📚 Partial Fraction Decomposition | 部分分式分解法
Partial fraction decomposition is a technique for rewriting a rational expression as a sum of simpler rational expressions. It is especially useful in calculus, differential equations, and series expansions because each simpler piece can be integrated or expanded separately.
部分分式分解法是将一个有理表达式改写为若干个更简单的有理表达式之和的方法。它在微积分、微分方程和级数展开中特别有用,因为每个更简单的部分都可以分别进行积分或展开。
1. What Is Partial Fraction Decomposition? | 什么是部分分式分解?
A rational expression has the form P(x)/Q(x), where P and Q are polynomials. Partial fraction decomposition rewrites this single expression into a sum of smaller rational expressions with simpler denominators.
有理表达式具有 P(x)/Q(x) 的形式,其中 P 和 Q 都是多项式。部分分式分解将这一个表达式改写成多个具有更简单分母的有理表达式之和。
For example, the expression (5x − 3) / ((x − 1)(x − 2)) can be rewritten as 2/(x − 1) + 3/(x − 2). This is easier to integrate, differentiate, or expand.
例如,表达式 (5x − 3) / ((x − 1)(x − 2)) 可以改写为 2/(x − 1) + 3/(x − 2)。这样更容易进行积分、求导或展开。
2. Proper and Improper Rational Expressions | 真分式与假分式
A rational expression P(x)/Q(x) is proper when the degree of P is strictly less than the degree of Q. If deg(P) ≥ deg(Q), it is improper and must be handled with polynomial division first.
当分子 P 的次数严格小于分母 Q 的次数时,称 P(x)/Q(x) 为真分式;若 deg(P) ≥ deg(Q),则它是假分式,必须先进行多项式除法。
For example, (x⁴ + 3)/(x² − 1) is improper because 4 ≥ 2. After dividing, we obtain a polynomial plus a proper fraction that can be decomposed normally.
例如,(x⁴ + 3)/(x² − 1) 是假分式,因为 4 ≥ 2。做除法后,会得到一个多项式加上一个真分式,然后再对这个真分式进行分解。
3. Case 1: Distinct Linear Factors | 情况一:互不相同的线性因式
When the denominator factorises into n different linear factors, we assign one constant numerator to each factor. This is the simplest type of partial fraction decomposition.
当分母能分解为 n 个互不相同的线性因式时,我们对每个因式指派一个常数作为分子。这是最简单的一类部分分式分解。
(3x + 5) / ((x + 1)(x − 2)) = A/(x + 1) + B/(x − 2)
Multiplying both sides by the denominator gives 3x + 5 = A(x − 2) + B(x + 1). Substituting x = −1 yields A, and x = 2 gives B.
两边同时乘以分母,得到 3x + 5 = A(x − 2) + B(x + 1)。代入 x = −1 可求出 A,代入 x = 2 可求出 B。
(3x + 5) / ((x + 1)(x − 2)) = −2/(3(x + 1)) + 11/(3(x − 2))
The key idea is to make the factor in the denominator equal to zero so that one unknown constant disappears.
关键思路是让分母中的某个因式等于零,从而消去一个未知常数。
4. The Cover-Up Rule | 遮盖法
The cover-up rule is a fast method for finding coefficients when all factors are distinct linear factors. To find A/(x − a), cover the factor (x − a) in the denominator and evaluate the remaining expression at x = a.
当所有因式都是互不相同的线性因式时,遮盖法可以快速求出系数。要求 A/(x − a) 中的 A,只需在分母中遮住 (x − a),再把 x = a 代入剩余表达式求值。
For example, consider (5x − 1) / ((x − 1)(x + 3)). Cover (x − 1) and set x = 1: (5 − 1)/(1 + 3) = 1. So A = 1.
