Edexcel A-Level Mathematics: Mastering Integration by Substitution

📚 Edexcel A-Level Mathematics: Mastering Integration by Substitution | 精通换元积分法

Integration by substitution is one of the most powerful techniques in Edexcel A-Level Pure Mathematics. It transforms a difficult integral into a standard form by changing the variable of integration. This method is essential for both Paper 1 and Paper 2 questions involving integration, especially when the integrand contains a function and its derivative.

换元积分法是 Edexcel A-Level 纯数学中最有力的技巧之一。它通过更换积分变量，把一个复杂的积分转化为标准形式。这个方法对于 Paper 1 和 Paper 2 中涉及积分的题目都至关重要，尤其是当被积函数中同时包含某个函数及其导数时。

1. What is Integration by Substitution? | 什么是换元积分法？

Integration by substitution is the reverse process of the chain rule used in differentiation. When a function is differentiated using the chain rule, the derivative contains an inner derivative factor. Substitution reverses this process by identifying that inner function and replacing it with a single variable u.

换元积分法是微分中链式法则的逆过程。当一个函数用链式法则求导时，导数中会包含一个内层函数的导数因子。换元法通过识别这个内层函数并用单一变量 u 替换它，从而逆转求导过程。

The key formula for substitution with u = g(x) is shown below. It states that if we choose u as the inner function, then du = g'(x) dx, and the integral can be rewritten entirely in terms of u.

当设 u = g(x) 时，换元法的核心公式如下。它表明如果我们选择 u 为内层函数，那么 du = g'(x) dx，积分就可以完全用 u 表示。

∫ f(g(x)) g′(x) dx = ∫ f(u) du

In Edexcel exam questions, you will usually be given the substitution, but you may also be asked to spot a suitable substitution yourself. Being confident with both forms is important.

在 Edexcel 考试题中，通常会给出换元变量，但有时也会要求你自己发现合适的换元。对这两种形式都足够熟练非常重要。

2. The Reverse Chain Rule | 反向链式法则

The reverse chain rule is a special case of substitution where the inner function appears with an exact constant multiple of its derivative. For example, in ∫ 2x (x² + 3)⁴ dx, the derivative of x² + 3 is 2x, which is already present in the integrand.

反向链式法则是换元法的一个特例，其中内层函数与它的导数正好以常数倍形式同时出现。例如，在 ∫ 2x (x² + 3)⁴ dx 中，x² + 3 的导数是 2x，它已经出现在被积函数中。

If we set u = x² + 3, then du/dx = 2x, so du = 2x dx. The integral becomes ∫ u⁴ du. This can be integrated directly using the power rule.

如果我们设 u = x² + 3，那么 du/dx = 2x，因此 du = 2x dx。积分就变成 ∫ u⁴ du。这可以用幂法则直接积分。

∫ 2x (x² + 3)⁴ dx = (x² + 3)⁵ / 5 + C

Recognising this pattern saves time, but you must check that the derivative factor is exact. If it is not exact, you may still use substitution by adjusting the constant.

识别这种模式可以节省时间，但你必须检查导数因子是否完全吻合。如果不完全吻合，你仍然可以通过调整常数来使用换元法。

3. Choosing the Substitution u | 选择换元变量 u

When the substitution is not given, a good rule is to choose u as the inside of a composite function. Look for brackets, roots, denominators, or arguments of sin, cos, ln, and exponential functions.

当题目没有给出换元变量时，一个有效的方法是选择复合函数的内部作为 u。寻找括号、根号、分母，或者 sin、cos、ln 和指数函数的自变量。

For example, if the integrand contains √(3x + 1), then letting u = 3x + 1 is often useful. If the integrand contains ln(x), then u = ln(x) may simplify the problem.

例如，如果被积函数含有 √(3x + 1)，那么设 u = 3x + 1 通常是有效的。如果被积函数含有 ln(x)，那么设 u = ln(x) 可能会简化问题。

A standard strategy for choosing u is listed below. These are common inner functions that appear repeatedly in Edexcel integration questions.

