📚 Using Partial Fractions | 使用部分分式
Partial fractions provide a powerful algebraic technique for breaking down complicated rational expressions into a sum of simpler fractions. This method is essential for integration, series expansion, and solving differential equations. In Edexcel A-Level Mathematics, you are expected to decompose rational functions with linear, repeated, and quadratic factors, and then apply these decompositions to integration and binomial expansions. Mastering partial fractions not only boosts your algebraic fluency but also opens the door to advanced calculus topics.
部分分式是一种强大的代数技巧，能把复杂的有理式拆分成几个简单分式的和。这种方法在积分、级数展开和微分方程中至关重要。在 Edexcel A-Level 数学中，你需要掌握分母包含不同线性因子、重复线性因子和不可约二次因子的拆分，并能将拆分结果用于积分和二项式展开。熟练掌握部分分式不仅能提高代数流畅度，还为高等微积分的学习打下基础。
1. What Are Partial Fractions? | 什么是部分分式？
A rational function is the ratio of two polynomials. When the degree of the numerator is less than the degree of the denominator, the rational function is said to be proper. The method of partial fractions expresses a proper rational function as a sum of simpler fractions whose denominators are the factors of the original denominator. This decomposition makes it easier to work with the expression, particularly in calculus.
有理函数是两个多项式的比值。当分子的次数小于分母的次数时，该有理函数称为真分式。部分分式法就是把一个真分式表示为几个简单分式之和，这些分式的分母就是原分母的因式。这样的分解使得后续运算（尤其是微积分运算）变得简单得多。
For example, (3x+5)/(x²+x-2) can be expressed as A/(x+2) + B/(x-1). The constants A and B are found by algebraic manipulation. The core idea is to reverse the process of adding algebraic fractions.
例如，(3x+5)/(x²+x-2) 可以写成 A/(x+2) + B/(x-1)。常数 A 和 B 通过代数方法求出。其核心思想就是将代数分式相加的过程逆向进行。
2. Distinct Linear Factors | 不同的线性因子
₁x + b₁)(a₂x + b₂)...(aₙx + bₙ), we write a sum of n fractions, each with a constant numerator over one linear factor. The general form is P(x)/[(a₁x+b₁)(a₂x+b₂)] = A/(a₁x+b₁) + B/(a₂x+b₂).
最简单的情形是分母可以因式分解为不同的线性因子（无重复）。对于分母为 (a₁x + b₁)(a₂x + b₂)...(aₙx + bₙ) 的真分式，我们可以把它写成 n 个分式之和，每个分式都以一个常数作为分子、一个线性因子作为分母。一般形式为 P(x)/[(a₁x+b₁)(a₂x+b₂)] = A/(a₁x+b₁) + B/(a₂x+b₂)。
To determine the constants, multiply both sides by the original denominator, then expand and equate coefficients of powers of x, or substitute clever values of x that make some terms vanish. Substitution is often faster: for each linear factor, set x such that the factor equals zero, and solve for the corresponding constant.
为求出常数，可将等式两边同乘原分母，然后展开并比较 x 的各次幂系数，或者代入巧妙的 x 值使某些项归零。代入法通常更快：对每个线性因子，令 x 的取值使该因子为零，进而求出对应的常数。
Example: Express (2x+1)/((x+1)(x-2)) in partial fractions.
Let (2x+1)/((x+1)(x-2)) ≡ A/(x+1) + B/(x-2). Multiply by (x+1)(x-2): 2x+1 ≡ A(x-2) + B(x+1).
Substitute x = -1: 2(-1)+1 = A(-1-2) ⇒ -1 = -3A ⇒ A = 1/3.
Substitute x = 2: 2(2)+1 = B(2+1) ⇒ 5 = 3B ⇒ B = 5/3.
Thus, (2x+1)/((x+1)(x-2)) = 1/(3(x+1)) + 5/(3(x-2)).
