/(ax+b) + B/(cx+d), where A and B are constants to be found. The same principle extends to repeated linear factors and irreducible quadratic factors, which we will explore step by step.
对于类似 (px+q)/(ax+b)(cx+d) 的表达式，我们的目标是将其写成 A/(ax+b) + B/(cx+d) 的形式，其中 A 和 B 是待定常数。同样的原理也适用于重复线性因式和不可约二次因式，我们将逐步展开探讨。
2. Denominator with Distinct Linear Factors | 分母为不同线性因式的情况
The simplest case arises when the denominator can be factorised into distinct linear factors, such as (x+1)(x-3). For a proper fraction with this denominator, we can write a general decomposition with one constant numerator over each linear factor. The method then requires us to determine the unknown constants by either substituting suitable x-values or by equating coefficients of like terms.
最简单的情况出现在分母可以分解为不同的线性因式时，例如 (x+1)(x-3)。对于以此类分母构成的真分式，我们可以写出一般分解式，在每个线性因式上放置一个常数分子。然后我们需要通过代入适当的x值或比较同类项系数来求出这些未知常数。
For instance, to decompose 5x+1/(x-1)(x+2), we assume:
5x+1/(x-1)(x+2) ≡ A/(x-1) + B/(x+2)
Multiplying through by the common denominator (x-1)(x+2) gives 5x+1 ≡ A(x+2) + B(x-1) 1 eliminates B, leaving A = 2. Choosing x = -2 gives B = 3. The decomposition is therefore 2/(x-1) + 3/(x+2).
例如，要分解 5x+1/(x-1)(x+2)，我们假设：
5x+1/(x-1)(x+2) ≡ A/(x-1) + B/(x+2)
将等式两边同乘以公分母 (x-1)(x+2) 得到 5x+1 ≡ A(x+2) + B(x-1)。选取 x = 1 可消去 B，解出 A = 2。选取 x = -2 可得 B = 3。因此分解结果为 2/(x-1) + 3/(x+2)。
This substitution method is efficient because selecting the roots of each linear factor makes one of the unknowns disappear instantly. It is often quicker than equating coefficients, although equating coefficients remains a reliable alternative when the substitution yields awkward values.
这种代入法很高效，因为选取每个线性因式的根可以让其中一个未知数瞬间消失。尽管当代入值导致计算不便时，比较系数法仍是一个可靠的备用方案，但代入法通常更快捷。
3. Worked Example – Distinct Linear Factors | 实例演练 – 不同线性因式
Consider the decomposition of (7x-4)/(2x-1)(x+3). We set:
(7x-4)/(2x-1)(x+3) ≡ A/(2x-1) + B/(x+3)
Multiplying by (2x-1)(x+3) yields 7x-4 ≡ A(x+3) + B(2x-1) itute x = 1/2 to eliminate B: 7(1/2)-4 = -1/2 = A(1/2+3) = (7/2)A, giving A = -1/7. Hence:
(7x-4)/(2x-1)(x+3) ≡ -1/(7(2x-1)) + 25/(7(x+3))
We can simplify constants if required, but leaving the solution in this form is perfectly acceptable in Edexcel exams.
考虑分解 (7x-4)/(2x-1)(x+3)。我们设：
(7x-4)/(2x-1)(x+3) ≡ A/(2x-1) + B/(x+3)
两边乘 (2x-1)(x+3) 得 7x-4 ≡ A(x+3) + B(2x-1)。代入 x = -3 消去 A：7(-3)-4 = -25 = B(2(-3)-1) = -7B，故 B = 25/7。代入 x = 1/2 消去 B：7(1/2)-4 = -1/2 = A(1/2+3) = (7/2)A，得 A = -1/7。因此：
(7x-4)/(2x-1)(x+3) ≡ -1/(7(2x-1)) + 25/(7(x+3))
如有需要可以合并常数，但在Edexcel考试中保留这种形式完全可行。
4. The Cover-up Method | 遮盖法
The cover-up method is a brilliant shortcut for distinct linear factors. To find the numerator A over a factor (ax+b), you "cover" the factor (ax+b) in the denominator of the original expression and substitute the x-value that makes that factor zero into the remaining expression. This directly gives the constant A, without needing to write a full identity.