下面是选择 u 的标准策略。这些是 Edexcel 积分题中反复出现的常见内层函数。

u = expression inside a bracket raised to a power | u = 括号内并被乘方的表达式
u = expression under a square root | u = 根号内的表达式
u = denominator of a rational function | u = 有理函数的分母
u = angle inside a trigonometric function | u = 三角函数内的角度
u = argument of a logarithm or exponential | u = 对数或指数的自变量

After choosing u, always compute du/dx and rewrite dx in terms of du. Any remaining x terms must be replaced using the substitution equation.

选择 u 之后，一定要计算 du/dx，并把 dx 改写成 du 的形式。任何剩余的 x 项都必须利用换元等式替换掉。

4. Worked Example: Polynomial Substitution | 多项式换元示例

Consider the integral ∫ x √(2x + 1) dx. The square root contains the linear expression 2x + 1, so the natural substitution is u = 2x + 1.

考虑积分 ∫ x √(2x + 1) dx。根号内含有线性表达式 2x + 1，因此自然的换元是 u = 2x + 1。

If u = 2x + 1, then du/dx = 2, so dx = du/2. We also need to express x in terms of u: rearranging u = 2x + 1 gives x = (u – 1)/2.

如果 u = 2x + 1，那么 du/dx = 2，所以 dx = du/2。我们还需要用 u 表示 x：将 u = 2x + 1 移项得 x = (u – 1)/2。

Substituting everything into the integral gives ∫ ((u – 1)/2) √u (du/2). This simplifies to (1/4) ∫ (u – 1) u^(1/2) du.

把所有部分代入积分，得到 ∫ ((u – 1)/2) √u (du/2)。化简为 (1/4) ∫ (u – 1) u^(1/2) du。

Expanding gives (1/4) ∫ (u^(3/2) – u^(1/2)) du. Integrating term by term yields (1/4)( (2/5)u^(5/2) – (2/3)u^(3/2) ) + C.

展开得 (1/4) ∫ (u^(3/2) – u^(1/2)) du。逐项积分得到 (1/4)( (2/5)u^(5/2) – (2/3)u^(3/2) ) + C。

Finally, replace u with 2x + 1 and simplify the constants.

最后，将 u 替换回 2x + 1 并化简常数。

∫ x √(2x + 1) dx = (1/10)(2x + 1)^(5/2) – (1/6)(2x + 1)^(3/2) + C

This example shows that a linear substitution often allows all x terms to be eliminated, leading to a straightforward polynomial integral in u.

这个例子表明，线性换元通常可以消去所有 x 项，从而得到一个关于 u 的简单多项式积分。

5. Definite Integrals and Changing Limits | 定积分与更换上下限

When using substitution on a definite integral, you have two options. You can either convert everything back to the original variable after integrating, or you can change the limits of integration and evaluate directly in terms of u.

在定积分中使用换元法时，你有两种选择。可以在积分后把所有内容换回原变量，也可以直接更换积分上下限并在 u 变量下直接计算。

Changing the limits is usually cleaner and reduces algebraic errors. To change the limits, substitute the original x-values into u = g(x) to find the new lower and upper u-values.

更换上下限通常更简洁，并减少代数错误。要更换上下限，只需将原来的 x 值代入 u = g(x)，得到新的 u 下限和上限。

Example: evaluate ∫₀¹ 2x / (x² + 1)² dx. Let u = x² + 1, then du = 2x dx. The limits change as follows: when x = 0, u = 1; when x = 1, u = 2.

示例：计算 ∫₀¹ 2x / (x² + 1)² dx。设 u = x² + 1，则 du = 2x dx。上下限变化如下：当 x = 0 时，u = 1；当 x = 1 时，u = 2。

The integral becomes ∫₁² u⁻² du. Integrating gives [-u⁻¹] from 1 to 2, which equals -1/2 + 1 = 1/2.

