📚 PDF Joiner (4) - Question 192: Integration by Completing the Square and Substitution | 试卷合并(4)-第192题：用配方法和换元积分
This article provides a detailed, step-by-step solution to a classic Edexcel A-Level Pure Mathematics problem (from the PDF Joiner (4) collection, Question 192). The question combines the technique of completing the square with integration to evaluate a rational function. Mastering this method is essential for success in the Pure Mathematics 3 and 4 papers, where such integrals frequently appear.
本文为一道经典的爱德思A-Level纯数学问题（选自试卷合并(4)第192题）提供详细的分步解答。该题目将配平方技巧与积分相结合，用于计算有理函数的不定积分。掌握这一方法对于在纯数学3和4试卷中取得成功至关重要，因为此类积分经常出现。
1. Problem Overview | 题目概览
) instructs us to then use this result to find the indefinite integral ∫ (3x + 1)/(x² + 2x + 5) dx. This structure is typical of Edexcel exam questions — the first part provides a crucial hint for tackling the second part.
题目分为两部分。第(a)部分要求我们将二次分母 x² + 2x + 5 表示为完全平方形式 (x + a)² + b。第(b)部分则要求我们利用这一结果求不定积分 ∫ (3x + 1)/(x² + 2x + 5) dx。这种结构是爱德思考试题的典型特征——第一部分为解答第二部分提供了关键提示。
2. Completing the Square for the Denominator | 分母的配平方
² + 2x + 1) + 4 = (x + 1)² + 4. Therefore, a = 1 and b = 4. In exam marking schemes, showing this manipulation clearly earns full marks for part (a).
为了对 x² + 2x + 5 进行配方，我们首先取 x 的系数 2，将其减半得 1，再平方得 1。然后写出 x² + 2x + 5 = (x² + 2x + 1) + 4 = (x + 1)² + 4。因此，a = 1，b = 4。在考试评分方案中，清晰地展示这一操作过程可确保在(a)部分获得满分。
x² + 2x + 5 ≡ (x + 1)² + 4
3. Setting Up the Integral | 建立积分表达式
x + 1)/[(x + 1)² + 4] dx. The next challenge is to split the numerator so that one part becomes the derivative of the completed-square denominator, while the other leads to an arctangent form. The derivative of (x + 1)² + 4 is 2(x + 1), but our numerator is 3x + 1. We need to rewrite 3x + 1 in terms of (x + 1).
分母现可表示为 (x + 1)² + 4，积分变为 ∫ (3x + 1)/[(x + 1)² + 4] dx。下一个挑战是拆分分子，使得一部分成为配平方后分母的导数，而另一部分则导向反正切形式。(x + 1)² + 4 的导数为 2(x + 1)，但我们的分子是 3x + 1。我们需要用 (x + 1) 来表示 3x + 1。
4. Expressing the Numerator in Terms of the Derivative | 用分母的导数表示分子
g gives 2Ax + 2A + B. Equating coefficients of x: 2A = 3, so A = 3/2. Equating constants: 2A + B = 1 → 2(3/2) + B = 1 → 3 + B = 1 → B = −2. Hence, 3x + 1 = (3/2)(2x + 2) − 2.
注意到分母 x² + 2x + 5 的导数为 2x + 2。我们的目标是将分子 3x + 1 写成 A(2x + 2) + B 的形式，其中 A 和 B 为常数。展开得 2Ax + 2A + B。比较 x 的系数：2A = 3，因此 A = 3/2。比较常数项：2A + B = 1 → 2(3/2) + B = 1 → 3 + B = 1 → B = −2。因此，3x + 1 = (3/2)(2x + 2) − 2。
3x + 1 ≡ (3/2)(2x + 2) − 2
5. Splitting the Integral | 拆分积分
te integrals: (3/2)∫ (2x + 2)/[(x + 1)² + 4] dx − 2 ∫ 1/[(x + 1)² + 4] dx. The first integral will be handled by a simple substitution (which leads to a natural logarithm), and the second one will yield an inverse tangent function.
将此表达式代回积分，得到 ∫ [(3/2)(2x + 2) − 2]/[(x + 1)² + 4] dx。现在我们可将该积分拆分为两个独立积分：(3/2)∫ (2x + 2)/[(x + 1)² + 4] dx − 2 ∫ 1/[(x + 1)² + 4] dx。第一个积分可通过简单换元（导出自然对数）处理，第二个积分将得到反正切函数。
6. Evaluating the First Integral (Logarithmic Part) | 计算第一个积分（对数部分）
Then du/dx = 2(x + 1) = 2x + 2, so du = (2x + 2) dx. The integral becomes ∫ (1/u) du = ln|u| + C₁ = ln| (x + 1)² + 4 | necessary, but retaining them is acceptable. Therefore, (3/2) I₁ = (3/2) ln((x + 1)² + 4).