例题：将 (2x+1)/((x+1)(x-2)) 拆成部分分式。
设 (2x+1)/((x+1)(x-2)) ≡ A/(x+1) + B/(x-2)。两边同乘 (x+1)(x-2)：2x+1 ≡ A(x-2) + B(x+1)。
代入 x = -1：2(-1)+1 = A(-1-2) ⇒ -1 = -3A ⇒ A = 1/3。
代入 x = 2：2(2)+1 = B(2+1) ⇒ 5 = 3B ⇒ B = 5/3。
因此，(2x+1)/((x+1)(x-2)) = 1/(3(x+1)) + 5/(3(x-2))。
3. Worked Example: Cover-Up Method | 实例演练：遮盖法
The cover-up method is a shortcut for distinct linear factors. To find the numerator for a particular linear factor, cover up that factor in the original denominator, then substitute the x-value that makes that factor zero into the remaining expression. This directly yields the constant without solving equations.
遮盖法是处理不同线性因子时的一种快捷方式。要找出某个线性因子的分子，只需遮住原分母中的该因子，然后将使该因子为零的 x 值代入剩下的表达式。这样可以直接得出常数，无需解方程组。
For (3x-4)/((x+5)(2x-1)), using cover-up:
5, evaluate (3(-5)-4)/(2(-5)-1) = (-19)/(-11) = 19/11, so coefficient of 1/(x+5) is 19/11.
valuate (3(1/2)-4)/((1/2)+5) = (-5/2)/(11/2) = -5/11, so coefficient of 1/(2x-1) is -5/11.
Thus the decomposition is 19/(11(x+5)) - 5/(11(2x-1)).
对于 (3x-4)/((x+5)(2x-1))，用遮盖法：
对因子 (x+5)：遮住 (x+5)，设 x = -5，计算 (3(-5)-4)/(2(-5)-1) = (-19)/(-11) = 19/11，因此 1/(x+5) 的系数是 19/11。
对因子 (2x-1)：遮住 (2x-1)，设 x = 1/2，计算 (3(1/2)-4)/((1/2)+5) = (-5/2)/(11/2) = -5/11，因此 1/(2x-1) 的系数是 -5/11。
于是拆分为 19/(11(x+5)) - 5/(11(2x-1))。
Always check your answer by combining the partial fractions back to the original denominator or by testing a simple x-value. The cover-up method works only when all factors are linear and distinct, and the original fraction is proper.
务必验证答案：可以把部分分式合并回原分母，或者代入一个简单的 x 值检验。遮盖法仅适用于分母全是不同线性因子、且原分式为真分式的情况。
4. Repeated Linear Factors | 重复线性因子
When a linear factor appears raised to a power, e.g., (ax+b)ⁿ, the partial fraction decomposition must include that factor raised to each power from 1 to n. For (ax+b)², we set A/(ax+b) + B/(ax+b)². In general, for a repeated linear factor (ax+b)ⁿ, we include terms with denominators (ax+b), (ax+b)², ..., (ax+b)ⁿ, each with an unknown constant numerator.
当分母中出现线性因子的幂次，比如 (ax+b)ⁿ，部分分式的拆分必须包含该因子从 1 次方到 n 次方的所有形式。对于 (ax+b)²，应设为 A/(ax+b) + B/(ax+b)²。一般情况下，对于重复线性因子 (ax+b)ⁿ，我们需引入分母分别为 (ax+b), (ax+b)², ..., (ax+b)ⁿ 的项，每项分子为待定常数。
Example: Decompose (5x²-3x+2)/((x+1)²(2x-1)).
Write (5x²-3x+2)/((x+1)²(2x-1)) ≡ A/(x+1) + B/(x+1)² + C/(2x-1).
Multiply by denominator: 5x²-3x+2 ≡ A(x+1)(2x-1) + B(2x-1) + C(x+1)².
Substitute x = -1: 5+3+2 = B(-3) ⇒ 10 = -3B ⇒ B = -10/3.
Substitute x = 1/2: 5/4 -3/2 +2 = C(3/2)² ⇒ (5/4 -6/4 +8/4) = 7/4 = C(9/4) ⇒ C = 7/9.
compare coefficients of x², or choose another x, e.g., x = 0: 2 = A(1)(-1) + B(-1) + C(1)² ⇒ 2 = -A - B + C ⇒ 2 = -A + 10/3 + 7/9 ⇒ A = 10/3 + 7/9 - 2 = 30/9 + 7/9 - 18/9 = 19/9.
