Integration of f(ax+b) | f(ax+b)型函数的积分技巧

📚 Integration of f(ax+b) | f(ax+b)型函数的积分技巧

In IB Mathematics, functions of the form f(ax + b) appear frequently in integration. Whenever the argument of a function is a linear expression rather than simply x, a constant factor emerges from the derivative of ax + b. This pattern allows us to integrate without a full substitution once we recognise it.

在IB数学中，形如 f(ax + b) 的函数在积分中频繁出现。当函数的自变量不是单纯的 x，而是一个线性表达式 ax + b 时，求导过程中会产生常数因子 a。只要识别出这一模式，我们就能在没有完整换元步骤的情况下直接积分。

1. The Core Idea: Linear Substitution | 核心思想：线性替换

Consider an integral of the form ∫ f(ax + b) dx. Let u = ax + b. Then du/dx = a, so dx = du/a. The integral becomes (1/a) ∫ f(u) du.

考虑形如 ∫ f(ax + b) dx 的积分。令 u = ax + b，则 du/dx = a，因此 dx = du/a。原积分变为 (1/a) ∫ f(u) du。

In practice, we do not need to write u explicitly every time. Because the derivative of ax + b is the constant a, the reciprocal factor 1/a must appear in the antiderivative.

在实际操作中，我们不必每次都写出 u。因为 ax + b 的导数是常数 a，所以原函数中必须出现倒数因子 1/a。

∫ f(ax + b) dx = (1/a) F(ax + b) + C, where F'(x) = f(x)

This formula is sometimes called the “linearity of integration with respect to the variable” or the “inverse chain rule”. It is valid whenever a ≠ 0.

这个公式有时被称为“关于变量的线性积分”或“逆链式法则”。只要 a ≠ 0 它就成立。

2. The Inverse Chain Rule | 逆链式法则

When we differentiate a composite function F(ax + b), the chain rule gives a · F'(ax + b). Therefore, the integral of f(ax + b) must include the extra factor 1/a to cancel the factor a produced by differentiation.

当我们对复合函数 F(ax + b) 求导时，链式法则给出 a · F'(ax + b)。因此，f(ax + b) 的积分必须包含额外的因子 1/a，以抵消求导时产生的因子 a。

For example, the derivative of (3x + 2)⁵ is 5(3x + 2)⁴ · 3 = 15(3x + 2)⁴. Hence the integral of (3x + 2)⁴ must be (1/3) · (3x + 2)⁵/5 = (3x + 2)⁵/15 + C.

例如，(3x + 2)⁵ 的导数为 5(3x + 2)⁴ · 3 = 15(3x + 2)⁴。因此，(3x + 2)⁴ 的积分必须是 (1/3) · (3x + 2)⁵/5 = (3x + 2)⁵/15 + C。

The key is to check whether the integrand is exactly f(ax + b) with no extra x-dependent factor. If extra x terms appear, the simple formula does not apply.

关键在于要检查被积函数是否恰好是 f(ax + b) 而不带多余的含 x 因子。如果出现额外的 x 项，简单公式就不适用了。

3. Exponential Functions | 指数函数

For exponential functions, the rule is straightforward. Since the integral of eˣ is eˣ, the integral of eᵃˣ⁺ᵇ is (1/a) eᵃˣ⁺ᵇ + C.

对于指数函数，规则非常直接。由于 eˣ 的积分是 eˣ，eᵃˣ⁺ᵇ 的积分是 (1/a) eᵃˣ⁺ᵇ + C。

Example: ∫ e⁵ˣ⁻³ dx. Here a = 5, so the answer is (1/5) e⁵ˣ⁻³ + C.

例：∫ e⁵ˣ⁻³ dx。这里 a = 5，因此答案是 (1/5) e⁵ˣ⁻³ + C。

If the base is not e, use the formula ∫ aˣ dx = aˣ / ln a + C. For instance, ∫ 2^(3x+1) dx = (1/3) · 2^(3x+1) / ln 2 + C.