遮盖法是针对不同线性因式的一种巧妙快捷方法。要找出因式 (ax+b) 上的分子 A，你可以“遮盖”原分母中的因式 (ax+b)，然后将使该因式为零的 x 值代入剩下的表达式中。这样可以直接得到常数 A，无需写出完整的恒等式。
For example, decompose (4x+5)/(x-2)(x+1). To find A in A/(x-2), cover (x-2) and substitute x = 2 into (4x+5)/(x+1), giving (8+5)/(3) = 13/3, so A = 13/3. For B, cover (x+1) and substitute x = -1 into (4x+5)/(x-2), giving (-4+5)/(-3) = 1/(-3) = -1/3, so B = -1/3. The decomposition is 13/3(x-2) - 1/3(x+1).
例如，分解 (4x+5)/(x-2)(x+1)。要找出 A/(x-2) 中的 A，遮盖 (x-2) 并将 x = 2 代入 (4x+5)/(x+1)，得到 (8+5)/3 = 13/3，故 A = 13/3。对于 B，遮盖 (x+1) 并将 x = -1 代入 (4x+5)/(x-2)，得 (-4+5)/(-3) = -1/3，故 B = -1/3。分解结果是 13/3(x-2) - 1/3(x+1)。
The cover-up rule works only for distinct linear factors in the denominator, and the original fraction must be proper. It is highly recommended to verify the result by combining the partial fractions back, especially under exam pressure.
遮盖法仅适用于分母中的不同线性因式，且原分式必须为真分式。强烈建议在考试压力下通过重新合并部分分式来验算结果。
5. Denominator with a Repeated Linear Factor | 分母包含重线性因式
When a linear factor appears squared (or raised to a higher power), the decomposition must include all descending powers of that factor. If the denominator contains (ax+b)², the partial fractions take the form A/(ax+b) + B/(ax+b)². For a cube, we would have three terms, and so on.
当一个线性因式以平方（或更高次幂）出现时，分解式必须包含该因式的所有降幂形式。如果分母包含 (ax+b)²，部分分式将采取 A/(ax+b) + B/(ax+b)² 的形式。对于三次方，则需要三项，依此类推。
The presence of the repeated factor means we can still use substitution for the highest power constant, but we will generally need a second substitution or equate coefficients to find the remaining constants, because covering only gives the numerator for the highest power term.
重复因式的存在意味着我们仍然可以对最高次幂常数使用代入法，但通常需要第二次代入或比较系数来求出剩余的常数，因为遮盖法只能给出最高次幂项的分子。
For example, to decompose (2x+3)/(x-1)², we write A/(x-1) + B/(x-1)². Multiplying by (x-1)² gives 2x+3 ≡ A(x-1) + B es A: B = 2(1)+3 = 5. Choosing another value, say x = 0, gives 3 = -A + 5, so A = 2. Hence 2/(x-1) + 5/(x-1)².
例如，分解 (2x+3)/(x-1)²，我们写成 A/(x-1) + B/(x-1)²。乘以 (x-1)² 得 2x+3 ≡ A(x-1) + B。代入 x = 1 消去 A：B = 2(1)+3 = 5。再取另一个值，如 x = 0，得到 3 = -A + 5，故 A = 2。因此分解结果为 2/(x-1) + 5/(x-1)²。
6. Worked Example – Repeated Linear Factor | 实例演练 – 重线性因式
Decompose (3x²+5x+2)/(x+2)²(x-1). The denominator has a repeated factor (x+2)² and a distinct factor (x-1). The form is:
A/(x+2) + B/(x+2)² + C/(x-1)
Multiply through: 3x²+5x+2 ≡ A(x+2)(x-1) + B(x-1) + C(x+2)² = A(2)(-1) + (-4/3)(-1) + (10/9)(4) => 2 = -2A + 4/3 + 40/9. Combine: 4/3 + 40/9 = 12/9 + 40/9 = 52/9. So 2 = -2A + 52/9 => -2A = 2 - 52/9 = (18-52)/9 = -34/9 => A = 17/9. Thus decomposition is 17/9(x+2) - 4/3(x+2)² + 10/9(x-1).
分解 (3x²+5x+2)/(x+2)²(x-1)。分母含有一个重因式 (x+2)² 和一个不同因式 (x-1)。形式为：
A/(x+2) + B/(x+2)² + C/(x-1)
通分后乘开：3x²+5x+2 ≡ A(x+2)(x-1) + B(x-1) + C(x+2)²。代入 x = -2 求 B：3(4)-10+2=4，右边 B(-3)，得 B = -4/3。代入 x = 1 求 C：10 = C(3)² = 9C，得 C = 10/9。最后代入 x = 0 求 A：2 = A(2)(-1) + (-4/3)(-1) + (10/9)(4) => 2 = -2A + 4/3 + 40/9。合并 4/3 + 40/9 = 12/9 + 40/9 = 52/9，所以 2 = -2A + 52/9 => -2A = 2 - 52/9 = -34/9 => A = 17/9。因此分解为 17/9(x+2) - 4/3(x+2)² + 10/9(x-1)。
7. Denominator with an Irreducible Quadratic Factor | 分母含不可约二次因式
When the denominator contains a quadratic factor like ax²+bx+c that cannot be factorised into real linear factors (discriminant Δ = b²-4ac < 0), the corresponding partial fraction must have a linear numerator of the form Ax+B. Using merely a constant numerator is insufficient because the quadratic may contribute an x-term in the sum.
