📚 Example 5.6.1 – Integration by Substitution | 例题 5.6.1 – 换元积分法
This worked example demonstrates a core technique in A-Level Mathematics: integration by substitution. We will solve a typical AQA-style problem that requires a careful choice of substitution, and then verify the result by differentiation.
这个例题展示了 A-Level 数学中的一个核心技巧：换元积分法。我们将解决一个典型的 AQA 风格题目，需要仔细选择换元，并通过求导验证结果。
1. The Problem | 题目
Evaluate the following indefinite integral:
计算以下不定积分：
∫ 2x(x² + 1)⁵ dx
We aim to find the most efficient method to solve this integral, which is a classic example of when a simple substitution dramatically simplifies the expression.
我们的目标是找到解决这个积分最有效的方法，这是一个经典例子，说明简单的换元可以极大地简化表达式。
2. Identifying the Substitution | 确定换元
Look for a function and its derivative inside the integrand. Here we have x² + 1, and its derivative is 2x. This is exactly the factor outside the bracket.
寻找被积函数中一个函数及其导数。这里有 x² + 1，其导数是 2x。这正是括号外的因子。
Let us set:
我们设：
u = x² + 1
Then the differential is:
则微分为：
du/dx = 2x ⇒ du = 2x dx
3. Rewriting the Integral | 重写积分
Substitute u and du into the original integral:
将 u 和 du 代入原积分：
∫ 2x(x² + 1)⁵ dx = ∫ u⁵ du
Notice how the entire integrand has been reduced to a simple power of u. This is the essence of the substitution method.
注意整个被积函数如何被简化成 u 的简单幂。这是换元法的精髓。
4. Integrating with Respect to u | 对 u 积分
The integral of u⁵ with respect to u is straightforward:
u⁵ 对 u 的积分是直接的：
∫ u⁵ du = u⁶/6 + C
Remember to add the constant of integration C for indefinite integrals.
不要忘记为不定积分加上积分常数 C。
5. Substituting Back | 回代
Now replace u with the original expression x² + 1:
现在用原始表达式 x² + 1 替换 u：
(x² + 1)⁶ / 6 + C
This is the final answer to the integration problem.
这就是积分问题的最终答案。
6. Verification by Differentiation | 通过求导验证
To ensure our answer is correct, we differentiate the result:
为了确保答案正确，我们对结果求导：
d/dx [(x² + 1)⁶ / 6 + C] = (1/6) × 6(x² + 1)⁵ × 2x = 2x(x² + 1)⁵
This matches the original integrand, confirming our solution is correct.
这与原始被积函数一致，确认我们的解是正确的。
7. Common Mistakes | 常见错误
Here are typical errors students make with this type of question:
以下是学生在处理这类问题时常见的错误：
- Forgetting the constant of integration C.
- 不要忘记积分常数 C。
- Incorrectly computing du/dx, especially in more complex substitutions.
- 错误计算 du/dx，尤其是在更复杂的换元中。
- Failing to substitute back fully, leaving part of the answer in terms of u.
- 未能完全回代，使部分答案仍包含 u。
- Altering the coefficient 2x instead of recognizing it as exactly du/dx.
- 改变系数 2x，而不是识别出它正好是 du/dx。
8. When to Use Substitution | 何时使用换元法
Use this method when you spot a function and its derivative (up to a constant factor) in the integrand. For example:
当你在被积函数中发现一个函数及其导数（相差一个常数因子）时，使用这种方法。例如：
∫ f'(x) [f(x)]ⁿ dx
This pattern is a direct signal for the substitution u = f(x).
这种模式是换元 u = f(x) 的直接信号。
9. Definite Integrals with Substitution | 定积分的换元法
For definite integrals, we have two options: either integrate and substitute back, or change the limits. Here is the same example as a definite integral:
对于定积分，我们有两个选择：先积分再回代，或者更换积分限。下面是同一个例子的定积分版本：
∫₀¹ 2x(x² + 1)⁵ dx
Using u = x² + 1, when x = 0, u = 1; when x = 1, u = 2.
使用 u = x² + 1，当 x = 0 时，u = 1；当 x = 1 时，u = 2。
∫₁² u⁵ du = [u⁶/6]₁² = 64/6 − 1/6 = 63/6 = 10.5
10. Summary of Steps | 步骤总结
Follow these steps for any substitution problem:
对于任何换元问题，遵循以下步骤：
- Identify a suitable function u.
- 识别一个合适的函数 u。
- Compute du/dx and rearrange to find dx in terms of du.
- 计算 du/dx 并重新排列以用 du 表示 dx。
- Rewrite the entire integral in terms of u.
- 用 u 重写整个积分。
- Integrate with respect to u.
- 对 u 积分。
- Substitute back or change the limits for definite integrals.
- 回代或为定积分更换积分限。
11. Practice Problem | 练习题目
Try this on your own: evaluate ∫ 3x²(x³ + 2)⁴ dx using the same method.
请自己尝试：用同样的方法计算 ∫ 3x²(x³ + 2)⁴ dx。
Solution hint: let u = x³ + 2, then du = 3x² dx.
解题提示：设 u = x³ + 2，则 du = 3x² dx。
The answer is (x³ + 2)⁵/5 + C.
答案是 (x³ + 2)⁵/5 + C。
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