📚 Edexcel A-Level Maths: Question 63 – Integration by Substitution | 爱德思A-Level数学：第63题——换元积分法
This article works through Question 63 from an Edexcel A-Level Mathematics past-paper compilation. The question focuses on integration by substitution, a core skill assessed across Pure Mathematics 3 and 4. Mastering this technique will help you handle a wide range of integral problems confidently.
本文讲解爱德思A-Level数学真题汇编中的第63题。该题重点考查换元积分法，这是纯数学3和纯数学4的核心考点。掌握这一方法能够帮助你自信地应对各种积分问题。
1. The Question and Its Context | 题目与考点背景
A typical Question 63 asks: Find the exact value of the indefinite integral ∫ x√2x + 1 dx. This style of question appears frequently in Edexcel A-Level papers because it tests substitution, algebraic manipulation, and the ability to convert back to the original variable.
典型的第63题要求：求不定积分 ∫ x√2x + 1 dx。这类题目在爱德思A-Level试卷中经常出现，因为它同时考查换元、代数化简以及将结果回代为原变量的能力。
The integral contains a composite function, √2x + 1, multiplied by a linear factor x. Direct expansion is not practical because the square root is not of a simple polynomial. Substitution is the most efficient route.
该积分包含复合函数 √2x + 1，并且与一次因式 x 相乘。由于根号内不是简单多项式，直接展开并不现实，因此换元法是最有效的解题路径。
2. Understanding the Structure of the Integrand | 理解被积函数的结构
Write the integrand as x(2x + 1)1/2. The key is that the derivative of the inner function 2x + 1 is 2, which is a constant multiple of the outer factor x, after accounting for the linear relationship.
将被积函数写成 x(2x + 1)1/2。关键在于内层函数 2x + 1 的导数为 2，在考虑线性关系后，它与外层因式 x 只相差一个常数倍数。
it is not given, choosing u as the expression inside the square root is the standard first step.
在爱德思考试中，题目通常会给出代换 u = 2x + 1。如果题目没有给出，选择根号内的表达式作为 u 是标准的第一步。
= 2x + 1 gives x = (u - 1)/2, so the whole integrand can be expressed in terms of u only.
始终要检查所选代换是否能完全消去原变量。这里 u = 2x + 1 可推出 x = (u - 1)/2，因此整个被积函数都能只用 u 表示。
3. Choosing the Substitution | 选择代换变量
Let u = 2x + 1. This is a natural choice because the square root becomes √u, which is much easier to integrate using the power rule.
令 u = 2x + 1。这是一个自然的选择，因为根号部分变为 √u，使用幂法则积分会简单得多。
rentiate with respect to x to obtain du/dx = 2, or du = 2 dx. Rearranging gives dx = du/2.
由 u = 2x + 1，对 x 求导可得 du/dx = 2，即 du = 2 dx。整理后得到 dx = du/2。
ave x = (u - 1)/2. These three pieces are necessary before substituting into the integral.
同时把 x 用 u 表示：由 u = 2x + 1，可得 x = (u - 1)/2。在代入积分之前，这三个条件缺一不可。
4. Changing the Differential | 转换微分 dx
Substitute dx = du/2 into the integral. The original integral becomes ∫ x√2x + 1 dx = ∫ [(u - 1)/2] √u × (du/2).
将 dx = du/2 代入积分。原积分变为 ∫ x√2x + 1 dx = ∫ [(u - 1)/2] √u × (du/2)。
ine the constant factors: (1/2) × (1/2) = 1/4. The integral is therefore (1/4) ∫ (u - 1) u1/2 du.
合并常数因子：(1/2) × (1/2) = 1/4。因此积分为 (1/4) ∫ (u - 1) u1/2 du。
At this stage the original variable x has been fully removed. The new integral is purely in terms of u and can be expanded and integrated term by term.
此时原变量 x 已被完全消去。新积分完全用 u 表示，可以展开后逐项积分。
5. Expanding the Integrand | 展开被积函数
Expand (u - 1) u1/2 as u3/2 - u1/2. This step uses the index law am × an = am+n, where u1/2 multiplies both u and -1.
将 (u - 1) u1/2 展开为 u3/2 - u1/2。这一步利用指数律 am × an = am+n，其中 u1/2 分别乘以 u 和 -1。
The integral now becomes (1/4) ∫ [u3/2 - u1/2] du. This is a basic polynomial-type integral that can be solved with the power rule.
积分现在变为 (1/4) ∫ [u3/2 - u1/2] du。这是一个基本的多项式型积分，可以直接使用幂法则求解。
Writing out the expansion in full helps prevent sign errors and makes the subsequent integration steps clearer. Edexcel mark schemes reward this systematic presentation.
