A-Level Mathematics: Finding the Antiderivative from the Derivative | A-Level数学：由导函数求原函数的方法

📚 A-Level Mathematics: Finding the Antiderivative from the Derivative | A-Level数学：由导函数求原函数的方法

In A-Level Mathematics, reversing the process of differentiation is one of the most important skills you will learn. Given a function f′(x) that represents the gradient of some original function f(x), we aim to find f(x) itself. This process is called antidifferentiation, and the result is known as the general antiderivative or indefinite integral.

在A-Level数学中，逆转微分的过程是你将要学习的最重要技能之一。给定一个表示某个原函数 f(x) 之斜率的导函数 f′(x)，我们的目标是求出 f(x) 本身。这一过程称为“反微分”（antidifferentiation），其结果称为“一般原函数”或“不定积分”。

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1. The Basic Rule for Power Functions | 幂函数的基本法则

The most fundamental rule is the one for power functions. If we differentiate xⁿ, we get n·xⁿ⁻¹. Therefore, to find an antiderivative of xⁿ, we reverse this operation. For n ≠ −1:

最基本的法则是针对幂函数的法则。若对 xⁿ 求导，得到 n·xⁿ⁻¹。因此，要对 xⁿ 求原函数，我们逆向执行这一运算。当 n ≠ −1 时：

∫ xⁿ dx = xⁿ⁺¹/(n+1) + C

Here, C is an arbitrary constant. It appears because when we differentiate a constant, the result is zero. So any constant value could have been part of the original function.

这里的 C 是任意常数。它之所以出现，是因为对常数求导结果为零。因此，任何常数值都可能是原函数的一部分。

For example, the derivative of x³ is 3x². The antiderivative of 3x² is x³ + C. More generally, the antiderivative of k·xⁿ is k·xⁿ⁺¹/(n+1) + C.

例如，x³ 的导数是 3x²。欲求 3x² 的原函数，结果为 x³ + C。更一般地，k·xⁿ 的原函数是 k·xⁿ⁺¹/(n+1) + C。

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2. Handling Constant Factors and Addition/Subtraction | 常数因子与加减法的处理

When a function is multiplied by a constant, that constant simply carries through the integration. The rule is:

当一个函数乘以常数时，该常数直接保留在积分结果中。法则如下：

∫ k·f(x) dx = k·∫ f(x) dx

Similarly, when a function is the sum or difference of two simpler functions, we integrate each term separately. This is called term-by-term integration:

类似地，当一个函数是两个较简单函数的和或差时，我们分别对每一项积分。这称为逐项积分：

∫ [f(x) + g(x)] dx = ∫ f(x) dx + ∫ g(x) dx

For example, ∫ (4x³ − 2x + 5) dx = x⁴ − x² + 5x + C. Let us verify: differentiating x⁴ gives 4x³, differentiating −x² gives −2x, and differentiating 5x gives 5. The constant C covers any additive constant lost in differentiation.

例如，∫ (4x³ − 2x + 5) dx = x⁴ − x² + 5x + C。我们验证一下：对 x⁴ 求导得 4x³，对 −x² 求导得 −2x，对 5x 求导得 5。常数 C 涵盖了微分过程中丢失的任何加性常数。

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3. The Special Case: x⁻¹ and the Natural Logarithm | 特殊情况：x⁻¹ 与自然对数

The power rule fails when n = −1, because it would force us to divide by zero. Differentiating ln|x| gives 1/x, so we use the special result:

当 n = −1 时，幂法则失效，因为它要求我们除以零。而对 ln|x| 求导得到 1/x，因此我们使用特殊结果：

∫ x⁻¹ dx = ∫ (1/x) dx = ln|x| + C

The absolute value signs are essential because ln is only defined for positive inputs, but 1/x is defined for both positive and negative x (except x = 0). In exam questions, if the domain of x is positive, you may write ln x, but ln|x| is always safer.

绝对值符号至关重要，因为 ln 仅对正输入有定义，而 1/x 对正负 x 均有定义（x = 0 除外）。在考试题目中，若 x 的定义域为正数，你可以写 ln x，但写 ln|x| 总是更安全。

For example, ∫ (3/x) dx = 3 ln|x| + C, since the constant factor 3 is carried through.

