Substitution Method for ∫ f(ax + b) dx | 线性代换积分法

📚 Substitution Method for ∫ f(ax + b) dx | 线性代换积分法

he form ∫ f(ax + b) dx, where the inner function is a linear expression. This article unpacks the method step by step, provides worked examples, and links it directly to the chain rule, helping you build both confidence and speed for Paper 2 and Paper 3 questions.

初次接触积分时，微分的逆运算有时看起来像猜谜。然而，代换积分法是一种最有序、最可靠的技术。在IB数学考试中，最简单也最常见的案例就是线性代换，即处理形如 ∫ f(ax + b) dx 的积分，其中内部函数为一次表达式。本文会逐步解析这一方法，提供丰富的例题，并将其与链式法则直接联系起来，帮助你在Paper 2和Paper 3中获得信心和速度。

1. What Is the Substitution Method? | 什么是代换积分法？

e for differentiation. When we differentiate a composite function y = F(g(x)), we obtain y' = F'(g(x))·g' we can undo it by substituting u = g(x). In the linear case, g(x) = ax + b, which makes g' nstant. This constant factor is what makes linear substitution especially straightforward.

代换积分法是链式法则求导的逆过程。当我们对复合函数 y = F(g(x)) 求导时，得到 y' = F'(g(x))·g'(x)。反过来，如果被积函数看起来像是链式法则的结果，我们就可以通过设 u = g(x) 来还原积分。在线性情况下，g(x) = ax + b，从而 g'(x) = a 是一个常数。这个常数因子正是线性代换格外简单的原因。

Instead of memorising dozens of specific integral formulas, you can focus on recognising the pattern f(ax + b) and apply a single, unified method. This line of thinking aligns perfectly with the IB’s emphasis on conceptual understanding and problem solving. Once you master linear substitution, you will quickly adapt to more complex non-linear substitutions as well.

与其死记硬背几十个具体积分公式，不如专注于识别 f(ax + b) 这一模式，并应用统一的代换方法。这种思维方式与IB课程强调的概念理解与问题解决完全吻合。一旦掌握了线性代换，你也能迅速适应更复杂的非线性代换。

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2. The General Formula for ∫ f(ax + b) dx | ∫ f(ax + b) dx 的一般公式

Let u = ax + b. Then du/dx = a, so dx = du / a. Substituting into the integral gives:

令 u = ax + b，则 du/dx = a，因此 dx = du / a。代入积分得到：

∫ f(ax + b) dx = ∫ f(u) · (1/a) du = (1/a) ∫ f(u) du

If we know an antiderivative of f(u) is F(u), i.e. F' al answer is simply (1/a) F(u) + C. Replacing u with ax + b, we obtain the general result:

如果已知 f(u) 的一个原函数为 F(u)，即 F'(u) = f(u)，那么最终答案就是 (1/a) F(u) + C。将 u 换回 ax + b，即得一般结果：

∫ f(ax + b) dx = (1/a) F(ax + b) + C

This simple formula reveals why linear substitution is so powerful: the factor 1/a emerges naturally from integrating the inner derivative. You must never forget to divide by a, as this is the single most common mistake in IB scripts. The formula holds for any differentiable function f, including polynomial, trigonometric, exponential, and rational forms.

这个简单公式揭示了线性代换的强大之处：因子 1/a 源自对内部导数的积分。你绝不能忘记除以 a，这是IB答卷中最常见的错误。该公式适用于任何可微函数 f，包括多项式、三角函数、指数函数和有理函数等形式。

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3. Worked Example with Polynomials | 多项式函数的例子

n is 3x − 2, so a = 3. Let u = 3x − 2, then du = 3 dx and dx = du/3. The integral becomes:

考虑积分 ∫ (3x − 2)⁵ dx。这里内部函数是 3x − 2，所以 a = 3。设 u = 3x − 2，则 du = 3 dx，dx = du/3。积分变为：

∫ u⁵ · (1/3) du = (1/3) ∫ u⁵ du = (1/3) · (u⁶ / 6) + C = u⁶ / 18 + C

Re-substituting u = 3x − 2 yields the final answer:

将 u = 3x − 2 代回，得最终答案：

∫ (3x − 2)⁵ dx = (3x − 2)⁶ / 18 + C

ly the original integrand. This quick check is always recommended in the exam if time permits.

为了验证结果，对 (3x − 2)⁶ / 18 求导。利用链式法则，导数为 (6/18)(3x − 2)⁵ · 3 = (18/18)(3x − 2)⁵ = (3x − 2)⁵，正好是原被积函数。如果时间允许，考试中随时推荐这种快速检查。

x)⁵/10 + C. Notice how the negative a is handled naturally by the substitution process. Many IB questions deliberately include a negative coefficient to test your attention to detail.

