📚 IB Math: Application of Partial Fraction Decomposition in Integration | IB数学：部分分式分解在积分中的应用
A rational function is a quotient of two polynomials, written as P(x)/Q(x). Partial fraction decomposition rewrites this quotient as a sum of simpler rational expressions, each with a denominator that is a linear factor or an irreducible quadratic factor. This is particularly useful before integration because each simple term can be integrated directly using standard rules.
有理函数是两个多项式的商，记为 P(x)/Q(x)。部分分式分解将这个商改写为若干个更简单的有理式之和，其中每个分母都是线性因子或不可约二次因子。这在积分前尤其有用，因为每个简单项都可以直接用标准积分法则处理。
1. What Are Partial Fractions? | 什么是部分分式？
A partial fraction is one of the simpler fractions that appears when a complicated rational expression is split into a sum. For example, the fraction (5x-4)/(x²-4) can be decomposed as
部分分式是当一个复杂的有理表达式被拆分为和式时所出现的较简单的分式。例如，分式 (5x-4)/(x²-4) 可以分解为：
(5x-4)/(x²-4) = (3/2)/(x-2) + (7/2)/(x+2)
To verify, combine the two terms on the right over the common denominator (x-2)(x+2). The numerator becomes (3/2)(x+2) + (7/2)(x-2) = 5x-4, exactly the original numerator.
验证方法：将右边两项通分到公共分母 (x-2)(x+2) 上，分子变为 (3/2)(x+2) + (7/2)(x-2) = 5x-4，恰好等于原分子。
2. Why Use Partial Fractions in Integration? | 为什么在积分中使用部分分式？
Integrating a complicated rational function directly is often difficult. After partial fraction decomposition, the integrand becomes a sum of terms such as A/(x-a), A/(x-a)ⁿ, or (Ax+B)/(x²+px+q). These simpler forms match standard integration formulas.
直接积分一个复杂的有理函数通常很困难。经过部分分式分解后，被积函数变成若干项之和，例如 A/(x-a)、A/(x-a)ⁿ 或 (Ax+B)/(x²+px+q)。这些简单形式正好对应标准积分公式。
In IB Mathematics, especially in Analysis and Approaches Higher Level, partial fractions are commonly tested with integration techniques such as the natural logarithm rule and the arctangent rule. Mastering this method saves time and reduces mistakes.
在IB数学中，尤其是分析与方法（AA）高级水平课程，部分分式常与自然对数积分法和反正切积分法一起考查。掌握这个方法可以节省时间并减少错误。
3. Case 1: Distinct Linear Factors | 情况1：不同线性因子
If the denominator factors completely into distinct linear factors, say Q(x) = (x-a₁)(x-a₂)…(x-aₙ), then write
如果分母可以完全分解为互不相同的线性因子，即 Q(x) = (x-a₁)(x-a₂)…(x-aₙ)，则写成：
P(x)/Q(x) = A₁/(x-a₁) + A₂/(x-a₂) + … + Aₙ/(x-aₙ)
Find the constants by multiplying both sides by Q(x) and then either substituting x = aᵢ or equating coefficients of powers of x.
求解常数的方法是两边同乘 Q(x)，然后代入 x = aᵢ 或比较 x 的幂系数。
Example 1. Evaluate ∫ (5x-4)/(x²-4) dx.
例1. 计算 ∫ (5x-4)/(x²-4) dx。
Since x²-4 = (x-2)(x+2), set (5x-4)/(x²-4) = A/(x-2) + B/(x+2). Multiply through by (x-2)(x+2):
因为 x²-4 = (x-2)(x+2)，设 (5x-4)/(x²-4) = A/(x-2) + B/(x+2)。两边同乘 (x-2)(x+2)：
5x-4 = A(x+2) + B(x-2)
Substitute x = 2: 6 = 4A, so A = 3/2. Substitute x = -2: -14 = -4B, so B = 7/2. Therefore
代入 x = 2：6 = 4A，所以 A = 3/2。代入 x = -2：-14 = -4B，所以 B = 7/2。因此
∫ (5x-4)/(x²-4) dx = (3/2) ln|x-2| + (7/2) ln|x+2| + C
4. Case 2: Repeated Linear Factors | 情况2：重复线性因子
If the denominator contains a repeated factor (x-a)ⁿ, then include one partial fraction for each power from 1 to n:
如果分母含有重复因子 (x-a)ⁿ，则必须对从 1 到 n 的每一个幂都写出对应部分分式：
P(x)/Q(x) = A₁/(x-a) + A₂/(x-a)² + … + Aₙ/(x-a)ⁿ
The term with denominator (x-a)ⁿ cannot be reduced to a single logarithm. Its integral follows the power rule for functions.
