Integration by Substitution | 换元积分法

📚 Integration by Substitution | 换元积分法

Integration by substitution is one of the most important techniques in A-Level Mathematics. It allows you to turn a complicated integral into a simpler standard form by replacing part of the integrand with a new variable u. This method appears frequently in Edexcel Pure Mathematics papers, especially in questions on polynomials, roots, fractions, exponentials and trigonometric functions.

换元积分法是 A-Level 数学中最重要的技巧之一。它通过将被积表达式中的某一部分替换为一个新变量 u，把一个复杂的积分转化为更简单的标准形式。这种方法在 Edexcel 纯数学试卷中非常常见，尤其是在涉及多项式、根式、分式、指数函数和三角函数的题目中。

1. The Core Idea: Reverse Chain Rule | 核心思想：逆链式法则

When you differentiate a composite function y = f(g(x)), the chain rule gives dy/dx = f'(g(x)) g'(x). Therefore, integrating an expression of the form f'(g(x)) g'(x) gives f(g(x)) + C. Substitution is the systematic way to identify the inner function g(x) as u, then rewrite the whole integral in terms of u and du.

当你对复合函数 y = f(g(x)) 求导时，链式法则给出 dy/dx = f'(g(x)) g'(x)。因此，对形如 f'(g(x)) g'(x) 的表达式积分，结果是 f(g(x)) + C。换元法就是一种系统化的方法：先把内层函数 g(x) 设为 u，然后把整个积分用 u 和 du 重新表示。

∫ f'(g(x)) g'(x) dx = f(g(x)) + C

∫ f'(g(x)) g'(x) dx = f(g(x)) + C

2. Choosing the Right u | 如何选择替换变量 u

A useful rule is to let u be the function inside brackets, under a root, in a denominator, or in an exponent. You also need the derivative du/dx to appear in the integrand, possibly up to a constant factor. The table below shows common choices.

一个实用的法则是：令 u 为括号内、根号下、分母中或指数中的函数。同时，你需要确保被积函数中能找到 du/dx 本身或它与某个常数倍的乘积。下表列出了一些常见的选择。

3. Worked Example 1: Power of a Binomial | 例题 1：二项式的幂

Find ∫ 6x(3x² + 5)⁴ dx. First let u = 3x² + 5. Then du/dx = 6x, so du = 6x dx. The integral becomes ∫ u⁴ du, which is easy to integrate. Finally, replace u with 3x² + 5.

求 ∫ 6x(3x² + 5)⁴ dx。首先令 u = 3x² + 5。则 du/dx = 6x，因此 du = 6x dx。原积分变为 ∫ u⁴ du，这很容易积分。最后把 u 换回 3x² + 5。

∫ 6x(3x² + 5)⁴ dx = ∫ u⁴ du = u⁵/5 + C = (3x² + 5)⁵/5 + C

∫ 6x(3x² + 5)⁴ dx = ∫ u⁴ du = u⁵/5 + C = (3x² + 5)⁵/5 + C

4. Worked Example 2: Missing Constant Factor | 例题 2：缺少常数倍

Consider ∫ x√(x² + 9) dx. Let u = x² + 9, so du/dx = 2x, giving du = 2x dx. Since the integrand contains x dx, write x dx = du/2. The integral becomes ½ ∫ √u du. Integrating √u gives (2/3)u√u, so the final result is ⅓ (x² + 9)√(x² + 9) + C.

考虑 ∫ x√(x² + 9) dx。令 u = x² + 9，因此 du/dx = 2x，即 du = 2x dx。因为被积函数中含有 x dx，可以写成 x dx = du/2。积分变为 ½ ∫ √u du。对 √u 积分得到 (2/3)u√u，所以最终结果是 ⅓ (x² + 9)√(x² + 9) + C。

∫ x√(x² + 9) dx = ½ ∫ √u du = ⅓ u√u + C = ⅓ (x² + 9)√(x² + 9) + C

∫ x√(x² + 9) dx = ½ ∫ √u du = ⅓ u√u + C = ⅓ (x² + 9)√(x² + 9) + C

5. Definite Integrals and Changing Limits | 定积分与换限

For definite integrals, you have two options. You can substitute back to x and then use the original x limits, or you can convert the limits to u values and never return to x. The second method is usually faster and reduces the chance of sign errors.

对于定积分，你有两种选择。你可以在换回 x 后再代入原来的 x 上下限，也可以把上下限直接转换为 u 的值，不再回到 x。第二种方法通常更快，也能减少符号错误。

Find ∫₀¹ 4x(x² + 1)³ dx. Let u = x² + 1, so du = 2x dx, which means 4x dx = 2 du. When x = 0, u = 1; when x = 1, u = 2. The integral becomes ∫₁² 2u³ du. This evaluates to 2[u⁴/4] from 1 to 2, giving ½(16 − 1) = 15/2.

求 ∫₀¹ 4x(x² + 1)³ dx。令 u = x² + 1，则 du = 2x dx，因此 4x dx = 2 du。当 x = 0 时，u = 1；当 x = 1 时，u = 2。积分变为 ∫₁² 2u³ du。计算结果为 2[u⁴/4] 从 1 到 2，得到 ½(16 − 1) = 15/2。

∫₀¹ 4x(x² + 1)³ dx = ∫₁² 2u³ du = [u⁴/2]₁² = 16/2 − 1/2 = 15/2

∫₀¹ 4x(x² + 1)³ dx = ∫₁² 2u³ du = [u⁴/2]₁² = 16/2 − 1/2 = 15/2

6. Fractions and Logarithmic Integrals | 分式与对数积分

When the numerator of a fraction is exactly the derivative of the denominator,

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