例如，∫ (3/x) dx = 3 ln|x| + C，因为常数因子 3 会被保留。

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4. Trigonometric Functions | 三角函数

Antidifferentiation of trigonometric functions follows directly from reversing known derivatives. The five standard results are:

三角函数的原函数直接来自对已知导数的逆运算。五个标准结果是：

• ∫ cos x dx = sin x + C because d/dx (sin x) = cos x

• ∫ sin x dx = −cos x + C because d/dx (cos x) = −sin x

• ∫ sec²x dx = tan x + C because d/dx (tan x) = sec²x

• ∫ sec x tan x dx = sec x + C because d/dx (sec x) = sec x tan x

• ∫ cosec²x dx = −cot x + C because d/dx (cot x) = −cosec²x

注意：中文说明：第一项因为 d/dx(sin x)=cos x，第二项因为 d/dx(cos x)=−sin x，第三项因为 d/dx(tan x)=sec²x，第四项因为 d/dx(sec x)=sec x tan x，第五项因为 d/dx(cot x)=−cosec²x。

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5. Exponential Functions | 指数函数

For natural exponential functions, the result is elegantly simple:

对于自然指数函数，其结果简洁优美：

∫ eˣ dx = eˣ + C

Because eˣ is its own derivative, it is also its own antiderivative. For a more general base a, we use:

因为 eˣ 是自身的导数，所以也是自身的原函数。对于更一般的底数 a，我们使用：

∫ aˣ dx = aˣ/ln a + C (a > 0, a ≠ 1)

For an exponential with a linear argument such as e^(kx) or a^(kx), we divide by the coefficient of x:

对于含线性自变量的指数函数，如 e^(kx) 或 a^(kx)，我们将结果除以 x 的系数：

∫ e^(kx) dx = (1/k)·e^(kx) + C

This works because the chain rule introduces a factor k when differentiating, so we must undo that factor. The same idea applies to sin(kx), cos(kx), sec²(kx), and so on.

这个法则成立，是因为链式法则在求导时引入因子 k，所以我们须消除该因子。同样的思路适用于 sin(kx)、cos(kx)、sec²(kx) 等等。

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6. Using Initial Conditions to Find the Constant | 利用初始条件求常数

Without additional information, C remains undetermined. However, if we are told the value of the original function at a specific point, we can substitute to find C. Suppose we know dy/dx = 2x + 1 and that y = 4 when x = 1.

若无额外信息，C 始终无法确定。然而，若告知原函数在某点的值，我们可以代入求出 C。假设 dy/dx = 2x + 1，且当 x = 1 时 y = 4。

Step 1: Integrate to find the general antiderivative.

步骤一：积分求一般原函数。

y = ∫ (2x + 1) dx = x² + x + C

Step 2: Substitute x = 1, y = 4.

步骤二：代入 x = 1, y = 4。

4 = 1² + 1 + C ⇒ C = 2

Step 3: Write the particular solution.

步骤三：写出特解。

y = x² + x + 2

This simple procedure turns a whole family of parallel curves into one specific curve, and it is regularly examined in both pure mathematics and applied contexts.

这个简单的方法将一族平行曲线转化为一条特定曲线，在纯数学与应用情境中都是常规考点。

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7. Worked Example: A Polynomial Combined with a Negative Power | 综合例题：多项式与负幂次结合

Find the antiderivative of x³ + x⁺² − (4/x²) + 2.

求 x³ + x⁺² − (4/x²) + 2 的原函数。

Rewrite the negative power in index form: −4/x² = −4x⁻². Then integrate term by term:

将负幂次改写为指数形式：−4/x² = −4x⁻²。然后逐项积分：

∫ x³ dx = x⁴/4，∫ x⁺² dx = x³/3，∫ −4x⁻² dx = −4x⁻¹/(−1) = 4x⁻¹ = 4/x，∫ 2 dx = 2x

Combining these results:

合并这些结果：

F(x) = x⁴/4 + x³/3 + 4/x + 2x + C

Always remember to include the + C. Many marks are lost in tests simply because students forget this constant, especially when the question does not provide an initial condition.

始终记得加上 + C。许多学生在考试中失分，仅仅是因为忘记了这个常数，尤其在题目未给出初始条件时。

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

The table below summarises the most frequent errors made when finding antiderivatives, together with the correct approach.