如果底数不是 e，使用公式 ∫ aˣ dx = aˣ / ln a + C。例如，∫ 2^(3x+1) dx = (1/3) · 2^(3x+1) / ln 2 + C。

Remember that for exponential functions, neither the coefficient nor the constant inside the exponent changes the exponent itself.

记住，对于指数函数，指数内部的系数和常数都不会改变指数本身的函数形式。

4. Trigonometric Functions | 三角函数

Integrals of sin(ax + b), cos(ax + b), and sec²(ax + b) follow the same pattern. Each produces a reciprocal factor 1/a in front of the corresponding standard antiderivative.

sin(ax + b)、cos(ax + b) 和 sec²(ax + b) 的积分遵循相同的模式。每一个都会在对应标准原函数前面产生倒数因子 1/a。

∫ sin(ax + b) dx = −(1/a) cos(ax + b) + C
∫ cos(ax + b) dx = (1/a) sin(ax + b) + C
∫ sec²(ax + b) dx = (1/a) tan(ax + b) + C
∫ sec(ax + b) tan(ax + b) dx = (1/a) sec(ax + b) + C

Example: ∫ sin(π x / 2 + 3) dx. Here a = π/2, so the integral is −(1/(π/2)) cos(π x / 2 + 3) + C = −(2/π) cos(π x / 2 + 3) + C.

例：∫ sin(π x / 2 + 3) dx。这里 a = π/2，因此积分为 −(1/(π/2)) cos(π x / 2 + 3) + C = −(2/π) cos(π x / 2 + 3) + C。

Always keep the angle in the same units (usually radians) and do not change the constant b when integrating.

始终使角度保持相同单位（通常是弧度），并且在积分时不要改变常数 b。

5. Rational Functions: 1/(ax + b) and logarithms | 有理函数：1/(ax + b) 与对数

The integral of 1/(ax + b) is a special case because the antiderivative involves a natural logarithm. Since d/dx ln(ax + b) = a/(ax + b), we have:

1/(ax + b) 的积分是一个特例，因为其原函数涉及自然对数。由于 d/dx ln(ax + b) = a/(ax + b)，我们有：

∫ 1/(ax + b) dx = (1/a) ln|ax + b| + C

The absolute value is essential because the logarithm is only defined for positive inputs, while ax + b may be negative for some x.

绝对值是必要的，因为对数只对正数有定义，而 ax + b 对某些 x 可能是负数。

Example: ∫ 3/(2x + 5) dx = 3 · (1/2) ln|2x + 5| + C = (3/2) ln|2x + 5| + C.

例：∫ 3/(2x + 5) dx = 3 · (1/2) ln|2x + 5| + C = (3/2) ln|2x + 5| + C。

This formula also applies to integrals such as ∫ 1/(3 − x) dx, where a = −1. The result is −ln|3 − x| + C.

该公式也适用于如 ∫ 1/(3 − x) dx 这样的积分，此时 a = −1。结果为 −ln|3 − x| + C。

6. Powers and Roots | 幂函数与根式

For (ax + b)ⁿ with n ≠ −1, the integral is (ax + b)ⁿ⁺¹ / (a(n + 1)) + C. This includes fractional exponents representing roots.

对于 n ≠ −1 的 (ax + b)ⁿ，其积分为 (ax + b)ⁿ⁺¹ / (a(n + 1)) + C。这包括表示根式的分数指数。

Example: ∫ √(4x + 1) dx. Write √(4x + 1) = (4x + 1)^(1/2). Then n = 1/2 and a = 4. The integral is (4x + 1)^(3/2) / (4 · 3/2) + C = (4x + 1)^(3/2) / 6 + C.

例：∫ √(4x + 1) dx。将 √(4x + 1) 写为 (4x + 1)^(1/2)。则 n = 1/2，a = 4。积分为 (4x + 1)^(3/2) / (4 · 3/2) + C = (4x + 1)^(3/2) / 6 + C。

For negative powers, use the same formula. For example, ∫ (x + 2)⁻³ dx = (x + 2)⁻² / (−2) + C = −1 / (2(x + 2)²) + C.