积分变为 ∫₁² u⁻² du。积分后得到 [-u⁻¹] 从 1 到 2 计算，结果为 -1/2 + 1 = 1/2。

∫₀¹ 2x / (x² + 1)² dx = 1/2

Remember: if the substitution changes the limits, you must not use the original x-limits in the u-integral. This is one of the most common errors in A-Level exams.

记住：如果换元改变了上下限，就不能在 u 积分中使用原来的 x 上下限。这是 A-Level 考试中最常见的错误之一。

6. Trigonometric Substitutions | 三角换元

Trigonometric integrals often require substitution when a function of sin or cos appears alongside its derivative. For example, in ∫ sin x cos² x dx, the derivative of cos x is -sin x, so u = cos x is a useful substitution.

三角积分常常需要换元，尤其是当 sin 或 cos 函数与其导数同时出现时。例如，在 ∫ sin x cos² x dx 中，cos x 的导数是 -sin x，因此 u = cos x 是一个有用的换元。

Let u = cos x. Then du/dx = -sin x, so sin x dx = -du. The integral becomes -∫ u² du = -u³/3 + C.

设 u = cos x。那么 du/dx = -sin x，所以 sin x dx = -du。积分变为 -∫ u² du = -u³/3 + C。

∫ sin x cos² x dx = -cos³ x / 3 + C

In other cases, you may need the Pythagorean identity sin² x + cos² x = 1 to rewrite the integrand before substituting. For example, with ∫ sin³ x dx, write sin³ x = sin x (1 – cos² x), then let u = cos x.

在其他情况下，你可能需要先用勾股恒等式 sin² x + cos² x = 1 改写被积函数，然后再进行换元。例如，对于 ∫ sin³ x dx，可写 sin³ x = sin x (1 – cos² x)，然后设 u = cos x。

Edexcel frequently tests trigonometric substitution together with exact values. Always express your final answer in simplest exact form unless the question asks otherwise.

Edexcel 经常将三角换元与精确值一起考查。除非题目另有要求，最终答案始终要写成最简精确形式。

7. Using Substitution with Rational Functions | 有理函数的换元法

Rational functions are fractions where the numerator and denominator are polynomials. Substitution is especially useful when the numerator is a constant multiple of the derivative of the denominator.

有理函数是分子和分母均为多项式的分式。当分子是分母导数的常数倍时，换元法尤其有效。

Consider ∫ (2x + 1) / (x² + x) dx. The derivative of x² + x is 2x + 1, which is exactly the numerator. Let u = x² + x, so du = (2x + 1) dx.

考虑 ∫ (2x + 1) / (x² + x) dx。x² + x 的导数是 2x + 1，正好是分子。设 u = x² + x，则 du = (2x + 1) dx。

The integral becomes ∫ 1/u du = ln |u| + C. Replacing u gives the final answer.

积分变为 ∫ 1/u du = ln |u| + C。将 u 代回即可得到最终答案。

∫ (2x + 1) / (x² + x) dx = ln |x² + x| + C

If the numerator is not exactly the derivative, you may need to separate the fraction or adjust by a constant. For example, x / (x² + 3) dx differs by a factor of 2, so write it as (1/2) ∫ 2x / (x² + 3) dx.

如果分子不恰好是导数，你可能需要拆分分式或通过常数调整。例如，x / (x² + 3) dx 相差一个因子 2，因此可以写成 (1/2) ∫ 2x / (x² + 3) dx。

8. Common Mistakes and How to Avoid Them | 常见错误与避免方法

Substitution questions look straightforward, but small slips can cost many marks. Understanding the most common errors will help you avoid them under exam pressure.

换元题看起来直接，但小失误可能损失很多分数。了解常见错误有助于你在考试压力下避免这些问题。

Forgetting to replace dx with du correctly. | 忘记正确地将 dx 替换为 du。
Leaving original x-limits in a definite integral after substitution. | 换元后在定积分中仍使用原来的 x 上下限。
Not replacing all x terms before integrating. | 积分前没有将所有 x 项替换为 u。
Dropping minus signs when du/dx is negative. | 当 du/dx 为负时丢失负号。
Forgetting the constant of integration in indefinite integrals. | 不定积分中忘记积分常数。
Choosing u poorly so that the integral becomes harder. | 选择 u 不当，导致积分变得更复杂。

A good habit is to write every substitution step clearly, including du = … dx and the new limits. Check your final answer by differentiating to see if you recover the original integrand.

