📚 Integration by Substitution | 代入积分法
Integration by substitution is the reverse process of the chain rule in differentiation and serves as one of the most powerful techniques for evaluating complex integrals. It simplifies an integral by changing the variable of integration to a new variable, usually denoted by u, making the integrand easier to handle. For Edexcel A-Level Mathematics, mastering this method is essential both for pure mathematics and for solving differential equations in mechanics.
代入积分法是微分的链式法则的逆过程，也是计算复杂积分最强大的技巧之一。它通过将积分变量替换为一个新变量（通常用 u 表示）来简化积分表达式，使被积函数更容易处理。对 Edexcel A-Level 数学而言，掌握这种方法不仅对纯数学至关重要，对解决力学中的微分方程也同样重要。
1. Reversing the Chain Rule | 链式法则的逆过程
act that if we differentiate a composite function y = f(g(x)), the derivative is dy/dx = f'(g(x)) • g'(x). Reversing this, an integral of the form ∫ f'(g(x)) • g'(x) dx can be written as f(g(x)) + C. When we make a substitution u = g(x), we replace 'g(x)' with u and 'g'(x) dx' with du, converting the integral into ∫ f'(u) du, which is straightforward to solve.
代入法建立在这样一个事实上：如果我们对复合函数 y = f(g(x)) 求导，得到 dy/dx = f'(g(x)) • g'(x)。逆向操作，形如 ∫ f'(g(x)) • g'(x) dx 的积分可以写成 f(g(x)) + C。当我们作代换 u = g(x) 时，我们用 u 替换 'g(x)'，并用 du 替换 'g'(x) dx'，从而将积分转化为 ∫ f'(u) du，计算起来就非常简单。
2. Choosing the Substitution u | 选择代换变量 u
e derivative also appears (up to a constant factor) in the integrand. For example, in ∫ 2x• if the derivative differs by a constant; we can always adjust with a constant multiplier.
成功运用此方法的关键在于选择一个合适的表达式作为 u。通常我们寻找一个内层函数，且其导数（可能相差一个常数倍）也出现在被积表达式中。例如，在 ∫ 2x•(x²+1)³ dx 中，选取 u = x²+1 则 du/dx = 2x，故 du = 2x dx，与多余的因子完全吻合。如果导数相差一个常数倍也不必担心，我们总可以用常数系数进行调整。
3. Step-by-Step Procedure | 分步求解步骤
Follow these steps when applying integration by substitution:
- Step 1: Identify a suitable substitution u = g(x) and compute du/dx.
- Step 2: Express dx in terms of du: dx = du / (du/dx).
- Step 3: Rewrite the entire integrand in terms of u only – cancel the x's if possible.
- Step 4: Integrate with respect to u.
- Step 5: final answer in terms of the original variable x. In the case of a definite integral, change the limits from x to u and evaluate without converting back.
在使用代入积分法时请遵循以下步骤：
- 步骤一：确定一个合适的代换 u = g(x) 并计算 du/dx。
- 步骤二：用 du 表示 dx：dx = du / (du/dx)。
- 步骤三：将整个被积函数仅用 u 表达——若可能则消去所有 x。
- 步骤四：对 u 进行积分。
- 步骤五：将 u = g(x) 代回，得到关于原变量 x 的最终答案。对于定积分，直接将 x 的上下限转化为 u 的上下限，计算后无需代回。
4. Worked Example: Polynomial Inner Function | 例题：多项式内层函数
nsider ∫ 6x (x² + 1)⁴ dx. Let u = x² + 1, so du/dx = 2x, dx = du/(2x). Substitute: 6x (u)⁴ • ives (3/5)(x²+1)⁵ + C. You can check by differentiating the result: the chain rule gives 5 • (x²+1)⁴ • 2x • (3/5) = 6x (x²+1)⁴, confirming correctness.
考虑积分 ∫ 6x (x² + 1)⁴ dx。令 u = x² + 1，则 du/dx = 2x，dx = du/(2x)。代入：6x (u)⁴ • du/(2x) = 3u⁴ du。积分得到 (3/5) u⁵ + C。代回 x 得 (3/5)(x²+1)⁵ + C。你可以通过求导验证：链式法则给出 5 • (x²+1)⁴ • 2x • (3/5) = 6x (x²+1)⁴，结果正确。
5. Handling Definite Integrals with Changed Limits | 定积分的换限操作
ntegration to the u-variable rather than converting back to x. For ∫ from x=0 to x=1 of 2x √x²+1 1. When x=0, u=1; when x=1, u=2. Since du = 2x dx, the integral becomes ∫ from u=1 to 2 of √u du = [ (2/3) u3/2 ] from 1 to 2 = (2/3)(2√2 - 1). There is no need to reintroduce x.
对于定积分，通常更高效的做法是将积分上下限转换为 u 变量的值，而不必代回 x。例如，计算 ∫ 从 x=0 到 x=1 的 2x √x²+1 dx，令 u = x²+1。当 x=0 时，u=1；当 x=1 时，u=2。由于 du = 2x dx，原积分变为 ∫ u=1 到 2 的 √u du = [ (2/3) u3/2 ] 从 1 到 2 = (2/3)(2√2 - 1)。无需再引入 x。
6. Substitutions Involving Trigonometric Functions | 含三角函数的代换
= (1/4) sin⁴ x + C. In cases like ∫ tan x dx, write as ∫ (sin x / cos x) dx and set u = cos x, du = -sin x dx, yielding -∫ 1/u du = -ln|u| + C = -ln|cos x| + C = ln|sec x| + C. Familiarity with derivatives of trig functions is key.