分母为 (x-a)ⁿ 的项不能化简为单个对数。它的积分遵循函数幂法则。
Example 2. Evaluate ∫ (2x+3)/(x+1)² dx.
例2. 计算 ∫ (2x+3)/(x+1)² dx。
Write (2x+3)/(x+1)² = A/(x+1) + B/(x+1)². Multiply by (x+1)²:
设 (2x+3)/(x+1)² = A/(x+1) + B/(x+1)²。两边同乘 (x+1)²：
2x+3 = A(x+1) + B
Substitute x = -1: 1 = B. Compare coefficients of x: A = 2. Thus
代入 x = -1：1 = B。比较 x 的系数：A = 2。因此
∫ (2x+3)/(x+1)² dx = 2 ln|x+1| – 1/(x+1) + C
Notice that ∫ 1/(x+1)² dx = -1/(x+1) + C, not a logarithm.
注意 ∫ 1/(x+1)² dx = -1/(x+1) + C，而不是对数。
5. Case 3: Irreducible Quadratic Factors | 情况3：不可约二次因子
If the denominator has a quadratic factor x²+px+q that has no real roots, the partial fraction for this factor has a linear numerator:
如果分母含有不可约二次因子 x²+px+q，即该二次式没有实数根，则其部分分式应具有一次分子：
(Ax+B)/(x²+px+q)
To integrate such a term, complete the square in the denominator and split the numerator into a multiple of the derivative of the denominator plus a constant. The result is a logarithm plus an arctangent.
要积分这样的项，先对分母配方，再把分子拆成“分母的导数的倍数”加上一个常数。结果包含一个对数项和一个反正切项。
Example 3. Evaluate ∫ (2x+3)/(x²+4x+5) dx.
例3. 计算 ∫ (2x+3)/(x²+4x+5) dx。
Complete the square: x²+4x+5 = (x+2)²+1. Rewrite the numerator as 2(x+2)-1:
配方得 x²+4x+5 = (x+2)²+1。将分子改写为 2(x+2)-1：
∫ (2(x+2)-1)/((x+2)²+1) dx
The first part gives ln(x²+4x+5), and the second part gives -arctan(x+2). Therefore
第一部分积分为 ln(x²+4x+5)，第二部分积分为 -arctan(x+2)。因此
∫ (2x+3)/(x²+4x+5) dx = ln(x²+4x+5) – arctan(x+2) + C
For a simpler quadratic term such as A/(x²+a²), the integral is (A/a) arctan(x/a) + C.
对于更简单的二次项 A/(x²+a²)，其积分为 (A/a) arctan(x/a) + C。
6. Dealing with Improper Fractions | 假分式的处理
When the degree of P(x) is greater than or equal to the degree of Q(x), the fraction is called improper. In this case, polynomial long division must be performed first, so that the integrand becomes a polynomial plus a proper rational expression.
当 P(x) 的次数大于或等于 Q(x) 的次数时，这个分式称为假分式。此时必须先进行多项式长除法，使被积函数变成一个多项式加上一个真分式。
Example 4. Evaluate ∫ x³/(x²-1) dx.
例4. 计算 ∫ x³/(x²-1) dx。
Long division gives x³/(x²-1) = x + x/(x²-1). Then decompose x/(x²-1):
长除法得 x³/(x²-1) = x + x/(x²-1)。然后分解 x/(x²-1)：
x/((x-1)(x+1)) = (1/2)/(x-1) + (1/2)/(x+1)
Thus ∫ x³/(x²-1) dx = x²/2 + (1/2) ln|x-1| + (1/2) ln|x+1| + C.
因此 ∫ x³/(x²-1) dx = x²/2 + (1/2) ln|x-1| + (1/2) ln|x+1| + C。
7. Integration Techniques for the Resulting Terms | 各类分式的积分技巧
The following table summarises the standard integrals needed for partial fraction terms. In this table, n > 1 and q – p²/4 > 0.
下表总结了部分分式各项所需的标准积分。表中 n > 1 且 q – p²/4 > 0。
| Term in Decomposition | Integral |
| A/(x-a) | A ln|x-a| + C |
| A/(x-a)ⁿ, n>1 | A / ((1-n)(x-a)^(n-1)) + C |
| A/(x²+a²) | (A/a) arctan(x/a) + C |
| (Ax+B)/(x²+px+q) | (A/2) ln|x²+px+q| + (B – Ap/2) · (1/r) arctan((x+p/2)/r) + C, where r = √(q – p²/4) |
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