一个好习惯是把每一步换元都写清楚，包括 du = … dx 和新的上下限。通过求导检查你的最终答案是否能还原出原始被积函数。

In Edexcel mark schemes, method marks are awarded for correct substitution and limit-changing even if the final answer is wrong. Therefore, always show your working.

在 Edexcel 评分标准中，即使最终答案错误，只要换元和更换上下限正确，也会给方法分。因此，一定要展示解题过程。

9. Exam Question Walkthrough | 考试真题解析

Let us work through an Edexcel-style exam question. Evaluate the definite integral ∫₀¹ 3x² (x³ + 2)⁴ dx. You are given the substitution u = x³ + 2.

让我们逐步完成一道 Edexcel 风格的考试题。计算定积分 ∫₀¹ 3x² (x³ + 2)⁴ dx。题目给出换元 u = x³ + 2。

First, differentiate u: du/dx = 3x², so du = 3x² dx. This matches exactly the 3x² dx in the integrand, so the substitution is straightforward.

首先对 u 求导：du/dx = 3x²，因此 du = 3x² dx。这与被积函数中的 3x² dx 完全匹配，所以换元非常直接。

Next, change the limits. When x = 0, u = 0³ + 2 = 2. When x = 1, u = 1³ + 2 = 3. The integral becomes ∫₂³ u⁴ du.

接下来，更换上下限。当 x = 0 时，u = 0³ + 2 = 2。当 x = 1 时，u = 1³ + 2 = 3。积分变为 ∫₂³ u⁴ du。

Integrating u⁴ gives u⁵/5. Evaluating from 2 to 3 gives (3⁵/5) – (2⁵/5) = 243/5 – 32/5 = 211/5.

对 u⁴ 积分得到 u⁵/5。从 2 到 3 计算得 (3⁵/5) – (2⁵/5) = 243/5 – 32/5 = 211/5。

∫₀¹ 3x² (x³ + 2)⁴ dx = 211/5

This question would typically be worth four or five marks. You earn marks for correct differentiation of u, correct limit change, correct integration, and correct evaluation.

这类题通常值 4 到 5 分。正确对 u 求导、正确更换上下限、正确积分以及正确求值都能得分。

10. Strategy Summary and Practice Tips | 策略总结与练习建议

To master integration by substitution, follow a consistent strategy. First, inspect the integrand for a candidate inner function. Then compute du and rewrite dx. Replace all x terms with u, and if the integral is definite, change the limits.

要掌握换元积分法，请遵循一套固定策略。首先，检查被积函数，寻找可能的内层函数。然后计算 du 并改写 dx。将所有 x 项替换为 u；如果是定积分，还要更换上下限。

After integrating, always remember to substitute back for indefinite integrals. For definite integrals with changed limits, no back-substitution is needed.

积分之后，对于不定积分，一定要记得回代原变量。对于已经更换上下限的定积分，则无需回代。

Practice with a mix of polynomial, trigonometric, rational, and square-root integrands. Edexcel often embeds substitution inside larger problems such as finding the area under a curve or solving a differential equation.

练习时要混合多项式、三角、有理式和根号被积函数。Edexcel 经常把换元法嵌入更大的问题中，例如求曲线下方面积或求解微分方程。

Finally, check your work by differentiating your result. For example, differentiate (x³ + 2)⁵ / 5 and confirm that you get 3x² (x³ + 2)⁴. This verification helps build confidence and sharpen algebraic skills.

最后，通过求导检查你的答案。例如，求 (x³ + 2)⁵ / 5 的导数，确认你是否得到 3x² (x³ + 2)⁴。这种验证有助于建立信心并提高代数技能。

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Edexcel A-Level Mathematics: Mastering Integration by Substitution | 精通换元积分法