Example 5.6.3: Integration by Substitution | 换元积分法示例

📚 Example 5.6.3: Integration by Substitution | 换元积分法示例

This worked example demonstrates how to evaluate a tricky-looking integral by making a clever substitution. The method is essential for AQA A-Level Mathematics, particularly in Paper 1 and Paper 2. We will break down the solution into clear, exam-ready steps and explain the reasoning behind each one.

本例题展示如何通过巧妙的换元来求解一个看似复杂的积分。该方法在 AQA A-Level 数学中至关重要，尤其在 Paper 1 和 Paper 2 中。我们将把解答拆分为清晰的、符合考试要求的步骤，并解释每一步背后的逻辑。

1. The Problem | 问题陈述

Evaluate the following indefinite integral:

求解以下不定积分：

∫ x√(x² + 1) dx

We are asked to find an antiderivative of the function f(x) = x√(x² + 1). At first glance, this expression appears to involve a product of two different functions, so you might think about integration by parts. However, a much simpler approach is to notice that the derivative of the inside of the square root is 2x, and we have an x sitting outside. This suggests substitution.

题目要求我们求函数 f(x) = x√(x² + 1) 的一个原函数。乍一看，这个表达式似乎涉及两个不同函数的乘积，因此你可能会想到分部积分法。然而，更简单的方法是注意到根号内部的导数是 2x，而外面恰好有一个 x。这提示我们可以使用换元法。

2. Choosing the Substitution | 选择换元

We aim to simplify the integrand by choosing a new variable u that captures the ‘inner’ function. Since √(x² + 1) contains x² + 1, we set:

我们的目标是通过选择一个新变量 u 来简化被积函数，让它捕捉‘内部’函数。由于 √(x² + 1) 包含 x² + 1，我们设：

u = x² + 1

Why this choice? Because the derivative of u with respect to x is du/dx = 2x. This derivative is almost exactly the x factor in the integrand, apart from a constant factor of 2. We can use this to replace x dx in the integral.

为什么这样选？因为 u 对 x 的导数为 du/dx = 2x。该导数几乎正好是被积函数中的 x 因子，只是差了一个常数因子 2。我们可以据此用 x dx 替换积分中的项。

3. Expressing dx in Terms of u | 用 u 表示 dx

From u = x² + 1, we differentiate both sides with respect to x:

由 u = x² + 1，对 x 两边求导：

du/dx = 2x

This can be rearranged to give:

这可以变形得到：

du = 2x dx

Therefore x dx = ½ du. This is the key algebraic step. We have now expressed the original ‘x dx’ part entirely in terms of u and du.

因此 x dx = ½ du。这是关键的代数步骤。我们现在已经将原始中的“x dx”部分完全用 u 和 du 表示。

4. Rewriting the Integral | 重写积分

Original integral: ∫ x√(x² + 1) dx. Substitute u = x² + 1 and x dx = ½ du:

原积分：∫ x√(x² + 1) dx。代入 u = x² + 1 以及 x dx = ½ du：

∫ x√(x² + 1) dx = ∫ √u · (½ du) = ½ ∫ u^{½} du

Notice that the variable x has completely disappeared inside the integral – everything is now in terms of u. This is the moment when the substitution has paid off.

注意，变量 x 已经从被积函数中完全消失——现在一切都以 u 表示。这正是换元法发挥效率的时刻。

5. Integrating with Respect to u | 对 u 积分

Now we use the standard power rule for integration. Recall that for any n ≠ −1:

现在我们使用标准幂规则进行积分。回顾对于任何 n ≠ −1：

∫ uⁿ du = uⁿ⁺¹/(n+1) + C

Here n = ½, so n + 1 = 3/2. Therefore:

这里 n = ½，因此 n + 1 = 3/2。所以：

∫ u^{½} du = u^{3/2}/(3/2) = (2/3) u^{3/2} + C

Multiplying by the factor ½ that we took out earlier gives:

乘上之前提取的因子 ½，得到：

½ · (2/3) u^{3/2} + C = (1/3) u^{3/2} + C

6. Substituting Back | 回代

We cannot leave the answer in terms of u; we must return to x. Since u = x² + 1, we substitute back:

我们不能将答案留成 u 的形式，必须返回 x。由于 u = x² + 1，我们回代：

∫ x√(x² + 1) dx = (1/3)(x² + 1)^{3/2} + C

This is the final answer for the indefinite integral. The constant C is included because the derivative of a constant is zero.

这就是该不定积分的最终答案。常数 C 必须包含，因为常数的导数为零。

7. Verification | 验证

We should always check our result by differentiating. Differentiate F(x) = (1/3)(x² + 1)^{3/2} + C. Use the chain rule:

我们总是应该通过求导来检查结果。对 F(x) = (1/3)(x² + 1)^{3/2} + C 求导。使用链式法则：

d/dx [(1/3)(x² + 1)^{3/2}] = (1/3) · (3/2)(x² + 1)^{1/2} · 2x

Simplify: (1/3) · (3/2) · 2x = x. The derivative is x√(x² + 1), exactly the original integrand. So our answer is correct.

化简： (1/3) · (3/2) · 2x = x。导数为 x√(x² + 1)，恰好是原被积函数。所以我们的答案是正确的。

8. Common Mistakes | 常见错误

Students often make the following errors when using substitution:

学生在使用换元法时常犯以下错误：

Forgetting to replace dx completely – you must express dx in terms of du, not just substitute the inner function.
缺漏 dx 的替换——必须将 dx 用 du 表达，而不仅仅是替换内部函数。
Forgetting the constant factor when du = k·g'(x) dx. Here, x dx = ½ du, not x dx = du.
忽略常数因子。本题中 x dx = ½ du，而不是 x dx = du。
Not substituting back to x at the end, leaving the answer in u.
最后忘记回代，将答案留于 u。
Omitting the constant of integration for an indefinite integral.
对于不定积分遗漏积分常数。

9. When to Use This Substitution | 何时使用这种换元

This type of substitution works when you see a function and its derivative (up to a constant factor) appearing elsewhere in the integrand. In the example, x² + 1 and its derivative 2x appear as the square root and as the factor x.

这种换元适用于当你看到某个函数及其导数（相差一个常数因子）同时出现在被积函数中时。本例中，x² + 1 及其导数 2x 分别以根号和因子 x 的形式出现。

General pattern: if you see g'(x) · f(g(x)), try substitution u = g(x). This is often called ‘reverse chain rule’ or ‘integration by substitution’.

一般模式：如果看到 g'(x) · f(g(x))，尝试设 u = g(x)。这通常被称为“反向链式法则”或“换元积分法”。

10. Further Practice | 拓展练习

Try these similar integrals on your own to master the technique:

请自行尝试以下类似积分以掌握该技巧：

∫ 2x e^{x²} dx
∫ 3x² (x³ + 2)⁴ dx
∫ cos x e^{sin x} dx
∫ (2x)/(x² + 3) dx

For each one, identify the ‘inner function’ u, then rewrite and integrate. You can verify your answers by differentiation.

对于每一题，找出‘内部函数’ u，然后重写并积分。你可以通过求导来验证答案。

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Example 5.6.3: Integration by Substitution | 换元积分法示例