Using Partial Fractions | 部分分式的运用

📚 Using Partial Fractions | 部分分式的运用

Partial fractions is a powerful algebraic technique that allows us to break down complex rational expressions into a sum of simpler fractions. In IB Mathematics, this method is essential for integration, solving differential equations, and summing series. Mastering partial fractions not only sharpens your algebraic manipulation skills but also opens doors to handling calculus problems that would otherwise be extremely difficult.

部分分式是一种强大的代数技巧,能将复杂的有理式拆分为几个简单分式之和。在 IB 数学中,这种方法对积分、解微分方程和级数求和至关重要。掌握部分分式不仅能提升代数运算能力,还能让你解决那些原本极其困难的微积分问题。

1. Introduction to Partial Fractions | 部分分式简介

A rational function is the ratio of two polynomials, such as (3x+5)/(x²+3x+2). When the denominator can be factorised, we can express the rational function as a sum of simpler fractions with linear or quadratic denominators. This decomposition is called the method of partial fractions.

有理函数是两个多项式的比,例如 (3x+5)/(x²+3x+2)。当分母可因式分解时,我们可以将该有理函数表示为几个分母为线性或二次式的简单分式之和。这种拆分方法就叫部分分式。


2. When and Why We Use Partial Fractions | 何时及为何使用部分分式

Partial fractions are used whenever we need to simplify a rational expression for further processing. The most common applications are integrating rational functions, finding inverse Laplace transforms, and evaluating telescoping sums. Without decomposition, integrating something like (2x+3)/(x²+x) directly would be far more challenging.

当需要简化有理式以便后续处理时,常使用部分分式。最常见的应用包括有理函数积分、求拉普拉斯逆变换以及计算裂项求和。如果不拆分,直接对 (2x+3)/(x²+x) 这样的式子积分会困难得多。


3. Prerequisites: Polynomial Long Division | 前提:多项式长除法

Before applying partial fractions, ensure the rational expression is proper, meaning the degree of the numerator is less than the degree of the denominator. If the numerator’s degree is greater than or equal to the denominator’s, perform polynomial long division first. The quotient will be a polynomial, and the remainder will form a proper fraction that can then be decomposed.

使用部分分式前,要确保有理式为真分式,即分子的次数低于分母的次数。若分子的次数大于或等于分母的次数,须先进行多项式长除法。商为多项式,余数则构成真分式,再对该真分式进行拆分。

Example: (x³+2x²+1)/(x²+1) = x + 2 + (-x-1)/(x²+1)

示例:(x³+2x²+1)/(x²+1) = x + 2 + (-x-1)/(x²+1)


4. Linear Factors in Denominator | 分母为线性因子

When the denominator factorises into distinct linear factors, we set up a sum of fractions with unknown constants. For instance, if the denominator is (x+2)(x-3), we write: A/(x+2) + B/(x-3). We then multiply through by the denominator and find A and B.

当分母可分解为不同的一次因子时,我们设有未知常数的分式之和。例如,分母为 (x+2)(x-3) 时,设 A/(x+2) + B/(x-3)。然后通分,解出 A 和 B。

1/(x+2)(x-3) = A/(x+2) + B/(x-3) → 1 = A(x-3) + B(x+2)

1/(x+2)(x-3) = A/(x+2) + B/(x-3) → 1 = A(x-3) + B(x+2)


5. Repeated Linear Factors | 重复线性因子

If a linear factor is repeated, say (x+1)², we need terms for each power up to the highest power. For a denominator with (x+1)³, we write: A/(x+1) + B/(x+1)² + C/(x+1)³. This ensures all possible numerator-induced cancellations are captured.

若一次因子重复出现,如 (x+1)²,需要为从一次到最高次幂的每一项设立分式。分母含 (x+1)³ 时,写为 A/(x+1) + B/(x+1)² + C/(x+1)³。这样可确保涵盖所有可能的分子抵消情况。


6. Quadratic Factors | 二次因子

When the denominator contains an irreducible quadratic factor like x²+1 (discriminant < 0), the corresponding numerator must be of the form Ax+B, one degree less than the denominator. So, for a quadratic factor (x²+4), we use (Ax+B)/(x²+4).