Thus decomposition is 19/(9(x+1)) - 10/(3(x+1)²) + 7/(9(2x-1)).
例题：拆分 (5x²-3x+2)/((x+1)²(2x-1))。
设 (5x²-3x+2)/((x+1)²(2x-1)) ≡ A/(x+1) + B/(x+1)² + C/(2x-1)。
乘以分母：5x²-3x+2 ≡ A(x+1)(2x-1) + B(2x-1) + C(x+1)²。
代入 x = -1：5+3+2 = B(-3) ⇒ 10 = -3B ⇒ B = -10/3。
代入 x = 1/2：5/4 -3/2 +2 = C(3/2)² ⇒ (5/4 -6/4 +8/4) = 7/4 = C(9/4) ⇒ C = 7/9。
求 A 时可以比较 x² 的系数，或另选 x 值，例如 x = 0：2 = A(1)(-1) + B(-1) + C(1)² ⇒ 2 = -A - B + C ⇒ 2 = -A + 10/3 + 7/9 ⇒ A = 10/3 + 7/9 - 2 = 30/9 + 7/9 - 18/9 = 19/9。
因此拆分结果为 19/(9(x+1)) - 10/(3(x+1)²) + 7/(9(2x-1))。
5. Quadratic Factors (Irreducible) | 二次因子（不可约）
Sometimes the denominator contains a quadratic factor that cannot be factorised into real linear factors, such as x²+1 or x²+x+1. For each distinct irreducible quadratic factor ax²+bx+c, the partial fraction includes a term with a linear numerator: (Ax+B)/(ax²+bx+c). If the quadratic factor is repeated, we follow a pattern similar to repeated linear factors.
分母中有时含有无法分解为实线性因子的二次因式，如 x²+1 或 x²+x+1。对于每一个不同的不可约二次因式 ax²+bx+c，部分分式中要包含一个分子为线性形式的项：(Ax+B)/(ax²+bx+c)。如果该二次因子重复出现，处理方法与重复线性因子类似。
Example: Decompose (3x²+2x+1)/((x²+1)(x-2)).
Form: (Ax+B)/(x²+1) + C/(x-2). Multiply: 3x²+2x+1 ≡ (Ax+B)(x-2) + C(x²+1).
Expand RHS: Ax² -2Ax + Bx -2B + Cx² + C = (A+C)x² + (-2A+B)x + (-2B+C).
Equate coefficients: x²: A+C = 3; x: -2A+B = 2; constant: -2B+C = 1.
Solving: from x² eq, C = 3-A. Substitute into constant eq: -2B+3-A = 1 ⇒ -2B - A = -2 ⇒ A + 2B = 2.
From x eq: -2A+B = 2. Solve simultaneously: A = 2-2B, substitute into A+2B=2 ⇒ (2-2B)+2B=2 ⇒ 2=2, infinite solutions? Need careful. Let' -2B, plug into -2(2-2B)+B=2 => -4+4B+B=2 => 5B=6 => B=6/5. Then A=2-12/5=-2/5. C=3-(-2/5)=17/5. All consistent. So decomposition: (-2/5 x + 6/5)/(x²+1) + (17/5)/(x-2).
例题：拆分 (3x²+2x+1)/((x²+1)(x-2))。
形式：(Ax+B)/(x²+1) + C/(x-2)。乘以分母：3x²+2x+1 ≡ (Ax+B)(x-2) + C(x²+1)。
展开右边：Ax² -2Ax + Bx -2B + Cx² + C = (A+C)x² + (-2A+B)x + (-2B+C)。
比较系数：x²：A+C=3；x：-2A+B=2；常数：-2B+C=1。
解方程组：由 x² 式得 C=3-A。代入常数式：-2B+3-A=1 ⇒ -2B-A=-2 ⇒ A+2B=2。
由 x 式：-2A+B=2。联立求解：A=2-2B，代入 A+2B=2 得恒等式，需重新代入第二个方程：-2(2-2B)+B=2 ⇒ -4+4B+B=2 ⇒ 5B=6 ⇒ B=6/5。A=2-12/5=-2/5。C=3-(-2/5)=17/5。因此拆分结果：(-2/5 x + 6/5)/(x²+1) + (17/5)/(x-2)。
When integrating, such terms lead to logs and arctan functions. The linear numerator over a quadratic can be split further for integration, but for partial fractions we leave it as (Ax+B)/(ax²+bx+c).