当分母包含形如 ax²+bx+c 且不能分解为实线性因式的二次因式时（判别式 Δ = b²-4ac < 0），对应的部分分式必须具有 Ax+B 形式的线性分子。仅用常数分子是不够的，因为二次因式在求和中可能贡献 x 的一次项。
A typical decomposition for (px²+qx+r)/(x²+1)(x-2) would be (Ax+B)/(x²+1) + C/(x-2) hen comparing coefficients of x², x, and the constant term to solve a small linear system.
对于 (px²+qx+r)/(x²+1)(x-2) 的典型分解为 (Ax+B)/(x²+1) + C/(x-2)。求解常数 A、B 和 C 的方法是通分后乘开，代入任何实根（如 x = 2）以简化，然后比较 x²、x 及常数项的系数来求解一个小型线性方程组。
8. Worked Example – Irreducible Quadratic Factor | 实例演练 – 不可约二次因式
Decompose (2x²+x+3)/(x²+1)(x-1). The factor x²+1 has no real roots, so we set:
(2x²+x+3)/(x²+1)(x-1) ≡ (Ax+B)/(x²+1) + C/(x-1)
Multiply by (x²+1)(x-1): 2x²+x+3 ≡ (Ax+B)(x-1) + C(x²+1). Substitute x = 1 to eliminate the first term: 2(1)+1+3 = 6 = C(2) => C = 3. Now expand: (Ax+B)(x-1) = Ax² - Ax + Bx - B. Add 3x²+3 qual 1, so B = 0. Constant term: -B + 3 = -0 + 3 = 3, which matches. Thus decomposition is (-x)/(x²+1) + 3/(x-1), or -x/(x²+1) + 3/(x-1).
分解 (2x²+x+3)/(x²+1)(x-1)。因式 x²+1 无实根，故设：
(2x²+x+3)/(x²+1)(x-1) ≡ (Ax+B)/(x²+1) + C/(x-1)
两边乘 (x²+1)(x-1)：2x²+x+3 ≡ (Ax+B)(x-1) + C(x²+1)。代入 x = 1 消去第一项：2+1+3=6 = C(2) => C = 3。现在展开：(Ax+B)(x-1) = Ax² - Ax + Bx - B。加上 C 部分的 3x²+3。合并同类项：x² 系数 A+3 须等于2，故 A = -1。x 系数：-A+B = 1+B 须等于1，故 B = 0。常数项：-B+3 = 3，吻合。因此分解结果为 (-x)/(x²+1) + 3/(x-1)，即 -x/(x²+1) + 3/(x-1)。
Note that the numerator over the quadratic can be a simple multiple of x or a constant, and sometimes both terms appear. Always check your result by combining the fractions to see if you recover the original numerator.
请注意，二次因式上的分子可以是 x 的简单倍数，也可以是常数，有时两个项都会出现。务必通过合并分式来检验是否能还原原始分子。
9. Handling Improper Fractions | 处理假分式
If the degree of the numerator is greater than or equal to the degree of the denominator, the rational expression is improper. Partial fractions can only be directly applied to proper fractions. Therefore, we must first carry out polynomial long division (or inspection) to express the fraction as a polynomial plus a proper rational part. The partial fraction decomposition is then performed on the proper remainder.
如果分子的次数大于或等于分母的次数，该有理表达式就是假分式。部分分式只能直接应用于真分式。因此，我们必须先进行多项式长除法（或凑配法），将分式表示为一个多项式与一个真有理部分的和，然后再对真余式进行部分分式分解。
For instance, (x³+2x+1)/(x²-1) Multiply back: x(x²-1) = x³ - x. Subtract from numerator: (x³+2x+1) - (x³ - x) = 3x+1. So (x³+2x+1)/(x²-1) = x + (3x+1)/(x²-1). Now decompose the proper part. The denominator (x-1)(x+1) gives (3x+1)/(x²-1) ≡ A/(x-1) + B/(x+1). Using cover-up or substitution yields A = 2, B = 1. Thus the full expression becomes x + 2/(x-1) + 1/(x+1).