下表总结了求原函数时最常出现的错误以及正确做法。

Common Mistake | 常见错误	Correct Approach | 正确做法
Forgetting the constant C | 忘记常数 C	Always append + C for an indefinite integral | 不定积分恒需加 + C
Applying the power rule to x⁻¹ | 对 x⁻¹ 应用幂法则	Use ∫ x⁻¹ dx = ln|x| + C | 使用 ∫ x⁻¹ dx = ln|x| + C
Forgetting to divide by the new exponent | 忘记除以新指数	After raising the power by 1, divide by n+1 | 幂次加 1 后除以 n+1
Sign errors with sin and cos | 对 sin 和 cos 出现符号错误	∫ sin x dx = −cos x + C; remember the minus | ∫ sin x dx = −cos x + C；牢记负号
Ignoring the chain rule factor for e^(kx) | 对 e^(kx) 忽略链式法则因子	Divide by k: ∫ e^(kx) dx = e^(kx)/k + C | 除以 k：∫ e^(kx) dx = e^(kx)/k + C

A systematic approach will help you avoid these pitfalls. Before integrating a term, first rewrite it in the standard form axⁿ, a·e^(kx), a·sin(kx), a·cos(kx) or a/x. Then identify the rule you need, perform the operation, and finally add the constant C.

系统性的方法能帮助你避免这些陷阱。在积分每一项之前，先将其改写为标准形式 axⁿ、a·e^(kx)、a·sin(kx)、a·cos(kx) 或 a/x。然后确定所需法则，进行运算，最后加上常数 C。

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9. Applications: Solving Simple Differential Equations | 应用：解简单微分方程

Finding an antiderivative is exactly the same as solving a differential equation of the form dy/dx = f(x). For example, if we are given:

求原函数正是解形如 dy/dx = f(x) 的微分方程。例如，若给出：

dy/dx = 6x² − 2x, with y = 3 when x = 0

We integrate first:

我们首先积分：

y = ∫ (6x² − 2x) dx = 2x³ − x² + C

Then substitute the initial condition:

然后代入初始条件：

3 = 0 − 0 + C ⇒ C = 3

Thus the particular solution is y = 2x³ − x² + 3. This is a common A-Level question type, often embedded in kinematics problems where acceleration is integrated to get velocity, and velocity is integrated to get displacement.

因此特解为 y = 2x³ − x² + 3。这是A-Level常见题型，常嵌入运动学问题中：加速度积分得到速度，速度积分得到位移。

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10. Practice Checklist for Exam Preparation | 备考自检清单

Use the following checklist before attempting any antiderivative question:

在解答任何原函数题目之前，请使用以下自检清单：

• Have you rewritten all radicals as fractional powers, and all reciprocals as negative powers? | 是否已将根号改写为分数次幂、倒数改写为负次幂？

• Does each term match one of the standard forms: xⁿ, x⁻¹, eˣ, aˣ, sin x, cos x, sec²x, sec x tan x, cosec²x? | 每一项是否匹配标准形式之一：xⁿ、x⁻¹、eˣ、aˣ、sin x、cos x、sec²x、sec x tan x、cosec²x？

• For linear arguments like 2x or 3x, have you divided by the coefficient? | 对于线性自变量如 2x 或 3x，是否已经除以系数？

• Have you included the constant C in indefinite integrals? | 不定积分中是否已包含常数 C？

• If a point is given, have you substituted it to find C? | 若给定了点坐标，是否已代入求出 C？

• Can you check your answer by differentiating it to obtain the original integrand? | 你是否能通过对结果求导来验证它还原为原被积函数？

Checking by differentiation is the most powerful self-validation tool. If you differentiate your proposed antiderivative and recover f(x), your answer is almost certainly correct.

通过求导进行验证是最强大的自我校验工具。若对你所得的原函数求导后能还原 f(x)，你的答案几乎必然正确。

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11. Summary | 总结

Reversing differentiation is a core skill in A-Level mathematics. The central rules are simple: add one to the power and divide, except for x⁻¹ which becomes ln|x|; copy sin and cos with the appropriate sign; copy exponential functions and divide by the coefficient of x; and always add the constant C. When an initial condition is supplied, use it to find the exact value of C.

逆转微分是A-Level数学的核心技能。核心规则简洁明了：幂次加一并除以新幂次，唯独 x⁻¹ 变成 ln|x|；对 sin 和 cos，注意相应符号；对指数函数，直接复制并除以 x 的系数；始终加上常数 C。当题目给出初始条件时，利用它求出 C 的具体值。

With careful practice and the habit of checking your answer by differentiation, you will master antidifferentiation and be fully prepared for related exam questions.

通过认真练习以及“求导回验”的习惯，你将掌握反微分，并可从容应对相关考试题目。

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