另一个常见的多项式例子是 ∫ (5 − 2x)⁴ dx。这里 a = −2。设 u = 5 − 2x，du = −2 dx，所以 dx = du/(−2)。结果是 (1/(−2)) · (u⁵/5) = −u⁵/10 + C = −(5 − 2x)⁵/10 + C。注意负系数的 a 在代换过程中如何自然地被处理。很多IB题目会刻意包含负系数，以考察你对细节的关注。

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4. Trigonometric Integrals | 三角函数积分

sin(4x + π) dx as an example. Set u = 4x + π, du = 4 dx, dx = du/4. Then the integral is:

具有线性参数的三角函数是线性代换最常见的应用对象之一。以 ∫ sin(4x + π) dx 为例。设 u = 4x + π，du = 4 dx，dx = du/4。那么积分为：

∫ sin u · (1/4) du = (1/4)(−cos u) + C = −(1/4) cos(4x + π) + C

dx = (1/2) sin(2x) + C. Notice that ∫ cos(ax + b) dx generalises to (1/a) sin(ax + b) + C.

因子 1/4 至关重要；忘记它会导致导数是正确值的四倍。对于余弦积分，模式是类似的：∫ cos(2x) dx = (1/2) sin(2x) + C。请注意，∫ cos(ax + b) dx 的通式为 (1/a) sin(ax + b) + C。

sin(2x), which is a non-linear substitution. Stick to pure f(ax + b) patterns for this method.

对于 ∫ sec²(5x − 1) dx，sec² u 的原函数是 tan u。于是积分变为 (1/5) tan(5x − 1) + C。当被积函数为 sin³(2x)cos(2x) 这样的乘积时，仅靠线性代换可能不够；你可能需要结合 u = sin(2x) 的非线性代换。请保持本方法仅用于纯 f(ax + b) 形式。

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5. Exponential and Logarithmic Integrals | 指数与对数积分

Exponential integrals of the form ∫ e^{ax + b} dx follow the same rule. Since ∫ e^u du = e^u + C, we have ∫ e^{ax + b} dx = (1/a) e^{ax + b} + C. For instance, ∫ e^{−3x + 5} dx = −(1/3) e^{−3x + 5} + C. The negative a yields a negative factor in front.

形如 ∫ e^{ax + b} dx 的指数积分遵循同样的规则。因为 ∫ e^u du = e^u + C，我们有 ∫ e^{ax + b} dx = (1/a) e^{ax + b} + C。例如，∫ e^{−3x + 5} dx = −(1/3) e^{−3x + 5} + C。负的 a 导致前面出现负因子。

thmic integrals, consider the rational function 1/(ax + b). Here the pattern ∫ 1/u du = ln |u| + C applies. So ∫ 1/(2x + 7) dx = (1/2) ln |2x + 7| absolute value, especially when the linear expression could be negative over the interval.

对于对数积分，考虑有理函数 1/(ax + b)。这里适用模式 ∫ 1/u du = ln |u| + C。因此 ∫ 1/(2x + 7) dx = (1/2) ln |2x + 7| + C。绝对值是必要的，因为原函数的定义域可能要求如此。在IB考试中，尤其当线性表达式在积分区间内可能为负时，你需要写出绝对值。

A slightly more involved example is ∫ (e^2x / (1 + e^2x is is not strictly of the form f(ax+b) at first glance, a secondary substitution v = 1 + e^2x works. However, you can also write e^2x as e^2x and note that the derivative of 1 + e^2x is 2e^2x, making it a natural log integral. The result would be (1/2) ln(1 + e^2x) + C. This highlights how linear substitution sometimes combines with other techniques.

一个稍微复杂的例子是 ∫ (e^2x / (1 + e^2x)) dx。虽然乍看不是严格的 f(ax+b) 形式，但可以进行二次代换 v = 1 + e^2x。不过，你也可以注意到 1 + e^2x 的导数是 2e^2x，从而将其视为自然对数积分。结果是 (1/2) ln(1 + e^2x) + C。这凸显了线性代换有时如何与其他技巧结合使用。

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6. The Link with the Chain Rule | 与链式法则的联系

The reason linear substitution works is precisely because differentiation and integration are inverse processes. If we differentiate F(ax + b), the chain rule gives F'(ax + b) · a. Therefore, when we integrate a f(ax + b), we are essentially integrating a F'(ax + b). To obtain just f(ax + b), we need to divide by a. This is why (1/a) appears in the integration formula.

线性代换之所以有效，正是因为微分与积分互为逆运算。如果我们对 F(ax + b) 求导，链式法则给出 F'(ax + b) · a。因此，当我们积分 a f(ax + b) 时，本质上是在积分 a F'(ax + b)。要得到单纯的 f(ax + b)，就需要除以 a。这就是为什么 (1/a) 会出现在积分公式中。

Imagine you are trying to reverse the derivative of sin(3x). The derivative is 3 cos(3x). So if you are asked to integrate cos(3x), you need to guess a function whose derivative is cos(3x). Since derivative of sin(3x) gives an extra factor of 3, you must compensate by writing (1/3) sin(3x). This compensation thinking is the heart of the method and deepens your understanding of calculus rather than promoting rote memorisation.