例如，(x³+2x+1)/(x²-1)。相除：x³ ÷ x² = x。回乘：x(x²-1) = x³ - x。从分子中减去：(x³+2x+1) - (x³ - x) = 3x+1。所以 (x³+2x+1)/(x²-1) = x + (3x+1)/(x²-1)。现在对真分式部分进行分解。分母 (x-1)(x+1) 给出 (3x+1)/(x²-1) ≡ A/(x-1) + B/(x+1)。使用遮盖法或代入法可得 A = 2，B = 1。因此完整表达式变为 x + 2/(x-1) + 1/(x+1)。
rees first; if deg(num) >= deg(den), do long division before attempting partial fractions.
忘记进行除法步骤是最常见的陷阱之一。务必先检查次数；如果分子次数 ≥ 分母次数，在尝试部分分式之前先做长除法。
10. Using Partial Fractions in Integration | 部分分式在积分中的应用
One of the main reasons for learning partial fractions is to integrate rational functions that would otherwise be intractable. Once decomposed, each term can be integrated individually, typically resulting in natural logarithms for linear denominators and arctangent forms for irreducible quadratics.
学习部分分式的主要原因之一是为了对原本难以处理的有理函数进行积分。一旦分解完毕，每一项可以单独积分，线性分母的积分通常得到自然对数，而不可约二次式则得到反正切等形式。
Consider ∫ (5x+1)/(x-1)(x+2) dx. From partial fractions we have 2/(x-1) + 3/(x+2). Integrating gives 2 ln|x-1| + 3 ln|x+2| + c. For a quadratic case like ∫ -x/(x²+1) dx, a substitution u = x²+1 yields -1/2 ln|x²+1|, while ∫ 3/(x-1) dx gives 3 ln|x-1| eavily examined in Edexcel Pure Mathematics 3, often combined with inverse trig functions.
考虑 ∫ (5x+1)/(x-1)(x+2) dx。由部分分式我们得到 2/(x-1) + 3/(x+2)。积分得到 2 ln|x-1| + 3 ln|x+2| + c。对于二次情况如 ∫ -x/(x²+1) dx，令 u = x²+1 可得到 -½ ln|x²+1|，而 ∫ 3/(x-1) dx 得到 3 ln|x-1|。这种技巧在Edexcel Pure Mathematics 3中考查较多，常常与反三角函数组合出现。
11. Common Mistakes to Avoid | 常见错误
Students often lose marks by misidentifying the form of the decomposition. Using a constant numerator over a quadratic factor, e.g., writing A/(x²+1) instead of (Ax+B)/(x²+1), will lead to an unsolvable system or incorrect expansion later. Always remember: linear numerator for irreducible quadratic denominators.
学生常因错误识别分解形式而丢分。在二次因式上使用常数分子，例如写成 A/(x²+1) 而非 (Ax+B)/(x²+1)，会导致方程组无解或后续展开出错。务必牢记：不可约二次分母的分子为线性形式。
Another frequent error is handling repeated factors inappropriately. Writing only A/(x+2)² when the denominator is (x+2)² misses the necessary A/(x+2) + B/(x+2)² form. Always include all lower powers. Additionally, failing to simplify the original fraction (cancelling common factors) or to perform long division for improper fractions will invalidate the partial fractions process.
另一个常见错误是处理重复因式不当。当分母为 (x+2)² 时，若只写 A/(x+2)²，就遗漏了必要的 A/(x+2) + B/(x+2)² 形式。务必包含所有较低的幂次。此外，未对原分式进行约分（消去公因式）或未对假分式进行长除法，都会导致部分分式过程无效。
Arithmetic slips when substituting or equating coefficients are also common. Double-check your work by combining the final partial fractions, or use a different x-value to verify the identity. Time invested in verification can save several marks.
代入或比较系数时的算术错误也很常见。通过重新合并最终的部分分式来复核，或使用不同的 x 值验证恒等式。花在验证上的时间可以挽回不少分数。
12. Exam Tips and Summary | 考试技巧与总结
In Edexcel exams, partial fractions questions often appear as the first part of a longer integration or series expansion problem. Read the question carefully: if it says "express in partial fractions", you need only produce the decomposition. If the next part asks for integration, use your answer directly. Always present the constants in their simplest rational form unless otherwise stated.
在Edexcel考试中，部分分式题通常以一个较长的积分或级数展开问题的第一部分出现。仔细审题：如果题目要求“用部分分式表示”，
Published by TutorHao | A-Level Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图