考虑 I₁ = ∫ (2x + 2)/[(x + 1)² + 4] dx。令 u = (x + 1)² + 4，则 du/dx = 2(x + 1) = 2x + 2，从而 du = (2x + 2) dx。积分变为 ∫ (1/u) du = ln|u| + C₁ = ln| (x + 1)² + 4 | + C₁。由于 (x + 1)² + 4 恒为正，绝对值符号并非严格要求，但保留亦可。因此，(3/2) I₁ = (3/2) ln((x + 1)² + 4)。
7. Evaluating the Second Integral (Arctangent Part) | 计算第二个积分（反正切部分）
(x + 1)/2)² + 1 ] dx. Let v = (x + 1)/2, so dv = (1/2) dx → dx = 2 dv. Substituting gives I₂ = (1/4) ∫ 1/(v² + 1) · 2 dv = (1/2) ∫ 1/(v² + 1) dv = (1/2) arctan(v) + C₂ = (1/2) arctan((x + 1)/2) + C₂. Consequently, the term −2 I₂ becomes −2 × (1/2) arctan((x + 1)/2) = − arctan((x + 1)/2).
现在计算 I₂ = ∫ 1/[(x + 1)² + 4] dx。从分母中提出因子 4：(x + 1)² + 4 = 4[ ((x + 1)/2)² + 1 ]。那么 I₂ = ∫ 1/(4[ ((x + 1)/2)² + 1 ]) dx = (1/4) ∫ 1/[ ((x + 1)/2)² + 1 ] dx。令 v = (x + 1)/2，则 dv = (1/2) dx → dx = 2 dv。代换后得 + 1) · 2 dv = (1/2) ∫ 1/(v² + 1) dv = (1/2) arctan(v) + C₂ = (1/2) arctan((x + 1)/2) + C₂。因此，项 −2 I₂ 变为 −2 × (1/2) arctan((x + 1)/2) = − arctan((x + 1)/2)。
8. Combining the Results and Final Answer | 组合结果并给出最终答案
2) ln((x + 1)² + 4) − arctan((x + 1)/2) + C. This can also be written in terms of the original quadratic: (3/2) ln|x² + 2x + 5| − arctan((x + 1)/2) + C. Both forms are equivalent and acceptable in the exam.
将以上计算所得的积分结果合并，并加上一个积分常数 C，完整的不定积分为：∫ (3x + 1)/(x² + 2x + 5) dx = (3/2) ln((x + 1)² + 4) − arctan((x + 1)/2) + C。该结果也可用原来的二次式表示：(3/2) ln|x² + 2x + 5| − arctan((x + 1)/2) + C。两种形式等价，考试中均可接受。
∫ (3x + 1)/(x² + 2x + 5) dx = (3/2) ln|x² + 2x + 5| − arctan((x + 1)/2) + C
9. Verification and Checking | 验证与检查
to differentiate the result and confirm we obtain the original integrand. Let y = (3/2) ln|x² + 2x + 5| 2x+5) − (1/2)·4/(x²+2x+5) = (3x+3 − 2)/(x²+2x+5) = (3x+1)/(x²+2x+5). This matches exactly, confirming the correctness of our integration.
一种强有力的验证方法是对结果进行微分，确认能否得到原被积函数。令 y = (3/2) ln|x² + 2x + 5| − arctan((x + 1)/2)。则 4 ]·(1/2) = (3x+3)/(x²+2x+5) − (1/2)·4/(x²+2x+5) = (3x+3 − 2)/(x²+2x+5) = (3x+1)/(x²+2x+5)。完全匹配，从而确认了积分结果的正确性。
10. Common Mistakes and How to Avoid Them | 常见错误及其避免方法
Students often make algebraic errors when writing the numerator in terms of the derivative. A recurring mistake is forgetting to adjust the constant term after identifying A. Another common pitfall is mishandling the coefficient when integrating 1/(u² + a²) — the factor 1/a must be included correctly. Lastly, always remember to add the constant of integration C; omitting it can cost a mark.
学生常犯的错误包括在用导数表示分子时出现代数错误。一个反复出现的错误是在确定 A 后忘记调整常数项。另一个常见陷阱是积分 1/(u² + a²) 时系数处理不当——必须正确包含因子 1/a。最后，务必记得加上积分常数 C；遗漏它会丢掉分数。
∫ 1/(u² + a²) du = (1/a) arctan(u/a) + C
11. Extension: Similar Exam Style Questions | 拓展：类似考题风格
x² + 4x + 13) dx. The denominator completes to (x + 2)² + 9, and the numerator can be expressed as (1/2)(2x + 4) + 1, leading to a combination of log and arctan. Practising such variations builds fluency in recognising the correct substitution patterns.
在掌握这道题之后，你可通过修改分子进行拓展练习。例如，求 ∫ (x + 3)/(x² + 4x + 13) dx。分母配方为 (x + 2)² + 9，分子可表示为 (1/2)(2x + 4) + 1，从而得到对数和反正切函数的组合。练习此类变式有助于提高识别正确换元模式的熟练度。
12. Key Takeaways and Exam Tips | 核心要点与考试技巧
To summarise, when integrating a rational function with an irreducible quadratic denominator, always complete the square first. Then split the numerator so that one part aligns with the derivative of the denominator (for a ln integral) and the remainder fits the 1/(u² + a²) pattern. In an exam, show your completing the square step clearly and use the 'hence' command to save time. With these skills, questions like PDF Joiner (4) - Question 192 become straightforward marks.
总结来说，在对含有不可约二次分母的有理函数进行积分时，务必先进行配平方。然后拆分分子，使其一部分与分母的导数对齐（用于 ln 积分），而剩余部分符合 1/(u² + a²) 模式。在考试中，应清晰地展示配方步骤，并利用“并由此”的提示来节省时间。掌握这些技能后，像试卷合并(4)第192题这样的题目将成为你稳拿的分数。
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