在积分时，这样的项会导出对数和反正切函数。分子为线性的二次分式在积分时还可以进一步拆分，但作为部分分式拆分的结果，保留 (Ax+B)/(ax²+bx+c) 的形式即可。
6. Handling Improper Fractions | 处理假分式
An improper rational function has numerator degree greater than or equal to the degree of the denominator. Before applying partial fractions, you must perform polynomial long division to obtain a polynomial quotient plus a proper rational remainder. The partial fraction decomposition is then applied only to that proper remainder.
假分式是指分子次数大于或等于分母次数的有理函数。在进行部分分式拆分之前，必须先通过多项式长除法，得到一个多项式商式加上一个真分式余式。然后再对这个真分式余式进行部分分式分解。
Example: Express (x³+x²+2)/(x²-1) in partial fractions.
²+2) ÷ (x²-1). Quotient is x+1, remainder (x+3). So (x³+x²+2)/(x²-1) = x+1 + (x+3)/(x²-1).
Now decompose the proper part: (x+3)/((x-1)(x+1)) = A/(x-1) + B/(x+1).
Multiply: x+3 = A(x+1) + B(x-1). Substitute x=1: 4=2A ⇒ A=2. x=-1: 2 = -2B ⇒ B=-1.
Thus the final expression is x+1 + 2/(x-1) - 1/(x+1).
例题：将 (x³+x²+2)/(x²-1) 写成部分分式和整式之和。
先做除法：(x³+x²+2) ÷ (x²-1)。商式 x+1，余式 x+3。因此原式 = x+1 + (x+3)/(x²-1)。
再拆分真分式：(x+3)/((x-1)(x+1)) = A/(x-1) + B/(x+1)。
乘开：x+3 = A(x+1) + B(x-1)。代 x=1：4=2A ⇒ A=2。x=-1：2=-2B ⇒ B=-1。
最终表达式为 x+1 + 2/(x-1) - 1/(x+1)。
Many exam questions deliberately give an improper fraction to test whether you remember to divide first. Always check the degrees before starting partial fractions.
很多考题会故意给出假分式，以检验你是否记得先做除法。开始拆分前，务必先检查分子与分母的次数。
7. Partial Fractions in Integration | 部分分式在积分中的应用
One of the most important applications of partial fractions is evaluating integrals of rational functions. Once the expression is decomposed, each term becomes integrable using standard methods: linear denominators give natural logarithms, repeated linear factors give powers and logs, and irreducible quadratic factors give arctan and log terms after completing the square or substitution.
部分分式最重要的应用之一就是求解有理函数的积分。一旦表达式被拆分开，每一项都可以用标准方法积分：线性分母给出自然对数，重复线性因子给出幂函数和对数，不可约二次因子则通过配平方或代换得到反正切与对数项。
For example, ∫ (2x+1)/( (x+1)(x-2) ) dx = ∫ (1/3)/(x+1) dx + ∫ (5/3)/(x-2) dx = (1/3)ln|x+1| + (5/3)ln|x-2| + C.
For a repeated factor: ∫ (1/(x+1)²) dx = -1/(x+1) + C.
/(x²+a²) dx + ∫ B/(x²+a²) dx. The first is (A/2)ln(x²+a²), the second is (B/a)arctan(x/a).
例如，∫ (2x+1)/( (x+1)(x-2) ) dx = ∫ (1/3)/(x+1) dx + ∫ (5/3)/(x-2) dx = (1/3)ln|x+1| + (5/3)ln|x-2| + C。
对于重复因子：∫ (1/(x+1)²) dx = -1/(x+1) + C。
对于二次因子：∫ (Ax+B)/(x²+a²) dx 可拆分为 ∫ Ax/(x²+a²) dx + ∫ B/(x²+a²) dx。第一个为 (A/2)ln(x²+a²)，第二个为 (B/a)arctan(x/a)。
Edexcel often combines partial fractions with definite integration or area under a curve, requiring you to find exact values using ln properties. Always remember the absolute value inside the log for indefinite integrals.
