📚 Integrating f(ax + b) | 积分形如 f(ax+b) 的函数
When you progress through A-Level calculus, you meet integrals that are more involved than simple powers of x. One of the most common patterns is integrating a function of a linear expression, written as f(ax + b). Mastering this technique unlocks a wide range of problems, from exponentials and trigonometry to rational functions. This article breaks down the method, provides the key formula, and walks through worked examples to build your confidence.
在 A-Level 微积分的学习过程中，你会遇到比 x 的简单幂次更复杂的积分。最常见的模式之一，就是积分形如 f(ax+b) 的线性表达式函数。掌握这项技巧能够解锁指数、三角、有理函数等各种问题。本文将拆解这种方法，给出关键公式，并通过详细例题帮助你建立信心。
1. The Reverse Chain Rule | 反向链式法则
The chain rule for differentiation tells us that the derivative of a composite function F(ax + b) is a · f(ax + b), where f is the derivative of F. Integration is the reverse process: to integrate f(ax + b), we need to reverse the chain rule by dividing by the coefficient a.
微分的链式法则告诉我们，复合函数 F(ax+b) 的导数是 a · f(ax+b)，其中 f 是 F 的导函数。积分是这个过程的逆运算：要积分 f(ax+b)，我们需要通过除以系数 a 来反向使用链式法则。
d/dx [F(ax + b)] = a f(ax + b) ⇔ ∫ f(ax + b) dx = (1/a) F(ax + b) + C
tuting u = ax+b. This reverse chain rule is the cornerstone for all the cases that follow.
因此，积分 f(ax+b) 的基本公式就是在代换 u = ax+b 后，对 f(u) 积分再乘以 1/a。这个反向链式法则是随后所有情形的基础。
2. Standard Formula for f(ax+b) | f(ax+b) 的标准积分公式
If you can integrate a function f(u) to get F(u) + c, then you can immediately write down the integral of f(ax+b) by replacing u with ax+b and dividing by a. This one-step rule greatly speeds up integration when the inner function is linear.
如果你能将函数 f(u) 积分得到 F(u)+c，那么你可以直接将 u 换成 ax+b，并除以 a，立即写出 f(ax+b) 的积分。当内层函数为线性时，这一步骤可以极大加快积分速度。
∫ f(ax + b) dx = (1/a) F(ax + b) + C
≠ -1. The special case n = -1 leads to the natural logarithm, which is treated separately.
例如，若 f(u)=uⁿ，则 ∫ (ax+b)ⁿ dx = 1/a · (ax+b)ⁿ⁺¹/(n+1) + C，其中 n ≠ -1。n = -1 的特殊情况会导致自然对数，需要单独处理。
3. Integrating Polynomial Expressions | 多项式表达式的积分
When the function is a power of a linear term, you simply add one to the exponent and divide by both the new exponent and the coefficient of x. Always check that the inner derivative is exactly the coefficient you are dividing by.
当函数是线性项的幂次时，只需将指数加 1，然后除以新指数和 x 的系数。始终要确保内层导数正好是你所除的那个系数。
∫ (3x+2)⁴ dx = (1/3) · (3x+2)⁵/5 + C = (3x+2)⁵/15 + C
Here the inner function is 3x+2, its derivative is 3, so we divide by 3. The power increases from 4 to 5 and we divide by 5. The same method works for fractional powers, such as square roots.
这里内层函数是 3x+2，其导数为 3，所以除以 3。指数从 4 增加到 5，并除以 5。同样的方法也适用于分数指数，例如平方根。
∫ √2x-1 dx = ∫ (2x-1)1/2 dx = (1/2) · (2x-1)3/2/(3/2) + C = (1/3)(2x-1)3/2 + C
inner derivative is -4, so we divide by -4, giving -(1/4)·(1-4x)⁷/7 + C = -(1-4x)⁷/28 + C.
另一个常见例子是 ∫ (1-4x)⁶ dx。内层导数为 -4，因此除以 -4，得到 -(1/4)·(1-4x)⁷/7 + C = -(1-4x)⁷/28 + C。
4. Integrating Exponential Functions | 指数函数的积分
Exponential functions of the form eax+b are particularly straightforward. Since the integral of eu is eu, you just divide by the coefficient of x.
形如 eax+b 的指数函数特别直接。由于 eu 的积分还是 eu，你只需要除以 x 的系数即可。
∫ eax+b dx = (1/a) eax+b + C
For instance, ∫ e5x-3 dx = (1/5) e5x-3 + C. If a constant multiplier is present, keep it in front: ∫ 3e2x+1 dx = 3·(1/2) e2x+1 + C = (3/2) e2x+1 + C.
例如，∫ e5x-3 dx = (1/5) e5x-3 + C。如果存在常数乘子，将其保留在前面：∫ 3e2x+1 dx = 3·(1/2) e2x+1 + C = (3/2) e2x+1 + C。
This rule also extends to bases other than e, such as ∫ akx+b dx, which gives (1/(k ln a)) akx+b + C, but at A-Level the focus is predominantly on the natural exponential function.
该法则也可扩展到 e 以外的底数，例如 ∫ akx+b dx 会给出 (1/(k ln a)) akx+b + C，但在 A-Level 阶段，重点主要放在自然指数函数上。
5. Integrating Trigonometric Functions | 三角函数的积分
For sine and cosine, recall that differentiation with a linear argument produces an extra factor of a. Integration therefore introduces division by a and must respect the standard sign changes.
对于正弦和余弦，请记住具有线性参数的微分会产生额外的因子 a，因此积分需要除以 a，并且要遵守标准的符号变化。
∫ sin(ax+b) dx = - (1/a) cos(ax+b) + C
∫ cos(ax+b) dx = (1/a) sin(ax+b) + C
h a constant multiple, ∫ 2 cos(0.5x-1) dx = 2·(1/0.5) sin(0.5x-1) + C = 4 sin(0.5x-1) + C.
示例：∫ sin(3x+π/4) dx = -(1/3) cos(3x+π/4) + C。带有常数倍乘时，∫ 2 cos(0.5x-1) dx = 2·(1/0.5) sin(0.5x-1) + C = 4 sin(0.5x-1) + C。
The same pattern applies
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