完整地写出展开过程有助于避免符号错误，并使后续积分步骤更加清晰。爱德思评分标准对这种系统性的书写给予加分。
6. Integrating Term by Term | 逐项积分
Apply the power rule for integration: ∫ un du = un+1/(n+1) + C, for n ≠ -1. For u3/2, n = 3/2, so the integral is (2/5) u5/2.
应用幂法则积分：∫ un du = un+1/(n+1) + C，其中 n ≠ -1。对于 u3/2，n = 3/2，因此积分为 (2/5) u5/2。
For u1/2, n = 1/2, so the integral is (2/3) u3/2. The overall result inside the brackets is (2/5)u5/2 - (2/3)u3/2.
对于 u1/2，n = 1/2，因此积分为 (2/3) u3/2。括号内的整体结果是 (2/5)u5/2 - (2/3)u3/2。
Multiplying by the outside factor 1/4 gives (1/10) u5/2 - (1/6) u3/2 + C. Do not forget the constant of integration, as this is an indefinite integral.
乘以外部的系数 1/4，得到 (1/10) u5/2 - (1/6) u3/2 + C。不要忘记积分常数，因为这是不定积分。
7. Substituting Back to x | 回代为 x 的表达式
Replace u with 2x + 1 in the result. The antiderivative becomes (1/10)(2x + 1)5/2 - (1/6)(2x + 1)3/2 + C.
将结果中的 u 替换为 2x + 1。原函数变为 (1/10)(2x + 1)5/2 - (1/6)(2x + 1)3/2 + C。
This is a valid final answer, but Edexcel mark schemes often accept factorised forms. We can simplify by factorising (2x + 1)3/2 from both terms.
这已经是有效的最终答案，但爱德思评分标准通常接受因式分解后的形式。我们可以从两项中提取公因式 (2x + 1)3/2 进行化简。
Factorising gives (2x + 1)3/2 plify the bracket to (2x + 1)/10 - 1/6 = (6(2x + 1) - 10)/60 = (12x - 4)/60 = (3x - 1)/15.
因式分解得到 (2x + 1)3/2 [(1/10)(2x + 1) - 1/6] + C。化简括号内为 (2x + 1)/10 - 1/6 = (6(2x + 1) - 10)/60 = (12x - 4)/60 = (3x - 1)/15。
Thus the simplified antiderivative is (2x + 1)3/2 (3x - 1)/15 + C. This form is neat and often preferred in final answers.
因此化简后的原函数为 (2x + 1)3/2 (3x - 1)/15 + C。这种形式简洁，通常是最终答案的首选。
8. Verification by Differentiation | 通过求导验证结果
To check the answer, differentiate (2x + 1)3/2 (3x - 1)/15 with respect to x. Use the product rule or expand first, then differentiate.
为了验证答案，对 (2x + 1)3/2 (3x - 1)/15 关于 x 求导。可以使用乘积法则，也可以先展开再求导。
Let y = (2x + 1)3/2 (3x - 1)/15. Using the product rule with u = (2x + 1)3/2 and v = (3x - 1)/15 gives y' = u'v + uv'.
设 y = (2x + 1)3/2 (3x - 1)/15。使用乘积法则，令 u = (2x + 1)3/2，v = (3x - 1)/15，则 y' = u'v + uv'。
Compute u' = (3/2)(2x + 1)1/2 × 2 = 3(2x + 1)1/2, and v' = 3/15 = 1/5. Substituting gives y' = 3(2x + 1)1/2 × (3x - 1)/15 + (2x + 1)3/2 × 1/5.
计算 u' = (3/2)(2x + 1)1/2 × 2 = 3(2x + 1)1/2，且 v' = 3/15 = 1/5。代入得到 y' = 3(2x + 1)1/2 × (3x - 1)/15 + (2x + 1)3/2 × 1/5。
Factorise (2x + 1)1/2/15 from both terms: y' = (2x + 1)1/2/15 [3(3x - 1) + 3(2x + 1)] = (2x + 1)1/2/15 [9x - 3 + 6x + 3] = (2x + 1)1/2 × (15x)/15 = x√2x + 1.
从两项中提取公因式 (2x + 1)1/2/15：y' = (2x + 1)1/2/15 [3(3x - 1) + 3(2x + 1)] = (2x + 1)1/2/15 [9x - 3 + 6x + 3] = (2x + 1)1/2 × (15x)/15 = x√2x + 1。
Differentiation recovers the original integrand exactly, confirming that the antiderivative is correct. This verification step is useful in exam conditions to gain confidence.