含三角函数的积分常能通过代换简化。对 ∫ sin³ x cos x dx，令 u = sin x，则 du = cos x dx。积分变为 ∫ u³ du = (1/4)u⁴ + C = (1/4) sin⁴ x + C。对于类似 ∫ tan x dx 的情形，写作 ∫ (sin x / cos x) dx 并令 u = cos x，du = -sin x dx，得到 -∫ 1/u du = -ln|u| + C = -ln|cos x| + C = ln|sec x| + C。熟悉三角函数的导数是关键。
7. The ‘Hidden’ Substitution: Factor Inside a Function | 隐藏的代换：函数内部的因子
= 3x² + 5, du/dx = 6x. We have x dx but need 6x dx, so scale by 1/6: x dx = (1/6) du. The integral becomes (1/6) ∫ u⁶ du = (1/42) u⁷ + C = (1/42)(3x²+5)⁷ + C. Always remember to adjust with a constant multiplier when the derivative does not match exactly.
有时内层函数的导数不那么显眼，因为差了一个常系数。例如 ∫ x (3x² + 5)⁶ dx。令 u = 3x² + 5，du/dx = 6x。我们有 x dx 但需要 6x dx，所以乘以 1/6：x dx = (1/6) du。积分变为 (1/6) ∫ u⁶ du = (1/42) u⁷ + C = (1/42)(3x²+5)⁷ + C。当导数不完全匹配时，务必记得用常系数进行调整。
8. Substitution for Rational and Radical Expressions | 有理式与根式表达式的代换
unctions and those involving square roots are prime candidates for substitution. For ∫ x / √2x+1 dx, set u = 2x+1, so x = (u-1)/2 and dx = du/2. The integrand becomes [(u-1)/2] / √u • (du/2) = (1/4) (u-1) u-1/2 du = (1/4)(u1/2 - u-1/2) du. Integration yields (1/4)[ (2/3)u3/2 - 2u1/2 ] + C, and finally substituting back gives the answer in x. This technique requires careful algebraic manipulation but is extremely reliable.
有理函数和含有平方根的表达式是代换法的理想应用对象。对于 ∫ x / √2x+1 dx，令 u = 2x+1，则 x = (u-1)/2，dx = du/2。被积函数变为 [(u-1)/2] / √u • (du/2) = (1/4)(u-1) u-1/2 du = (1/4)(u1/2 - u-1/2) du。积分得 (1/4)[ (2/3)u3/2 - 2u1/2 ] + C，最后将 u 代回即可得到关于 x 的答案。这一技巧需要仔细的代数处理，但极为可靠。
9. Using Standard Patterns and Recognition | 利用标准模式与识别
With practice, many integrals of the form ∫ k•f '(x)•[f(x)]n dx can be integrated directly by inspection using the pattern [f(x)]n+1 / (n+1) + C, avoiding explicit substitution. For example, ∫ cos x esin x dx fits ∫ f 'x) e^(f(x)) dx = e^(f(x)) + C with f(x)=sin x. Likewise, ∫ (2x)/(x²+4) dx is of the form f '(x)/f(x) → ln|f(x)| exam questions often test the ability to spot these patterns quickly under time pressure.
通过练习，许多形如 ∫ k•f '(x)•[f(x)]n dx 的积分可以通过模式识别直接积分，使用公式 [f(x)]n+1 / (n+1) + C，从而避免显式代换。例如，∫ cos x esin x dx 符合 ∫ f '(x) e^(f(x)) dx = e^(f(x)) + C，其中 f(x)=sin x。同理，∫ (2x)/(x²+4) dx 形如 f '(x)/f(x) → ln|f(x)| + C。Edexcel 试题经常考查学生在时间压力下快速识别这些模式的能力。
10. Common Pitfalls and How to Avoid Them | 常见陷阱与应对策略
tive. Another mistake is leaving the answer in terms of u for an indefinite integral. Always differentiate your final expression mentally to check if it matches the original integrand. If a factor of x remains after substitution, reassess your choice of u – the integral is probably not suitable for that substitution.
学生常忘记在定积分中变换上下限，或错误地替换 dx。一个常见错误是写成 du = dx 而忽略导数。另一个错误是在不定积分中将答案留在 u 变量中。永远在心里对最终表达式求导，检查是否与原被积函数一致。若代换后仍有 x 的因子残留，请重新考虑 u 的选择——该积分可能不适合那个代换。
11. Substitution in the Context of Differential Equations | 微分方程背景下的代换法
tution v = x + y transforms the equation into a separable one. These problems require you to differentiate the substitution with respect to x, sometimes implicitly, and replace dy/dx. This links the technique directly to pure integration skills and highlights the interconnectedness of the syllabus.
在 Edexcel A-Level 中，代换法也出现在求解一阶微分方程时。例如，要求解 dy/dx = (x + y)²，作代换 v = x + y 能将方程转化为可分离变量的形式。此类问题需要你对代换式关于 x 求导（有时是隐式求导）并替换 dy/dx。这就将本技巧与纯积分能力直接联系起来，突显了考纲各部分的相互关联。
12. Practice and Mastery | 巩固练习与精通掌握
ove to trigonometric and exponential variations, and finally tackle definite integrals with limit changes. Use official exam past papers: typical examples include ∫ x√2x-1 dx, ∫ sin(3x) dx, and ∫ e2x+1 dx. As confidence grows, attempt the 'reverse thinking' approach where you differentiate a function and then write its inverse integration problem.
要在 Edexcel 积分题中稳拿高分，最有效的方法是进行有条理的持续练习。从简单的多项式练习入手，再过渡到三角函数和指数函数的变体，最后攻克带换限的定积分。使用官方真题考卷：典型的例题包括 ∫ x√2x-1 dx、∫ sin(3x) dx 以及 ∫ e2x+1 dx。随着自信的提升，尝试“逆向思维”法——先对一个函数求导，然后写出对应的逆积分题目。
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