例如,考虑 (5x − 1) / ((x − 1)(x + 3))。遮住 (x − 1) 并令 x = 1,得到 (5 − 1)/(1 + 3) = 1,所以 A = 1。
(5x − 1) / ((x − 1)(x + 3)) = 1/(x − 1) + 4/(x + 3)
Cover (x + 3) and set x = −3: (−15 − 1)/(−3 − 1) = 4. The method works only when each linear factor is different.
遮住 (x + 3) 并令 x = −3,得到 (−15 − 1)/(−3 − 1) = 4。这种方法只适用于每个线性因式互不相同的情况。
5. Case 2: Repeated Linear Factors | 情况二:重复的线性因式
If a linear factor appears with power n, say (x − a)ⁿ, the decomposition must include terms for every power from 1 to n.
若某个线性因式以 n 次幂出现,例如 (x − a)ⁿ,则分解中必须包含从 1 次到 n 次的所有对应项。
(2x + 3) / (x(x − 1)²) = A/x + B/(x − 1) + C/(x − 1)²
Multiplying through and comparing coefficients leads to a unique set of constants. The term with the highest power can often be found by the cover-up rule; the others require substitution or coefficient comparison.
两边乘以后通过比较系数可以得到唯一一组常数。最高次幂对应的项通常可以用遮盖法求得;其余项则需要代入特殊值或比较系数。
For this example, C = 5, A = 3, and B = −3. Always include all lower powers, or the decomposition will be incomplete.
在这个例子中,C = 5,A = 3,B = −3。一定要包含所有较低次幂的项,否则分解将不完整。
6. Case 3: Irreducible Quadratic Factors | 情况三:不可约二次因式
When a quadratic factor has no real roots, such as x² + 1 or x² + x + 1, it cannot be broken into real linear factors. In that case, the numerator assigned to that factor must be a linear expression Bx + C.
当二次因式没有实根,例如 x² + 1 或 x² + x + 1,它无法在实数范围内分解为线性因式。此时该因式对应的分子必须是线性表达式 Bx + C。
(3x² + x + 1) / ((x² + 1)(x − 1)) = (Ax + B)/(x² + 1) + C/(x − 1)
To find the constants, multiply by the denominator and compare coefficients. The quadratic factor receives a linear numerator, not just a constant.
求常数时,两边乘以分母并比较系数。二次因式对应的分子必须是线性式,而不能只是一个常数。
7. Improper Fractions: Divide First | 假分式:先做除法
If the numerator has degree greater than or equal to the denominator, perform polynomial long division first. The result is a polynomial plus a proper rational expression.
如果分子的次数大于或等于分母的次数,先进行多项式长除法。结果是多项式加上一个真分式。
(x³ + 2)/(x² − 1) = x + (x + 2)/(x² − 1)
Now decompose the remaining proper fraction: (x + 2)/((x − 1)(x + 1)) = 3/(2(x − 1)) − 1/(2(x + 1)).
现在分解剩余的真分式:(x + 2)/((x − 1)(x + 1)) = 3/(2(x − 1)) − 1/(2(x + 1))。
Never attempt to decompose an improper fraction directly; polynomial division first is essential.
千万不要直接分解假分式;必须先做多项式除法。
8. Determining Coefficients: Two Reliable Methods | 求系数:两种可靠方法
There are two standard ways to solve for unknown constants after multiplying through by the denominator: substitution and equating coefficients.
两边乘以分母后,有两种标准方法求解未知常数:代入法和比较系数法。
(x + 1) / ((x − 2)(x + 3)) = A/(x − 2) + B/(x + 3)
Substitution method: set x = 2 and x = −3. Equating coefficients method: rewrite the numerator as x + 1 = A(x + 3) + B(x − 2).
代入法:令 x = 2 和 x = −3。比较系数法:将分子改写为 x + 1 = A(x + 3) + B(x − 2)。
Expanding gives x + 1 = (A + B)x + (3A − 2B). Therefore A + B = 1 and 3A − 2B = 1. Solving gives A = 3/5 and B = 2/5.