求导后完全还原为原被积函数，证明该原函数是正确的。在考试中进行这一验证步骤有助于增强信心。
9. Common Mistakes in Edexcel Exams | 爱德思考试中的常见错误
One frequent mistake is forgetting to change dx to du. Students sometimes write du instead of dx/2, which introduces an incorrect factor of 2.
一个常见错误是忘记将 dx 转换为 du。学生有时会写成 du 而不是 dx/2，从而错误地引入了系数 2。
Another common error is failing to express x in terms of u before integrating. Leaving x in the integrand after substitution makes the integral impossible to evaluate in u.
另一个常见错误是在积分前没有把 x 表示成 u。代换后仍保留 x 会使积分无法用 u 求出。
Sign errors when expanding (u - 1) u1/2 also occur. Always expand carefully and check that u3/2 - u1/2 is correct, not u3/2 + u1/2.
展开 (u - 1) u1/2 时也容易出现符号错误。务必仔细展开，确认结果是 u3/2 - u1/2，而不是 u3/2 + u1/2。
Finally, forgetting the constant of integration C loses marks in indefinite integral questions. Edexcel A-Level mark schemes explicitly require C.
最后，忘记积分常数 C 会在不定积分题中丢分。爱德思A-Level评分标准明确要求写出 C。
10. Exam Technique and Time Management | 考试技巧与时间管理
Allocate about 6 to 8 minutes for a question of this type in an Edexcel A-Level Pure Mathematics paper. Start by identifying the composite function and writing the substitution clearly.
在爱德思A-Level纯数学试卷中，这类题目建议分配大约6到8分钟。首先识别复合函数，并清楚地写出代换关系。
ps. Edexcel mark schemes award method marks for these substitutions even if a later arithmetic slip occurs.
展示所有中间步骤，尤其是 dx = du/2 和 x = (u - 1)/2 这两步。即使之后出现算术错误，爱德思评分标准也会为这些代换步骤给予方法分。
If the question asks for a definite integral, remember to change the limits or substitute back before evaluating. For indefinite integrals, always add C.
如果题目要求求定积分，记得转换积分上下限，或者在求值前先回代为 x。对于不定积分，一定要加上常数 C。
Use a quick differentiation check if time permits. This can catch algebraic errors before you move on to the next question.
如果时间允许，可以快速求导验证。这能在进入下一题之前发现代数错误。
11. Practice Variation and Extension | 练习变式与拓展
To deepen understanding, try integrating x(3x - 2)4 with u = 3x - 2. The same method applies, but the algebraic expansion is slightly longer.
为了加深理解，可以尝试用 u = 3x - 2 计算 ∫ x(3x - 2)4 dx。方法相同，但代数展开稍长。
d dx = du/2, turning the integrand into (u - 1)/(4u), which simplifies before integration.
另一个有用的变式是 ∫ x/(2x + 1) dx。这里令 u = 2x + 1，得到 x = (u - 1)/2 和 dx = du/2，被积函数变为 (u - 1)/(4u)，积分前需要先化简。
For definite integrals, such as ∫ from 0 to 1 of x√2x + 1 imits: when x = 0, u = 1; when x = 1, u = 3. Then evaluate the u-integral between 1 and 3.
对于定积分，例如 ∫ 从 0 到 1 的 x√2x + 1 dx，转换上下限：当 x = 0 时 u = 1；当 x = 1 时 u = 3。然后在 1 到 3 之间计算关于 u 的积分。
Practising these variations builds flexibility with substitution and prepares you well for mixed integration problems in Edexcel A-Level Maths.
练习这些变式能够提高换元法的灵活性，帮助你更好地应对爱德思A-Level数学中的混合积分问题。
12. Summary and Key Takeaways | 总结与核心要点
Integration by substitution is essential for Edexcel A-Level Mathematics. For integrands of the form x√ax + b, let u = ax + b, express x in terms of u, and convert dx to du.
换元积分法是爱德思A-Level数学的核心内容。对于形如 x√ax + b 的被积函数，令 u = ax + b，将 x 用 u 表示，并将 dx 转换为 du。
The solved example ∫ x√2x + 1 dx yields (2x + 1)3/2 (3x - 1)/15 + C. The key stages are substitution, expansion, term-by-term integration, and back-substitution.
示例 ∫ x√2x + 1 dx 的结果为 (2x + 1)3/2 (3x - 1)/15 + C。关键步骤包括代换、展开、逐项积分和回代。
Always verify your answer by differentiating. A correct antiderivative must reproduce the original integrand exactly under differentiation.
务必通过求导来验证答案。正确的原函数求导后必须完全还原为原被积函数。
Keep practising past-paper questions such as Question 63 to build speed and accuracy for the Edexcel A-Level Mathematics examinations.
持续练习像第63题这样的真题，能够提高你在爱德思A-Level数学考试中的解题速度和准确率。
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