IB Math: Integration by Partial Fractions | IB数学：利用部分分式法求积分

📚 IB Math: Integration by Partial Fractions | IB数学：利用部分分式法求积分

When you are asked to integrate a rational function, direct techniques such as substitution or recognition often fail because the fraction is too complicated. Partial fractions solve this problem by rewriting the rational function as a sum of simpler fractions, each of which can be integrated using standard results.

当你需要对一个有理函数进行积分时，直接代入法或直接识别公式往往无法奏效，因为分式过于复杂。部分分式法将有理函数改写为若干个更简单的分式之和，其中每一个都可以使用标准积分公式处理。

---

1. What Are Partial Fractions? | 什么是部分分式？

A rational function is a ratio of two polynomials, written as P(x)/Q(x). Partial fraction decomposition expresses P(x)/Q(x) as a sum of simpler rational expressions.

有理函数是两个多项式的比，记作 P(x)/Q(x)。部分分式分解将 P(x)/Q(x) 表示成若干个更简单的有理表达式之和。

For example, the fraction (5x − 7)/((x − 3)(x + 1)) can be decomposed as 2/(x − 3) + 3/(x + 1). Rewriting the original fraction in this way turns a difficult integral into two easy logarithmic integrals.

例如，分式 (5x − 7)/((x − 3)(x + 1)) 可以分解为 2/(x − 3) + 3/(x + 1)。这样改写之后，原先较难的积分就变成了两个简单的对数积分。

P(x)/Q(x) = F₁(x) + F₂(x) + … + Fₙ(x)

Each Fᵢ(x) is called a partial fraction. The goal is to choose partial fractions whose denominators are linear factors or irreducible quadratic factors of Q(x).

每一个 Fᵢ(x) 称为一个部分分式。我们的目标是选取分母为 Q(x) 的一次因式或不可约二次因式的部分分式。

---

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

Before applying partial fractions, you must check whether the rational function is proper. It is proper if the degree of P(x) is less than the degree of Q(x). If the numerator has degree greater than or equal to the denominator, the fraction is improper.

在使用部分分式法之前，必须先判断该有理函数是否为真分式。若 P(x) 的次数小于 Q(x) 的次数，则为真分式；若分子的次数大于或等于分母的次数，则为假分式。

For an improper fraction, you must first perform polynomial long division to obtain a polynomial plus a proper rational function. Only then can partial fractions be applied to the proper remainder.

对于假分式，必须先进行多项式长除法，将其化成一个多项式加上一个真分式。之后才能对这个真分式进行部分分式分解。

For example:

例如：

x³/(x² + 1) = x − x/(x² + 1)

This makes the integral easy: ∫ x³/(x² + 1) dx = x²/2 − ½ ln(x² + 1) + C.

这样积分就变得简单：∫ x³/(x² + 1) dx = x²/2 − ½ ln(x² + 1) + C。

---

3. The General Decomposition Rules | 分解的一般规则

To decompose a proper rational function completely, factor the denominator into linear factors of the form ax + b and irreducible quadratic factors of the form ax² + bx + c. The form of the partial fraction depends on the type and multiplicity of each factor.

要完整地分解一个真分式，需要将分母分解为 ax + b 形式的一次因式和 ax² + bx + c 形式的不可约二次因式。部分分式的形式取决于每个因式的类型和重数。

Denominator factor 分母因式	Partial fraction form 部分分式形式
ax + b	A/(ax + b)
(ax + b)ⁿ	A₁/(ax + b) + A₂/(ax + b)² + … + Aₙ/(ax + b)ⁿ
ax² + bx + c	(Ax + B)/(ax² + bx + c)
(ax² + bx + c)ⁿ	(A₁x + B₁)/(ax² + bx + c) + (A₂x + B₂)/(ax² + bx + c)² + … + (Aₙx + Bₙ)/(ax² + bx + c)ⁿ

For a linear factor, the numerator is a constant. For an irreducible quadratic factor, the numerator must be a linear expression Ax + B.

对于一次因式，分子是常数；对于不可约二次因式，分子必须是 Ax + B 这样的线性表达式。

---

4. Case 1: Distinct Linear Factors | 情形一：互异一次因式

Suppose the denominator can be written as a product of different linear factors. Then each factor contributes one partial fraction with a constant numerator.

