F – Integrating f(ax+b) | F – 积分 f(ax+b)

📚 F – Integrating f(ax+b) | F – 积分 f(ax+b)

Integration of functions where the variable appears inside a linear expression, such as sin(5x+2) or (3x-1)^4, is a core skill in IB Mathematics. Mastering this pattern saves time and reduces errors compared to full substitution every time. This article explains the reverse chain rule approach, provides worked examples for all common function types, and highlights the most frequent pitfalls.

在 IB 数学中，积分被积函数变量出现在线性表达式内部的形式（例如 sin(5x+2) 或 (3x-1)^4）是一项核心技能。掌握这一模式比每次都进行完整代换更省时且不易出错。本文将解释反向链式法则的思路，针对所有常见函数类型给出实例，并重点指出最容易犯的错误。

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1. Recognising the Linear Inside Function | 识别线性内部函数

An integral of the form ∫ f(ax+b) dx has a linear inner function g(x) = ax+b, where a and b are constants and a ≠ 0. The outer function f can be anything whose antiderivative you already know – a power, sine, cosine, exponential, or reciprocal. The key is that the entire argument of f is exactly ax+b, not something more complicated like x²+3.

形式为 ∫ f(ax+b) dx 的积分，其内部函数 g(x) = ax+b 是线性的，其中 a 和 b 为常数且 a ≠ 0。外层函数 f 可以是任何你已知其反导数的函数——幂函数、正弦、余弦、指数或倒数。关键在于 f 的整个输入恰好是 ax+b，而不是像 x²+3 这样更复杂的表达式。

For example, ∫ (2x+7)^5 dx, ∫ e^(4x-3) dx, and ∫ cos(0.5x+π) dx all fall into this category. The linear inner part makes the integration extremely predictable once you reverse the differentiation chain rule.

例如，∫ (2x+7)^5 dx、∫ e^(4x-3) dx 和 ∫ cos(0.5x+π) dx 都属于这一类。一旦你反向运用微分的链式法则，线性的内部部分会让积分变得极易预测。

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2. The Reverse Chain Rule Concept | 反向链式法则概念

When you differentiate F(ax+b), where F’ = f, the chain rule gives d/dx [F(ax+b)] = f(ax+b) · a. Therefore, if you want the derivative to be only f(ax+b) without the extra factor a, you must divide by a. Integrating both sides yields ∫ f(ax+b) dx = (1/a) F(ax+b) + C.

当你对 F(ax+b)（其中 F’ = f）求导时，链式法则给出 d/dx [F(ax+b)] = f(ax+b) · a。因此，如果你希望导数仅仅是 f(ax+b) 而没有额外的因子 a，就必须除以 a。对两边积分即得 ∫ f(ax+b) dx = (1/a) F(ax+b) + C。

In essence, you integrate the outer function as normal but then divide by the coefficient of x inside the brackets. This is the reverse chain rule for linear functions – the simplest and most common case of integration by substitution.

本质上，你先正常积分外层函数，但随后除以括号内 x 的系数。这就是线性函数的反向链式法则——积分代换法中最简单也最常见的情形。

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3. The General Formula | 通用公式

The standard result can be stated concisely: if F is an antiderivative of f (i.e. F'(x) = f(x)), then for constants a ≠ 0 and b,

∫ f(ax+b) dx = (1/a) · F(ax+b) + C.

该标准结果可简明陈述为：若 F 是 f 的一个反导数（即 F'(x) = f(x)），则对于常数 a ≠ 0 和 b，有 ∫ f(ax+b) dx = (1/a) · F(ax+b) + C。

This formula is valid whenever the corresponding basic integral ∫ f(x) dx = F(x) + C exists. In practice, you simply write down F(ax+b) and multiply by 1/a. The constant C accounts for the fact that differentiating a constant gives zero, preserving the family of antiderivatives.

只要相应的基本积分 ∫ f(x) dx = F(x) + C 存在，该公式即有效。实践中，你只需写出 F(ax+b) 并乘以 1/a。常数 C 则用来体现常数的导数为零这一事实，保留了反导数族。

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4. Power Functions: (ax+b)ⁿ | 幂函数：(ax+b)ⁿ

For any real exponent n ≠ -1, the basic rule ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C extends to ∫ (ax+b)ⁿ dx. Raise the power by 1, divide by the new power, and divide by a:

∫ (ax+b)ⁿ dx = (1/a) · (ax+b)ⁿ⁺¹ / (n+1) + C.

对于任何实数指数 n ≠ -1，基本规则 ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C 可推广至 ∫ (ax+b)ⁿ dx。将幂指数加 1，除以新指数，再除以 a： ∫ (ax+b)ⁿ dx = (1/a) · (ax+b)ⁿ⁺¹ / (n+1) + C。

Example: ∫ (3x-5)² dx. Here a = 3, b = -5, n = 2. The antiderivative is (1/3) · (3x-5)³ / 3 + C = (1/9)(3x-5)³ + C. You can verify by differentiation: the derivative brings down a factor 3 from the chain rule, cancelling with the 1/9 to give (3x-5)².