Edexcel 经常将部分分式与定积分或曲线下面积结合起来，要求利用对数性质求出精确值。对于不定积分，切记对数内要加绝对值。
8. Partial Fractions for Binomial Expansion | 部分分式用于二项式展开
Partial fractions can also simplify binomial expansions of rational functions. If a rational function can be expressed as a sum of terms of the form A/(ax+b)ⁿ, each term can be expanded using the binomial theorem (1+u)ⁿ form, provided |u| <1. This is particularly useful when the denominator factorises easily.
部分分式还可用来简化有理函数的二项式展开。如果一个有理函数可以表示为形如 A/(ax+b)ⁿ 的若干项之和，那么每一项都可以用二项式定理（(1+u)ⁿ 形式）展开，前提是 |u| <1。当分母容易因式分解时，这种方法尤其有用。
For example, expand (3x+2)/((1-x)(1+2x)) in ascending powers of x up to x³.
First decompose: (3x+2)/((1-x)(1+2x)) ≡ ver-up, A at x=1: (3+2)/(1+2)=5/3; B at x=-1/2: ( -3/2+2 )/(1 - (-1/2)) = (1/2)/(3/2)=1/3.
Thus function = (5/3)(1-x)⁻¹ + (1/3)(1+2x)⁻¹.
Expand: (1-x)⁻¹ = 1 + x + x² + x³ + ... ; (1+2x)⁻¹ = 1 - 2x + 4x² - 8x³ + ...
x+4x²-8x³) = (5/3+1/3) + (5/3 -2/3)x + (5/3+4/3)x² + (5/3 -8/3)x³ = 2 + x + 3x² - x³ + ...
Validity: |x|<1 and |2x|<1 ⇒ |x|<1/2. The final validity is |x|<1/2.
例如，将 (3x+2)/((1-x)(1+2x)) 按 x 的升幂展开至 x³。
先拆分：(3x+2)/((1-x)(1+2x)) ≡ A/(1-x) + B/(1+2x)。遮盖法，A 取 x=1：(3+2)/(1+2)=5/3；B 取 x=-1/2：( -3/2+2 )/(1 - (-1/2)) = (1/2)/(3/2)=1/3。
所以函数 = (5/3)(1-x)⁻¹ + (1/3)(1+2x)⁻¹。
展开：(1-x)⁻¹ = 1 + x + x² + x³ + ...；(1+2x)⁻¹ = 1 - 2x + 4x² - 8x³ + ...
因此表达式 = (5/3)(1+x+x²+x³) + (1/3)(1-2x+4x²-8x³) = 2 + x + 3x² - x³ + ...
收敛范围：|x|<1 且 |2x|<1 ⇒ |x|<1/2。最终有效性为 |x|<1/2。
Exam questions often require stating the range of validity for the expansion. Remember to take the most restrictive condition from all individual expansions.
考题常常要求写出展开式的有效范围。记得取所有单项展开条件中最严格的那个。
9. Matching Coefficients vs Substitution | 比较系数法与代入法
You have two main algebraic techniques to determine the unknown constants: equating coefficients and substituting strategic x-values. Substituting values that zero out factors is fast and reduces simultaneous equations. However, for factors that are not easily zeroed (like quadratics) or when too many unknowns exist, equating coefficients after expanding is systematic.
求待定常数有两种主要的代数方法：比较系数法和代入特殊 x 值。代入可以使因子为零的 x 值速度很快，并能减少联立方程的数量。然而，对不容易归零的因子（如二次式）或当未知数较多时，展开后比较系数则更为系统化。
Often a combination is best: use substitution to find as many constants as possible, then compare coefficients for the remaining ones. If you have n unknown constants, you need n independent equations. Choose the method that minimises the risk of algebraic errors.
