📚 How to Solve Example 7.4.2: Integration by Substitution in AQA A-Level Maths | 如何解决AQA A-Level数学例题7.4.2：换元积分法
In this worked example, we will tackle a classic AQA A-Level Mathematics problem that requires the method of integration by substitution. The example is designed to test your ability to recognise when a substitution is helpful, to transform the integral correctly, and to evaluate it carefully. We will break the solution down into clear steps so that you can follow the reasoning and apply it to similar problems in your own revision.
在这个例题中，我们将处理一个典型的AQA A-Level数学问题，它需要运用换元积分法。该例题旨在测试你是否能识别何时使用换元法、正确转换积分并仔细求值。我们将把解答分解成清晰的步骤，帮助你理解推理过程，并能在复习中将其应用于类似问题。
1. Problem Statement | 题目陈述
We are asked to evaluate the definite integral:
∫₀⁴ x/(2x+1) dx
我们要求计算以下定积分：
∫₀⁴ x/(2x+1) dx
This integral is not an immediate standard formula, so we need to use a substitution to make it easier to handle.
这个积分不是直接的标准公式，所以我们需要使用换元法使其更容易处理。
2. Recognising When to Use Substitution | 识别何时使用换元法
Integration by substitution is useful when the integrand contains a composite function, or when a linear denominator with a non-constant numerator suggests that simplifying the variable will help. Here the denominator is 2x + 1, which is linear, but the numerator contains x, so the fraction is not immediately integrable as a simple logarithm.
当被积函数包含复合函数，或者当线性分母与非常数分子组合使变量化简有帮助时，换元积分法很有用。这里分母是 2x + 1，它是线性的，但分子包含 x，因此该分式不能直接作为简单对数积分。
The presence of a linear expression such as 2x + 1 suggests that setting u = 2x + 1 could simplify the integral, because the derivative of 2x + 1 is a constant, which will pair neatly with the dx term.
出现如 2x + 1 这样的线性表达式，提示我们设 u = 2x + 1 可以简化积分，因为 2x + 1 的导数是常数，会与 dx 项很好地配合。
3. Choosing the Substitution | 选择换元
Let us set:
我们设：
u = 2x + 1
It is essential to express x in terms of u, and also to find dx in terms of du. From the substitution we have:
必须用 u 表示 x，同时用 du 表示 dx。由换元可得：
x = (u − 1)/2
dx = du/2
These two relations will allow us to replace every occurrence of x and dx in the original integral.
这两个关系式使我们能够替换原积分中所有出现的 x 和 dx。
4. Rewriting the Integral | 改写积分
Substituting x = (u − 1)/2 and dx = du/2 into the integral gives:
将 x = (u − 1)/2 和 dx = du/2 代入积分，得到：
∫ x/(2x+1) dx = ∫ [ (u − 1)/2 ] / u · du/2
Simplifying the fraction inside the integral:
简化积分内的分式：
= ∫ (u − 1)/(4u) du
We can split the fraction into two separate terms:
我们可以将分式拆成两项：
= (1/4) ∫ (1 − 1/u) du
This new form is much easier to integrate.
这个新形式更容易积分。
5. Changing the Limits | 转换积分限
Because we are working with a definite integral, the limits of integration must be changed from x to u. Using the substitution u = 2x + 1:
由于我们处理的是定积分，积分限必须从 x 转换为 u。利用换元 u = 2x + 1：
When x = 0: u = 2(0) + 1 = 1
当 x = 0 时：u = 2(0) + 1 = 1
When x = 4: u = 2(4) + 1 = 9
当 x = 4 时：u = 2(4) + 1 = 9
Therefore the definite integral becomes:
因此定积分变为：
∫₁⁹ (1/4)(1 − 1/u) du
Notice that we do not need to switch back to x at the end because the limits are now in terms of u.
注意我们最后不需要再换回 x，因为上下限现在是用 u 表示的。
6. Integrating the Simplified Expression | 对简化表达式积分
Now we integrate each term separately. Recall that the integral of 1 with respect to u is u, and the integral of 1/u with respect to u is ln|u| (with the absolute value). Since our limits are positive, we can drop the absolute value in this case.
现在我们分别对每一项积分。回顾一下，1 对 u 的积分是 u，1/u 对 u 的积分是 ln|u|（带绝对值）。由于我们的积分限为正，此处可以去掉绝对值。
(1/4) ∫ (1 − 1/u) du = (1/4)(u − ln u) + C
For a definite integral, we will apply the limits directly to this antiderivative.
对于定积分，我们将直接对原函数应用上下限。
7. Applying the Limits | 应用上下限
We now evaluate the antiderivative F(u) = (1/4)(u − ln u) between u = 1 and u = 9. By the fundamental theorem of calculus:
我们现在计算原函数 F(u) = (1/4)(u − ln u) 在 u = 1 到 u = 9 之间的值。根据微积分基本定理：
∫₁⁹ (1/4)(1 − 1/u) du = F(9) − F(1)
First compute F(9):
首先计算 F(9)：
F(9) = (1/4)(9 − ln 9)
Then compute F(1):
然后计算 F(1)：
F(1) = (1/4)(1 − ln 1) = (1/4)(1 − 0) = 1/4
Therefore:
因此：
F(9) − F(1) = (1/4)(9 − ln 9) − 1/4 = (1/4)(8 − ln 9)
So the exact value of the integral is:
所以积分的精确值为：
2 − (1/4) ln 9
Using the logarithmic identity ln 9 = 2 ln 3, we can also write this as:
使用对数恒等式 ln 9 = 2 ln 3，我们也可以写为：
2 − (1/2) ln 3
8. Final Answer and Verification | 最终答案与验证
Thus the solution to Example 7.4.2 is:
因此例题7.4.2的解为：
∫₀⁴ x/(2x+1) dx = 2 − (1/2) ln 3 ≈ 1.4507
To verify, we can differentiate 2 − (1/2) ln 3? Wait, that is the final constant value, not the antiderivative. Instead, differentiate the antiderivative (1/4)(u − ln u) with respect to u to confirm it gives the integrand in u. We have d/du [ (1/4)(u − ln u) ] = (1/4)(1 − 1/u), which matches the simplified integrand. This confirms our integration is correct.