示例：∫ (3x-5)² dx。这里 a=3, b=-5, n=2。反导数为 (1/3) · (3x-5)³ / 3 + C = (1/9)(3x-5)³ + C。你可以通过求导验证：链式法则会带下一个因子 3，与 1/9 相消得到 (3x-5)²。

Another common case is the square root, which is a power of ½. ∫ √(2x+1) dx = ∫ (2x+1)^(½) dx. Using the formula with a=2, n=½ gives (1/2) · (2x+1)^(3/2) / (3/2) + C = (1/2)·(2/3)(2x+1)^(3/2) + C = (1/3)(2x+1)^(3/2) + C.

另一个常见情形是平方根，即 ½ 次幂。∫ √(2x+1) dx = ∫ (2x+1)^(½) dx。使用公式，a=2, n=½，得到 (1/2) · (2x+1)^(3/2) / (3/2) + C = (1/2)·(2/3)(2x+1)^(3/2) + C = (1/3)(2x+1)^(3/2) + C。

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5. Trigonometric Functions | 三角函数

The integrals of sin(ax+b) and cos(ax+b) follow directly from the general rule, remembering that the antiderivative of sin x is -cos x, and that of cos x is sin x.

对于 sin(ax+b) 和 cos(ax+b) 的积分可直接由通用公式得出，记住 sin x 的反导数是 -cos x，cos x 的反导数是 sin x。

∫ sin(ax+b) dx = -(1/a) cos(ax+b) + C

∫ cos(ax+b) dx = (1/a) sin(ax+b) + C

Example: ∫ sin(2x + π/4) dx. Here a = 2, so the integral is -(1/2) cos(2x + π/4) + C. If you ever forget the sign, differentiate mentally: the derivative of cos(2x+π/4) is -2 sin(2x+π/4), so we need the negative to cancel the minus.

示例：∫ sin(2x + π/4) dx。这里 a = 2，因此积分为 -(1/2) cos(2x + π/4) + C。如果忘了符号，可以在脑中求导：cos(2x+π/4) 的导数是 -2 sin(2x+π/4)，因此我们需要负号来抵消这个减号。

For ∫ cos(-3x+1) dx, a = -3, so the integral is (1/(-3)) sin(-3x+1) + C = -(1/3) sin(-3x+1) + C. It is perfectly acceptable to leave the negative coefficient, but you can also use the odd/even properties to simplify if desired.

对于 ∫ cos(-3x+1) dx，a = -3，因此积分为 (1/(-3)) sin(-3x+1) + C = -(1/3) sin(-3x+1) + C。保留负系数完全没问题，若想化简也可利用奇偶性质。

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6. The Exponential Function e^(ax+b) | 指数函数 e^(ax+b)

Since the derivative of eˣ is eˣ, the exponential function is its own antiderivative. With a linear inner function, we simply divide by a:

∫ e^(ax+b) dx = (1/a) e^(ax+b) + C.

因为 eˣ 的导数是 eˣ，指数函数自身的反导数就是它本身。对于线性内部函数，我们只需除以 a：∫ e^(ax+b) dx = (1/a) e^(ax+b) + C。

Example: ∫ e^(5x-3) dx = (1/5) e^(5x-3) + C. This is extremely quick – no need for u-substitution once you trust the pattern. Be careful with negative a: ∫ e^(2-7x) dx has a = -7, so the result is -(1/7) e^(2-7x) + C.

示例：∫ e^(5x-3) dx = (1/5) e^(5x-3) + C。这非常快捷——一旦你熟悉这个模式就无需进行 u 代换。注意负的 a：∫ e^(2-7x) dx 的 a = -7，因此结果是 -(1/7) e^(2-7x) + C。

Exponentials with bases other than e, such as 2^(ax+b), can be handled by first rewriting as e^(ln 2 · (ax+b)) and then applying the same rule. However, IB questions usually concentrate on base e.

对于底数不是 e 的指数，例如 2^(ax+b)，可以先重写为 e^(ln 2 · (ax+b)) 再套用规则。不过 IB 试题通常集中在以 e 为底的指数上。

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7. Reciprocal and Rational Functions | 倒数与有理函数

The special case n = -1 for the power rule corresponds to the reciprocal function. Because ∫ x⁻¹ dx = ln|x| + C, we obtain:

∫ 1/(ax+b) dx = (1/a) ln|ax+b| + C.

幂法则中 n = -1 的特殊情况对应于倒数函数。由于 ∫ x⁻¹ dx = ln|x| + C，我们得到：∫ 1/(ax+b) dx = (1/a) ln|ax+b| + C。

Absolute value bars are essential to keep the logarithm defined for negative arguments. For example, ∫ 1/(2x-6) dx = (1/2) ln|2x-6| + C. Without the absolute value, the domain would be unnecessarily restricted.