展开得 x + 1 = (A + B)x + (3A − 2B)。因此 A + B = 1,3A − 2B = 1。解得 A = 3/5,B = 2/5。
Substitution is faster when factors are distinct; equating coefficients is safer when fractions are complex or repeated.
当因式互不相同时,代入法更快;当分式较复杂或含有重因式时,比较系数法更稳妥。
9. Application: Integration | 应用:积分
Once a fraction is decomposed, integration becomes straightforward because each partial fraction has a known antiderivative.
一旦分式被分解,积分就会变得直接,因为每个部分分式都有已知的原函数。
∫ (3x + 5)/((x + 1)(x − 2)) dx = ∫ [−2/(3(x + 1)) + 11/(3(x − 2))] dx
Then the integral is −(2/3) ln|x + 1| + (11/3) ln|x − 2| + C.
因此积分结果为 −(2/3) ln|x + 1| + (11/3) ln|x − 2| + C。
For a quadratic factor such as x² + 1, the integral often involves arctan x. For example, ∫ 1/(x² + 1) dx = arctan x + C.
对于 x² + 1 这样的二次因式,积分通常涉及 arctan x。例如,∫ 1/(x² + 1) dx = arctan x + C。
10. Application: Binomial Expansion | 应用:二项式展开
Partial fractions can convert a rational expression into terms of the form A(1 + kx)⁻¹ or A(1 + kx)⁻², to which the binomial series can be applied.
部分分式可以将有理表达式转换为 A(1 + kx)⁻¹ 或 A(1 + kx)⁻² 的形式,从而可以应用二项式级数展开。
For example, (5x − 1)/((x − 1)(x + 3)) = 1/(x − 1) + 4/(x + 3). This can be rewritten as −1/(1 − x) + (4/3)/(1 + x/3).
例如,(5x − 1)/((x − 1)(x + 3)) = 1/(x − 1) + 4/(x + 3)。它可以改写为 −1/(1 − x) + (4/3)/(1 + x/3)。
Each term can then be expanded using the binomial series, provided |x| < 1 for the example above.
然后每一项都可以用二项式级数展开,但要求上述例子中 |x| < 1。
11. Common Mistakes and Tips | 常见错误与技巧
Many errors come from forgetting the structure of the decomposition, not from algebra itself.
许多错误来自忘记分解的结构,而不是代数计算本身。
- Forgetting repeated factors: include all powers up to the highest one. / 忘记重因式:必须包含从最低次到最高次的所有项。
- Using A instead of Bx + C for a quadratic factor. / 对于二次因式,使用常数 A 而不是线性式 Bx + C。
- Failing to divide when the fraction is improper. / 当分式为假分式时,没有先做多项式除法。
- Sign errors when substituting negative roots. / 代入负数根时出现符号错误。
- Cover-up rule used for repeated factors without justification. / 不加判断地对重因式使用遮盖法。
A reliable strategy is to multiply out carefully and choose convenient values of x first, then compare coefficients for the remaining unknowns.
一个可靠的策略是:先仔细展开,选择方便代入的 x 值求部分系数,再用比较系数法求剩余未知数。
12. Summary and Practice | 总结与练习
Partial fraction decomposition follows a clear order: check whether the fraction is proper, factorise the denominator, choose the correct template, and solve for the constants.
部分分式分解遵循一个清晰的步骤:先检查分式是否为真分式,再分解分母,选择正确的分解形式,最后求解常数。
With distinct linear factors use the cover-up rule. With repeated linear factors include every power. With irreducible quadratic factors use a linear numerator. For improper fractions, divide first.
对于互不相同的线性因式使用遮盖法;对于重复线性因式,包含每个次幂;对于不可约二次因式,使用线性分子;对于假分式,先做除法。
Practising these four cases will build fluency and prepare you for integration, binomial expansion, and Laplace transforms.
反复练习这四类情况,可以帮助你熟练掌握该方法,并为积分、二项式展开和拉普拉斯变换做好准备。
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