当分母含有不可约二次因子,如 x²+1(判别式 < 0),对应的分子须为一次式 Ax+B,即次数比分母低一次。也就是说,对于二次因子 (x²+4),我们使用 (Ax+B)/(x²+4)。

(2x+1)/[(x+1)(x²+1)] = A/(x+1) + (Bx+C)/(x²+1)

(2x+1)/[(x+1)(x²+1)] = A/(x+1) + (Bx+C)/(x²+1)


7. Finding Constants: Substitution Method | 求常数:代入法

The substitution method involves choosing clever x-values to simplify the equation. After clearing denominators, substitute the roots of the denominators (where factors become zero). This immediately gives the constant for that factor. For example, substitute x = -2 to find the coefficient A in a term with denominator (x+2).

代入法就是选择巧妙的 x 值来简化等式。通分后,代入分母的根(因子为零的点),可立刻求得该因子对应的常数。例如,代入 x = -2 就能求出分母为 (x+2) 的项中 A 的值。

1 = A(x-3) + B(x+2):代入 x=3 得 B=1/5,代入 x=-2 得 A=-1/5。

1 = A(x-3) + B(x+2): substitute x=3 → B=1/5, substitute x=-2 → A=-1/5.


8. Finding Constants: Equating Coefficients | 求常数:比较系数法

After expanding the equation, we can compare the coefficients of corresponding powers of x on both sides. This gives a system of linear equations. This method is particularly useful when dealing with repeated or quadratic factors, where substitution roots are limited.

展开等式后,比较两边 x 的同次幂系数,可得到一组线性方程。在处理重复因子或二次因子时,由于可用的根有限,比较系数法格外有用。

1 = Ax – 3A + Bx + 2B → coeff of x: A+B=0, constant: -3A+2B=1

1 = Ax – 3A + Bx + 2B → 比较 x 的系数:A+B=0,常数:-3A+2B=1


9. Improper Fractions | 假分式

An improper rational function has a numerator degree ≥ denominator degree. We must perform polynomial division first to obtain a polynomial plus a proper fraction. Only the proper fraction part can be decomposed using partial fractions. The polynomial part remains untouched.

假分式指分子次数 ≥ 分母次数的有理函数。必须先做多项式除法,得到一个多项式加一个真分式。只有真分式部分可用部分分式拆分,多项式部分保持不变。

(x³+2)/(x²-1) = x + (x+2)/(x²-1),然后拆分 (x+2)/(x²-1)。

(x³+2)/(x²-1) = x + (x+2)/(x²-1), then decompose (x+2)/(x²-1).


10. Applications: Integration of Rational Functions | 应用:有理函数的积分

One of the most direct uses of partial fractions is integration. After decomposing the integrand, we obtain simpler fractions that integrate to logarithmic or arctangent forms. For example, ∫ 1/(x²-1) dx becomes ∫ [1/2(x-1) – 1/2(x+1)] dx, giving (1/2) ln|(x-1)/(x+1)| + C.

部分分式最直接的应用之一是积分。拆分被积函数后,得到的简单分式可积分为对数或反正切形式。例如,∫ 1/(x²-1) dx 变为 ∫ [1/2(x-1) – 1/2(x+1)] dx,结果为 (1/2) ln|(x-1)/(x+1)| + C。

∫ (2x+1)/[(x+1)(x²+1)] dx = ∫ [1/(x+1) + (x)/(x²+1)] dx,然后分别积分。

∫ (2x+1)/[(x+1)(x²+1)] dx = ∫ [1/(x+1) + (x)/(x²+1)] dx, then integrate separately.


11. Applications: Summation of Series | 应用:级数求和

Partial fractions can turn a series into a telescoping sum. By rewriting terms like 1/[n(n+1)] as 1/n – 1/(n+1), consecutive terms cancel, leaving only the first and last parts. This technique is frequently examined in IB HL papers.

部分分式可将级数转化为裂项相消形式。将 1/[n(n+1)] 写成 1/n – 1/(n+1) 后,相邻项抵消,仅剩首尾部分。这种技巧在 IB HL 试卷中经常出现。

Σ (1/[r(r+1)]) from r=1 to n = 1 – 1/(n+1)

∑_{r=1}^{n} (1/[r(r+1)]) = 1 – 1/(n+1)


12. Summary and Tips | 总结与提示

Partial fractions require careful algebraic work. Always check that the fraction is proper first. Choose the appropriate form based on denominator factors. Use substitution to find constants quickly, and equate coefficients when needed. Always double-check your decomposition by combining the partial fractions back to the original expression.

部分分式要求细致的代数操作。首先检查分式是否为真分式。根据分母因子的类型选择正确的拆分形式。用代入法快速求常数,必要时用比较系数法。最后,务必将拆得的分式合并还原,以检查分解是否正确。

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