对于负指数，使用相同公式。例如，∫ (x + 2)⁻³ dx = (x + 2)⁻² / (−2) + C = −1 / (2(x + 2)²) + C。

Do not forget that the exponent increases by exactly 1, and the denominator includes the constant a from inside the linear expression.

不要忘记指数恰好增加 1，且分母包含线性表达式内部的常数 a。

7. Definite Integrals: Changing Limits | 定积分：换限

For definite integrals of the form ∫ₐᵇ f(ax + b) dx, we can either find the antiderivative and evaluate at x = a and x = b, or perform the substitution u = ax + b and change the limits to u = aa + b and u = ab + b.

对于形如 ∫ₐᵇ f(ax + b) dx 的定积分，我们可以先求原函数，再在 x = a 和 x = b 处取值；或者进行替换 u = ax + b，并将上下限改为 u = aa + b 和 u = ab + b。

Example: Evaluate ∫₀¹ (2x + 1)³ dx. Using the antiderivative, we have [(2x + 1)⁴ / 8]₀¹ = (3⁴ − 1⁴)/8 = (81 − 1)/8 = 10.

例：计算 ∫₀¹ (2x + 1)³ dx。使用原函数，有 [(2x + 1)⁴ / 8]₀¹ = (3⁴ − 1⁴)/8 = (81 − 1)/8 = 10。

If instead we substitute u = 2x + 1, then du = 2 dx, so dx = du/2. When x = 0, u = 1; when x = 1, u = 3. The integral becomes ∫₁³ u³ · (1/2) du = (1/2) · [u⁴/4]₁³ = (1/8)(81 − 1) = 10.

如果改用替换 u = 2x + 1，则 du = 2 dx，所以 dx = du/2。当 x = 0 时，u = 1；当 x = 1 时，u = 3。积分变为 ∫₁³ u³ · (1/2) du = (1/2) · [u⁴/4]₁³ = (1/8)(81 − 1) = 10。

Both methods are valid; the substitution method often reduces algebraic complexity when the limits become simple numbers.

两种方法都有效；当换元后的上下限变成简单数字时，替换法往往能降低代数复杂度。

8. Integrals with Absolute Values | 含绝对值的积分

When the integrand contains |ax + b|, split the interval at the point where ax + b = 0. On each side, remove the absolute value sign with the appropriate sign.

当被积函数包含 |ax + b| 时，在 ax + b = 0 的点处分段。在每一侧，以适当的符号去掉绝对值符号。

Example: ∫₀³ |2x − 4| dx. The expression 2x − 4 changes sign at x = 2. Therefore, split the integral at x = 2:

例：∫₀³ |2x − 4| dx。表达式 2x − 4 在 x = 2 处变号。因此，在 x = 2 处分段：

∫₀³ |2x − 4| dx = ∫₀² (4 − 2x) dx + ∫₂³ (2x − 4) dx

Evaluating: [4x − x²]₀² + [x² − 4x]₂³ = (8 − 4) + (9 − 12 − 4 + 8) = 4 + 1 = 5.

计算：[4x − x²]₀² + [x² − 4x]₂³ = (8 − 4) + (9 − 12 − 4 + 8) = 4 + 1 = 5。

This technique is essential in IB questions that ask for area, since area is always positive even when the function is negative.

这一技巧在IB求面积问题中至关重要，因为面积总是正的，即使函数为负。

9. Common Mistakes and Pitfalls | 常见错误与陷阱

One frequent error is forgetting the factor 1/a. For example, ∫ cos(3x) dx is (1/3) sin(3x) + C, not sin(3x) + C. Differentiating sin(3x) gives 3 cos(3x), so the constant is necessary.