假设分母可以写成若干个互不相同的一次因式之积。此时每个因式贡献一个分子为常数的部分分式。

Follow these steps:

按以下步骤操作：

• Write the partial fraction form with unknown constants A, B, C, …

  写出带有待定常数 A、B、C、… 的部分分式形式。

• Multiply both sides by the full denominator to eliminate all fractions.

  等式两边同乘整个分母，以消去所有分式。

• Substitute the roots of the denominator to solve for the constants, or compare coefficients.

  代入分母的根来求解常数，或比较等式两边的系数。

For example, suppose you need to decompose (2x + 3)/((x − 1)(x − 2)).

例如，需要分解 (2x + 3)/((x − 1)(x − 2))。

(2x + 3)/((x − 1)(x − 2)) = A/(x − 1) + B/(x − 2)

Multiplying through by (x − 1)(x − 2) gives 2x + 3 = A(x − 2) + B(x − 1).

两边同乘 (x − 1)(x − 2)，得到 2x + 3 = A(x − 2) + B(x − 1)。

Substitute x = 1: 5 = −A, so A = −5. Substitute x = 2: 7 = B, so B = 7.

代入 x = 1：5 = −A，故 A = −5；代入 x = 2：7 = B，故 B = 7。

This method of substituting roots is often called the cover-up rule and is very fast for distinct linear factors.

这种代入根的方法通常称为“遮盖法”，对互异一次因式来说非常快捷。

---

5. Case 2: Repeated Linear Factors | 情形二：重一次因式

If a linear factor appears n times, you must include n partial fractions for that factor, with denominators of increasing powers.

如果一个一次因式出现 n 次，则必须为该因式写出 n 个部分分式，其分母的幂次依次增大。

For a factor (x − a)ⁿ, the partial fractions are:

对于因式 (x − a)ⁿ，对应部分分式为：

A₁/(x − a) + A₂/(x − a)² + … + Aₙ/(x − a)ⁿ

Consider (3x − 1)/(x(x − 1)²). Its decomposition must include terms for x, x − 1, and (x − 1)².

考虑 (3x − 1)/(x(x − 1)²)。其分解必须包含 x、x − 1 和 (x − 1)² 对应项。

(3x − 1)/(x(x − 1)²) = A/x + B/(x − 1) + C/(x − 1)²

Multiplying by x(x − 1)² gives:

两边同乘 x(x − 1)² 得到：

3x − 1 = A(x − 1)² + Bx(x − 1) + Cx

Expanding and comparing coefficients gives A = −1, B = 1, C = 2.

展开并比较系数，得 A = −1，B = 1，C = 2。

(3x − 1)/(x(x − 1)²) = −1/x + 1/(x − 1) + 2/(x − 1)²

The integral is therefore −ln|x| + ln|x − 1| − 2/(x − 1) + C.

所以积分为 −ln|x| + ln|x − 1| − 2/(x − 1) + C。

---

6. Case 3: Irreducible Quadratic Factors | 情形三：不可约二次因式

A quadratic factor ax² + bx + c is irreducible if its discriminant b² − 4ac is negative. In that case it cannot be factored into real linear factors.

若二次因式 ax² + bx + c 的判别式 b² − 4ac 小于零，则它是不可约的，不能分解为实一次因式。

Each distinct irreducible quadratic factor contributes a partial fraction of the form (Ax + B)/(ax² + bx + c).

每个互异的不可约二次因式对应一个 (Ax + B)/(ax² + bx + c) 形式的部分分式。

For example, consider (2x + 1)/(x(x² + 1)).

例如，考虑 (2x + 1)/(x(x² + 1))。

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

Multiplying through:

两边同乘分母：

2x + 1 = A(x² + 1) + (Bx + C)x

This gives A = 1, C = 2, and B = −1. Therefore:

由此得 A = 1，C = 2，B = −1。因此：

(2x + 1)/(x(x² + 1)) = 1/x + (−x + 2)/(x² + 1)

Integrating gives ln|x| −

Published by TutorHao | IB Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）