绝对值符号至关重要，它能保证对数在负输入时仍有定义。例如，∫ 1/(2x-6) dx = (1/2) ln|2x-6| + C。若不用绝对值，定义域会受到不必要的限制。

When the numerator is a constant multiple, factor it out: ∫ 3/(4x+5) dx = 3 · ∫ 1/(4x+5) dx = (3/4) ln|4x+5| + C. If the numerator contains x, the integrand is no longer of the simple form f(ax+b) because the outer function would depend on x in two places; such integrals often require algebraic division or partial fractions.

当分子为常数倍时，可将其提取出来：∫ 3/(4x+5) dx = 3 · ∫ 1/(4x+5) dx = (3/4) ln|4x+5| + C。若分子含有 x，被积函数就不再是简单形式 f(ax+b)，因为外层函数会在两处依赖于 x；这类积分通常需要代数除法或有理分式分解。

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8. Applying the Rule to Definite Integrals | 在定积分中的应用

For a definite integral ∫ₚᵠ f(ax+b) dx, you can either change the limits as you would with u = ax+b, or use the antiderivative directly and evaluate at the original x-values. The latter is often simpler:

∫ₚᵠ f(ax+b) dx = [ (1/a) F(ax+b) ]ₚᵠ.

对于定积分 ∫ₚᵠ f(ax+b) dx，你可以像用 u = ax+b 那样更换上下限，也可以直接运用反导数并在原来的 x 值处求值。后者通常更简单：∫ₚᵠ f(ax+b) dx = [ (1/a) F(ax+b) ]ₚᵠ。

Example: Evaluate ∫₀¹ e^(3x) dx. Reading a=3, b=0, F(3x) = e^(3x). So the definite integral is [ (1/3) e^(3x) ]₀¹ = (1/3)(e³ – e⁰) = (1/3)(e³ – 1).

例子：计算 ∫₀¹ e^(3x) dx。此处 a=3, b=0, F(3x) = e^(3x)。因此定积分为 [ (1/3) e^(3x) ]₀¹ = (1/3)(e³ – e⁰) = (1/3)(e³ – 1)。

For ∫₁² 1/(2x+1) dx, with a=2: the antiderivative is (1/2) ln|2x+1|. Evaluating from 1 to 2 gives (1/2)[ ln(5) – ln(3) ] = (1/2) ln(5/3). There is no need to adjust the limits when using this direct method, but you must remember to apply the 1/a factor before substituting the limits.

对于 ∫₁² 1/(2x+1) dx，a=2：反导数为 (1/2) ln|2x+1|。从 1 到 2 求值得 (1/2)[ ln(5) – ln(3) ] = (1/2) ln(5/3)。使用这种直接方法时无需调整上下限，但必须记得在代入上下限之前乘上 1/a 因子。

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9. Common Mistakes and How to Avoid Them | 常见错误与避免方法

One of the most frequent errors is forgetting to divide by a. Students often write ∫ sin(2x) dx = -cos(2x) + C. The correct answer is -(1/2) cos(2x) + C. Always mentally differentiate your result to check.

最常见的错误之一是忘记除以 a。学生经常写出 ∫ sin(2x) dx = -cos(2x) + C。正确答案是 -(1/2) cos(2x) + C。任何时候都应在心中对自己求出的答案求导以作检验。

Another trap is missing the sign when a is negative. For example, ∫ e^(5-x) dx: here ax+b = -x+5, so a = -1. The integral is -e^(5-x) + C. Writing +e^(5-x) would be incorrect because the derivative of e^(5-x) is -e^(5-x).

另一个陷阱是当 a 为负数时弄错符号。例如 ∫ e^(5-x) dx：这里 ax+b = -x+5，因此 a = -1。积分结果为 -e^(5-x) + C。若写成 +e^(5-x) 就错了，因为 e^(5-x) 的导数是 -e^(5-x)。

Students also occasionally try to integrate a product like x·(2x+1)⁴ by thinking of it as f(ax+b). This fails because the entire integrand is not a function of (2x+1) alone – the extra x breaks the pattern. Such integrals need expansion or substitution u=2x+1.

学生们偶尔会试图将 x·(2x+1)⁴ 这样的乘积当作 f(ax+b) 来积分。这行不通，因为整个被积函数并不仅仅依赖于 (2x+1)——多余的 x 破坏了该模式。这类积分需要先展开或使用代换 u=2x+1。

Finally, in definite integration, some incorrectly keep the 1/a factor inside the bracket when substituting limits, effectively dividing by a twice. Write the factor outside the bracket, e.g. (1/a)[…], to avoid confusion.

最后，在定积分中，有些人在代入上下限时错误地将 1/a 因子留在括号内，实际上就等于除了两次 a。将因子写在括号外面，例如 (1/a

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