为了验证，我们能否对 2 − (1/2) ln 3 微分？等等，那是最终常数值，不是原函数。正确的做法是对原函数 (1/4)(u − ln u) 关于 u 微分，确认得到的是 u 中的被积函数。我们有 d/du [ (1/4)(u − ln u) ] = (1/4)(1 − 1/u)，这与简化后的被积函数一致，这确认了我们的积分是正确的。
Additionally, a numerical approximation using a calculator gives 1.4507, which is consistent with the approximate area under the curve from x = 0 to x = 4.
此外，用计算器进行数值近似得到 1.4507，这与从 x = 0 到 x = 4 的曲线下面积近似值一致。
9. Common Pitfalls | 常见误区
There are several common mistakes that students make when solving problems like Example 7.4.2. Being aware of them can help you avoid losing marks.
学生在解决类似例题7.4.2的问题时常常会犯一些典型错误。了解这些错误有助于你避免失分。
- Forgetting to change the limits: Always convert the x-limits to u-limits using the substitution equation. If you forget, the final answer will be incorrect.
- Incorrectly solving for dx: From u = 2x + 1, we get du = 2 dx, so dx = du/2. Some students write dx = du or forget the factor of 1/2.
- Algebra errors when rewriting x: Since x = (u − 1)/2, substituting this into the numerator is a common source of mistakes. Double-check each step.
- 忘记转换积分限：一定要用换元方程将 x 的上下限转换为 u 的上下限。如果忘记，最终答案将是错误的。
- 错误求解 dx：由 u = 2x + 1 得 du = 2 dx，所以 dx = du/2。有些学生会写成 dx = du 或忘记系数 1/2。
- 改写 x 时的代数错误：因为 x = (u − 1)/2，将其代入分子时是常见错误来源。请仔细检查每一步。
10. Alternative Methods | 其他方法
Although substitution is the most direct method for this integral, it is worth noting that we could also use algebraic division first. Since the degree of the numerator is less than the degree of the denominator? Actually, the numerator is degree 1 and denominator degree 1, so we could rewrite the fraction as a constant plus a simpler rational expression.
虽然换元法是这个积分最直接的方法，但值得注意的是我们可以先进行代数除法。实际上分子的次数是1，分母的次数也是1，所以我们可以将分式化为一个常数加上一个更简单的有理式。
For example, we could write:
例如，我们可以写成：
x/(2x+1) = 1/2 − 1/(2(2x+1))
Check: 1/2 − 1/(2(2x+1)) = (2x+1 − 1)/(2(2x+1)) = x/(2x+1). This is correct.
检验：1/2 − 1/(2(2x+1)) = (2x+1 − 1)/(2(2x+1)) = x/(2x+1)。这是正确的。
Then integrating directly would give:
然后直接积分得到：
∫₀⁴ [1/2 − 1/(2(2x+1))] dx = [x/2 − (1/4)ln(2x+1)]₀⁴ = (2 − (1/4)ln9) − (0 − (1/4)ln1) = 2 − (1/4)ln9
This agrees with our previous answer, confirming the result. This alternative method is faster once you see the algebraic simplification.
这与我们之前的答案一致，验证了结果。这种替代方法在你看出代数化简后会更快捷。
11. Practice Questions | 练习题
To master this skill, try solving similar integrals yourself. Below are a few exercises that use the same technique of integration by substitution.
为了掌握这项技能，请尝试自己解决类似的积分。以下是一些使用相同换元积分技巧的练习。
- ∫₀² x/(3x+1) dx (Hint: Let u = 3x + 1)
- ∫₁⁵ x²/(2x+1) dx (Hint: Let u = 2x + 1, then express x² in terms of u)
- ∫₀¹ x/(√(x+1)) dx (Hint: Let u = x + 1)
- ∫₀² x/(3x+1) dx（提示：设 u = 3x + 1）
- ∫₁⁵ x²/(2x+1) dx（提示：设 u = 2x + 1，然后用 u 表示 x²）
- ∫₀¹ x/(√(x+1)) dx（提示：设 u = x + 1）
Work through each step carefully: choose the substitution, compute dx, change the limits, simplify, integrate, and evaluate. Then check your answers by differentiation or using a calculator.
请仔细完成每一步：选择换元、计算 dx、转换积分限、化简、积分并求值。然后通过微分或使用计算器检查答案。
12. Summary | 总结
In this article, we solved Example 7.4.2 by applying integration by substitution. We set u = 2x + 1, rewrote the integral, changed the limits, integrated, and obtained the exact answer 2 − (1/2) ln 3. We also discussed common pitfalls and verified the result using an alternative method.
在本文中，我们通过换元积分法解决了例题7.4.2。我们设 u = 2x + 1，改写积分，转换上下限，积分，并得到了精确答案 2 − (1/2) ln 3。我们还讨论了常见误区，并使用另一种方法验证了结果。
The key to success with substitution is to be systematic and careful with algebra. With enough practice, you will be able to recognise the right substitution quickly and apply it confidently in your exams.
换元法成功的关键是系统化并仔细处理代数运算。通过足够多的练习，你将能够快速识别正确的换元，并在考试中自信地应用它。
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