假设你试图反推 sin(3x) 的导数。它的导数是 3 cos(3x)。因此，如果要求积分 cos(3x)，就需要猜测一个函数，其导数恰好是 cos(3x)。由于 sin(3x) 的导数多了一个因子 3，你必须通过乘以 1/3 来补偿。这种补偿思维是本方法的核心，它加深了微积分的理解，而非鼓励死记硬背。

Understanding this connection enables you to spot errors immediately: if you have integrated to (1/2)F(ax+b) but your derivative check yields 2 f(ax+b), you know the factor is inverted. For IB Paper 3 analytical questions, explaining this relationship can also earn communication marks.

理解这层联系使你能够立即发现错误：如果你积分得到了 (1/2)F(ax+b)，但导数检查结果是 2 f(ax+b)，你就知道因子反了。对于IB Paper 3中的分析性问题，解释这一关系还能获得表述分。

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7. Common Pitfalls and How to Avoid Them | 常见错误及避免方法

e du = a dx but then mistakenly substitute dx = a du. Writing the substitution formally as 'du = a dx ⇒ dx = du/a' helps avoid this slip.

最具破坏性的错误是忘记除以 a。很多同学会写成 ∫ (2x+1)³ dx = (2x+1)⁴/4 + C，却漏掉了关键的 1/2。务必在心里对答案求导来发现这类问题。另一个错误是处理常数不当：当 u = ax + b 得到 dx = du/a 时，有些人推出 du = a dx 后却错误地代入 dx = a du。正式写出 'du = a dx ⇒ dx = du/a' 有助于避免这类差错。

d the linear expression inside the logarithm is a common technical error that can be penalised, especially in definite integration where the interval may cross a point where ax+b = 0.

同学们有时还会忘记将 u 换回原变量。答案保留 u 而非 x 会导致IB丢分，除非题目明确要求用 u 表示原函数。此外，在积分 1/(ax+b) 时，忽略对数内线性表达式的绝对值符号是常见的技术性错误，特别是在定积分中，积分区间可能跨过 ax+b = 0 的点时，会被扣分。

To build good habits, I recommend a three-step check after every linear substitution: (i) write down u and du clearly on the side, (ii) after integration, immediately replace u, and (iii) differentiate your final answer in your head or on scratch paper. This routine takes seconds but drastically reduces careless errors.

为了培养好习惯，我推荐每次线性代换后实行三步检查：(i) 在边上清晰地写出 u 和 du，(ii) 积分后立即代回 u，(iii) 在脑中或草稿纸上对最终答案求导。这一流程只需几秒钟，却能大幅减少粗心错误。

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8. Practice Tips for IB Exams | IB 考试备考建议

In IB Mathematics, linear substitution questions often appear in the non-calculator papers, where neat algebraic manipulation is tested. When you recognise an integrand like (5x − 3)⁻² or cos(πx/2), you should be able to write down the antiderivative almost instantly after a little practice. Use the formula (1/a)F(ax+b) directly once you are confident, but always be prepared to show the substitution steps if the markscheme requires it.

在IB数学考试中，线性代换题目经常出现在非计算器卷中，考查整洁的代数操作。当你认出如 (5x − 3)⁻² 或 cos(πx/2) 这样的被积函数后，经过一些练习，你应该能几乎立即写出原函数。一旦有信心，可以直接使用 (1/a)F(ax+b) 公式，但要随时准备展示代换步骤，以便满足评分标准的要求。

table of standard linear integrals: ∫ (ax+b)ⁿ dx = (ax+b)ⁿ⁺¹ / [a(n+1)] + C (n ≠ −1); ∫ e^ax+b dx = (1/a) e^ax+b dx = −(1/a) cos(ax+b) + C; ∫ sec²(ax+b) dx = (1/a) tan(ax+b) + C; ∫ 1/(ax+b) dx = (1/a) ln|ax+b| + C. These cover the vast majority of IB linear substitution tasks.

建立自己的标准线性积分表：∫ (ax+b)ⁿ dx = (ax+b)ⁿ⁺¹ / [a(n+1)] + C (n ≠ −1)；∫ e^ax+b dx = (1/a) e^ax+b + C；∫ cos(ax+b) dx = (1/a) sin(ax+b) + C；∫ sin(ax+b) dx = −(1/a) cos(ax+b) + C；∫ sec²(ax+b) dx = (1/a) tan(ax+b) + C；∫ 1/(ax+b) dx = (1/a) ln|ax+b| + C。这些涵盖了绝大部分IB线性代换题目。

When tackling more involved problems, such as definite integrals or area under a curve, apply the linear substitution, change the limits accordingly, and remember to report the final value in its simplest exact form. Practising with past papers from both time zones will expose you to the subtle variations examiners use, including negative coefficients, fractional constants, and combined function types.

在处理更复杂的问题时，如定积分或曲线下方面积，应用线性代换并相应变换积分限，并记住以最简精确形式报告最终值。练习不同时区的历年真题，将使你熟悉考官使用的细微变化，包括负系数、分数常数和组合函数类型。

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