通常将两者结合效果最好：先用代入法求出尽可能多的常数，再将剩下的常数通过比较系数来求解。如果有 n 个待定常数，就需要 n 个独立方程。选择代数错误风险最小的方法。
e original expression and your decomposed form. If they match, your constants are correct.
一定要通过代入某个简单的 x 值（比如 x=0）到原式和你的拆分结果中来验证。如果两边相等，就说明常数正确。
10. Common Pitfalls and How to Avoid Them | 常见陷阱与避错方法
Forgetting the proper fraction check: r degree, do the division first. Many students skip this step and get a partial fraction decomposition that can never match the original improper function.
忘记检查真分式：永远先比较分子分母的次数。如果分子次数 ≥ 分母次数，务必要先做除法。许多学生跳过这一步，得到的部分分式拆分根本无法还原为原假分式。
Incorrect setup for repeated factors: For (ax+b)², you must include both A/(ax+b) and B/(ax+b)². Omitting the lower power term is a frequent mistake.
重复因子的设定错误：对于 (ax+b)²，必须同时包含 A/(ax+b) 和 B/(ax+b)² 两项。漏掉较低次项是常见错误。
Misplacing linear numerators: Over a quadratic irreducible denominator, the numerator must be linear (Ax+B), not just a constant. Using a constant will usually lead to a contradiction or loss of solutions.
线性分子位置不当：在不可约二次分母上方，分子必须是线性形式 (Ax+B)，而不能只是一个常数。只用常数往往会导致矛盾或无解。
Ignoring validity ranges in expansion: When using partial fractions for series expansion, each binomial expansion has its own convergence interval. The overall expansion is valid only for the intersection of these intervals.
忽视展开的有效范围：在利用部分分式进行级数展开时，每个二项式展开都有各自的收敛区间。整体展开只在各区间的交集内有效。
11. Exam Tips for Edexcel Pure Maths | Edexcel 纯数考试技巧
In the Edexcel A-Level exam, partial fractions questions typically appear in the Pure Mathematics 3 or 4 paper. They are often worth 5-8 marks and may be combined with integration or binomial expansion. Show all your steps clearly: setting up the identity, stating the general form, multiplying by the denominator, and finding constants. Even if you make a mistake in solving for a constant, you might get method marks for the correct setup and subsequent integration/expansion.
在 Edexcel A-Level 考试中，部分分式题目通常出现在纯数 3 或纯数 4 的试卷里。这类题一般分值 5-8 分，并可能结合积分或二项式展开。务必清晰地写出所有步骤：建立恒等式、写出拆分的一般形式、两边同乘分母、求出常数。即便解常数时出错，正确的设定形式和之后的积分/展开仍可获得方法分。
For integration questions, remember to write the final answer with +C and absolute values inside logs. If the integral is definite, simplify the log terms using log laws to obtain exact values. For expansion questions, state the range of validity explicitly, and express coefficients in simplest fraction form.
对于积分题，记得在最后答案中加上 +C，对数内要有绝对值。如果是定积分，要利用对数运算律将对数项化简以得到精确值。对于展开题，要明确写明有效范围，并用最简分数给出系数。
12. Summary and Key Takeaways | 总结与核心要点
Mastering partial fractions hinges on recognizing the type of denominator factors and setting up the correct form. Practice moving between substitution and coefficient matching smoothly. Partial fractions are not an isolated topic; they are a crucial bridge to calculus and series work. Ensure you can handle distinct linear, repeated linear, and irreducible quadratic factors confidently. Always check for improper fractions first. With consistent practice, partial fraction decomposition becomes a reliable tool in your mathematical arsenal, allowing you to tackle complex integrals and expansions with confidence.
掌握部分分式的关键在于识别分母因子的类型并正确设定拆分形式。要在代入法和比较系数法之间流畅切换，需要多加练习。部分分式并非孤立的知识点，而是通往微积分和级数的重要桥梁。确保你能自信地处理不同线性因子、重复线性因子和不可约二次因子。永远先检查是否为假分式。通过持续的练习，部分分式拆分将成为你数学工具箱中一个可靠的工具，让你有信心应对复杂的积分和展开问题。
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