一个常见错误是忘记因子 1/a。例如，∫ cos(3x) dx 是 (1/3) sin(3x) + C，而不是 sin(3x) + C。对 sin(3x) 求导得到 3 cos(3x)，因此常数是必要的。

Another mistake is using the formula when the integrand is f(ax + b) multiplied by an x-dependent factor. For example, ∫ x(x + 1)² dx cannot be integrated directly as (x + 1)³/3; the extra x prevents the simple substitution.

另一个错误是在被积函数是 f(ax + b) 乘以某个含 x 因子时使用该公式。例如，∫ x(x + 1)² dx 不能直接积成 (x + 1)³/3；额外的 x 使得简单替换无法使用。

A third common pitfall is omitting the absolute value in the logarithm rule. Always write ln|ax + b| unless the question guarantees ax + b > 0.

第三个常见陷阱是对数法则中遗漏绝对值。除非题目保证 ax + b > 0，否则始终写 ln|ax + b|。

Finally, when handling definite integrals, be careful to adjust the limits if you use substitution, or to evaluate the antiderivative correctly at the original limits.

最后，处理定积分时，如果使用换元法要调整上下限，或者正确地将原函数在原始上下限处取值。

10. Combining with Other Techniques | 与其他技巧结合

Sometimes f(ax + b) appears inside a larger product. In that case, integration by parts may be required, with dv = f(ax + b) dx. The linear argument still obeys the 1/a rule for the v found by integration.

有时 f(ax + b) 出现在更大的乘积中。此时可能需要分部积分，令 dv = f(ax + b) dx。通过积分得到的 v 仍然遵循 1/a 规则来处理线性自变量。

Example: ∫ x e^(2x+1) dx. Choose u = x, dv = e^(2x+1) dx. Then v = (1/2)e^(2x+1). Integration by parts gives x·(1/2)e^(2x+1) − ∫ (1/2)e^(2x+1) dx = (1/2)x e^(2x+1) − (1/4)e^(2x+1) + C.

例：∫ x e^(2x+1) dx。令 u = x，dv = e^(2x+1) dx。则 v = (1/2)e^(2x+1)。分部积分得 x·(1/2)e^(2x+1) − ∫ (1/2)e^(2x+1) dx = (1/2)x e^(2x+1) − (1/4)e^(2x+1) + C。

In IB examinations, problems that combine the inverse chain rule with integration by parts or partial fractions are common at Higher Level.

在IB考试中，将逆链式法则与分部积分或部分分式结合的问题在高级水平（HL）中很常见。

11. Practice Questions | 练习题目

Work through these examples to build fluency with the technique.

完成以下例子以熟练该技巧。

Question	Answer
∫ (5x + 2)⁶ dx	(5x + 2)⁷ / 35 + C
∫ 3/(4 − 2x) dx	−(3/2) ln\|4 − 2x\| + C
∫ sin(π x + 1) dx	−(1/π) cos(π x + 1) + C
∫₁³ √(2x + 3) dx	(1/3) [ (9)^(3/2) − (5)^(3/2) ]

After attempting each, check that differentiation of your answer returns the original integrand.

尝试完每个题目后，检查对答案求导是否能还原原被积函数。

12. Summary | 总结

The integration of f(ax + b) is one of the most time-saving techniques in IB Mathematics. Recognising the linear argument immediately allows you to write down the antiderivative with a simple reciprocal factor 1/a.

f(ax + b) 的积分是IB数学中最节省时间的技巧之一。立即识别出线性自变量，你就能通过简单的倒数因子 1/a 直接写出原函数。

Always verify the condition a ≠ 0, remember absolute values for logarithms, and adjust limits carefully in definite integrals. With regular practice, this method becomes automatic.

始终检查条件 a ≠ 0，记住对数中的绝对值，并在定积分中仔细调整上下限。通过经常练习，这种方法会变得得心应手。

Mastering this rule not only saves time in exams but also builds a strong foundation for more advanced techniques such as substitution and integration by parts.

掌握这一法则不仅能节省考试时间，还能为换元法、分部积分等更高级的技巧奠定坚实基础。

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Integration of f(ax+b) | f(ax+b)型函数的积分技巧