Edexcel A-Level Maths: Integrating an Improper Algebraic Fraction – Worked Solution for pdfjoiner (4)-085 | 爱德思A-Level数学：有理假分式积分第085题精讲

📚 Edexcel A-Level Maths: Integrating an Improper Algebraic Fraction – Worked Solution for pdfjoiner (4)-085 | 爱德思A-Level数学：有理假分式积分第085题精讲

This worked solution covers a classic Edexcel A-Level Pure Mathematics question on integrating an improper algebraic fraction. It shows how algebraic division, partial fractions and logarithmic integration are combined into one high-value exam item.

本精讲解析覆盖爱德思A-Level纯数学中关于有理假分式积分的经典真题。题目将多项式长除法、部分分式与对数积分整合为一道高分考点。

---

1. The Question | 题目呈现

The question tagged as pdfjoiner (4)-085 asks you to find the indefinite integral of a rational function where the numerator and denominator have the same degree. The expression is deliberately written so that you cannot apply partial fractions immediately.

本题编号为 pdfjoiner (4)-085，要求求解一个分子与分母次数相同的有理函数的不定积分。该表达式经过设计，使你无法立即套用部分分式。

Find ∫ (3x² + 5x + 2) / (x² + x – 2) dx

This wording is typical of Edexcel Pure Mathematics 3 and Pure Mathematics 4 papers. The expression is not yet ready for partial fractions because it is an improper fraction, so the first task is to recognise that fact.

此类表述常见于爱德思纯数学3和纯数学4试卷。该表达式还不能直接使用部分分式，因为它是一个假分式，因此第一步是识别这一事实。

---

2. First Check: Is the Fraction Proper? | 第一步：判断是否为真分式

Both the numerator 3x² + 5x + 2 and the denominator x² + x – 2 have degree 2. When the degree of the numerator is equal to or greater than the degree of the denominator, the rational function is called improper.

分子 3x² + 5x + 2 与分母 x² + x – 2 的次数均为 2。当分子的次数大于或等于分母的次数时，该有理函数称为假分式。

Partial fractions can only be applied directly to a proper fraction, where the numerator degree is strictly less than the denominator degree. If you try to decompose an improper fraction straight away, the resulting constants will usually be inconsistent or impossible to find.

部分分式只能直接应用于真分式，即分子次数严格小于分母次数。如果直接对假分式进行分解，通常会导致常数矛盾或根本无法求解。

A quick degree comparison should always be the first step: degree of numerator = degree of denominator, so polynomial long division is required before any partial fraction work.

快速比较次数应始终是第一步：分子次数等于分母次数，因此在进行任何部分分式运算之前，必须先做多项式长除法。

---

3. Dividing the Numerator by the Denominator | 多项式长除法

Divide 3x² + 5x + 2 by x² + x – 2. Since the leading terms are 3x² and x², the first term of the quotient is 3. Multiply the entire denominator by 3 and subtract.

用 x² + x – 2 去除 3x² + 5x + 2。由于首项分别为 3x² 和 x²，商的第一个项为 3。将整个分母乘以 3 后相减。

3(x² + x – 2) = 3x² + 3x – 6

Subtracting this from the numerator gives:

从分子中减去该乘积得到：

(3x² + 5x + 2) – (3x² + 3x – 6) = 2x + 8

Therefore the division identity is:

因此长除法恒等式为：

3x² + 5x + 2 = 3(x² + x – 2) + (2x + 8)

Because (2x + 8) has degree 1, which is less than the degree 2 of the denominator, the division process stops here. The remainder is a proper rational expression.

因为 (2x + 8) 的次数为 1，小于分母的次数 2，所以长除法到此结束。余项是一个真分式。

(3x² + 5x + 2) / (x² + x – 2) = 3 + (2x + 8) / (x² + x – 2)

This rewritten form is much more useful because the constant 3 integrates directly, and the remaining fraction is now proper and suitable for partial fractions.

这个重写形式更为实用，因为常数 3 可以直接积分，而剩余的分式现在是真分式，适合使用部分分式。

---

4. Factorising the Denominator | 因式分解分母

The denominator x² + x – 2 factorises into two linear factors. You can find them by looking for two numbers that multiply to -2 and add to +1.

分母 x² + x – 2 可分解为两个一次因式。寻找两个数，使它们的乘积为 -2，和为 +1，即可得到因式。

x² + x – 2 = (x + 2)(x – 1)

Therefore, the remaining proper fraction becomes (2x + 8) / [(x + 2)(x – 1)]. The two linear factors are distinct, so we can use the simplest partial fraction form.

因此，剩余的真分式变为 (2x + 8) / [(x + 2)(x – 1)]。两个一次因式互不相同，因此可以使用最简单的部分分式形式。

Always factorise the denominator fully before setting up partial fractions. If the denominator had a repeated factor such as (x – 1)², you would need two terms instead of one.

在建立部分分式之前，务必先完整因式分解分母。如果分母含有重复因式，例如 (x – 1)²，则需要两项而不是一项。

---

5. Setting Up the Partial Fraction Decomposition | 建立部分分式分解式

Because the denominator has two distinct linear factors, write the decomposition with two unknown constants A and B.

由于分母有两个不同的一次因式，可以写出含两个待定常数 A 和 B 的分解式。

(2x + 8) / [(x + 2)(x – 1)] = A / (x + 2) + B / (x – 1)

Multiplying through by the common denominator (x + 2)(x – 1) gives the identity:

两边同乘公分母 (x + 2)(x – 1) 得到恒等式：

2x + 8 = A(x – 1) + B(x + 2)

This identity must hold for all values of x, which is why we are allowed

Published by TutorHao | A-Level Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）