📚 Integrating Standard Functions | 标准函数积分
f differentiation, and in A-Level Edexcel Mathematics you are expected to recognise and apply a set of standard integrals immediately, without deriving them from first principles every time. The general idea is simple: if F' re C is an arbitrary constant of integration. Every result in the standard table exists because some function, when differentiated, gives back the integrand.
积分是微分的逆运算。在 A-Level Edexcel 数学中，学生需要立刻识别并直接应用一组标准积分，而不必每次都从第一原理重新推导。其基本思想十分简单：如果 F'(x) = f(x)，那么 ∫ f(x) dx = F(x) + C，其中 C 是任意积分常数。标准积分表中的每一条结果，都来自某个函数求导后恰好等于被积函数。
This article summarises the standard integrals required for the Edexcel A-Level Mathematics course, explains where each one comes from, and shows how to use them with confidence in pure mathematics problems - including integration by substitution, integration by parts, and the solution of differential equations.
本文总结了 Edexcel A-Level 数学课程要求的标准积分，说明每一条的来源，并演示如何在纯数学问题中自信地使用它们——包括换元积分法、分部积分法以及微分方程的求解。
1. Introduction to Standard Integrals | 标准积分简介
derivative. For example, since d/dx(sin x) = cos x, it follows that ∫ cos x dx = sin x + C. This reverse relationship is the key to building a reliable table of standard integrals, and it means that every derivative you memorise simultaneously gives you an integral for free.
在 A-Level 数学中，标准积分是指你应该能够立刻写出的积分，通常因为这些积分正好是某个已知导数的逆运算。例如，因为 d/dx(sin x) = cos x，所以 ∫ cos x dx = sin x + C。这种逆关系是建立可靠标准积分表的关键，也意味着你每记住一个导数，就同时免费获得了一个积分。
You will be expected to use these results throughout pure mathematics: when evaluating definite integrals, when applying integration by substitution or integration by parts, and when solving first-order differential equations by separating the variables. Speed and accuracy with the standard table therefore pays off across the whole course.
在整个纯数学部分你都会用到这些结果：计算定积分时、使用换元积分法或分部积分法时，以及用分离变量法求解一阶微分方程时。因此，对标准积分表做到又快又准，会在整门课程中持续带来回报。
Make sure that you know both the basic forms and the forms involving linear functions of x, such as ax + b. The linear versions carry an extra factor of 1/a, and forgetting that factor is one of the most common sources of lost marks in examinations.
务必同时掌握基本形式以及含有 x 的线性函数形式，例如 ax + b。线性版本会额外带一个 1/a 因子，而漏掉这个因子是考试中最常见的失分原因之一。
2. The Power Rule | 幂函数积分法则
y the new index. This is the direct reverse of the differentiation rule d/dx(xⁿ) = n xⁿ⁻¹.
对于任何实数 n（除 -1 外），xⁿ 的积分通过将指数加一、再除以新的指数得到。这正好是求导法则 d/dx(xⁿ) = n xⁿ⁻¹ 的逆运算。
∫ xⁿ dx = xⁿ⁺¹ / (n + 1) + C, n ≠ −1
oduces a logarithm, which is covered in Section 5. Provided n ≠ −1, the rule works for positive integers, negative integers, and fractions alike.
限制条件 n ≠ −1 的存在是因为该公式会导致除以零。n = −1 的情形很特殊，其结果是对数，将在第 5 节讨论。只要 n ≠ −1，这条法则对正整数、负整数和分数同样适用。
Some typical applications are worth practising until they are automatic:
以下典型应用值得反复练习，直到能自动写出：
- ∫ x⁵ dx = x⁶ / 6 + C
- ∫ x dx = x² / 2 + C
- ∫ 1 dx = ∫ x⁰ dx = x + C
- ∫ x⁻² dx = −x⁻¹ + C = −1/x + C
- ∫ √x dx = ∫ x1/2 dx = (2/3) x3/2 + C
- ∫ 1/√x dx = ∫ x−1/2 dx = 2√x + C
Notice how the last two examples require you to rewrite roots and reciprocals as powers of x before integrating. This rewriting step is not optional - it is the first thing an examiner expects to see when a question contains surds or algebraic fractions.
注意最后两个例子需要先把根式和倒数改写成 x 的幂次，然后再积分。这一步不是可选的——当题目中出现根式或代数分式时，改写成幂次是阅卷人期望看到的第一个动作。
3. Preparing the Integrand | 被积函数的预处理
Many examination questions deliberately present an integrand that cannot be integrated term by term in its current form. The power rule applies to sums and differences of powers of x, so your first task is usually to expand brackets, divide through, or simplify before integrating.
许多考题会故意给出无法直接逐项积分的被积函数。幂函数法则适用于 x 的幂次的和与差，所以你的第一步通常是展开括号、逐项相除或先化简，然后再积分。
3x²/2 + C. Attempting to integrate the original product directly is not valid at A-Level.
当出现括号相乘时，先展开。例如 ∫ x(x + 3) dx 应先改写为 ∫ (x² + 3x) dx = x³/3 + 3x²/2 + C。在 A-Level 中，直接对原来的乘积积分是不成立的。
x, divide each term separately. So ∫ (x³ + 2x)/x² dx becomes ∫ (x + 2x⁻¹) dx = x²/2 + 2 ln|x| m and a logarithmic term, which is exactly the kind of mixed answer examiners like to set.
当某一项被 x 的幂次除时，应逐项相除。于是 ∫ (x³ + 2x)/x² dx 变成 ∫ (x + 2x⁻¹) dx = x²/2 + 2 ln|x| + C。这里的化简同时揭示了一个幂次项和一个对数项，这正是命题人喜欢设置的混合型答案。
A useful checklist before you integrate anything:
在积分之前，一个实用的检查清单是：
- Expand all brackets and remove all parentheses.
- Split fractions with a single denominator into separate terms.
- Rewrite every root as a fractional power.
- Rewrite every reciprocal as a negative power, unless it is 1/x or 1/(ax + b).
- Collect like terms so that you integrate the fewest possible pieces.
4. Integrals of Exponential Functions | 指数函数的积分
The exponential function eˣ is its own derivative, and therefore its own integral. This self-reproducing property makes it the simplest entry in the entire standard table.
指数函数 eˣ 的导数等于它本身，因此它的积分也等于它本身。这种"自我复制"的性质使它成为整张标准积分表中最简单的一条。
∫ eˣ dx = eˣ + C
When the exponent is a linear function of x, the extra factor of 1/a appears again:
当指数是 x 的线性函数时，那个额外的 1/a 因子再次出现：
∫ eᵃˣ⁺ᵇ dx = (1/a) eᵃˣ⁺ᵇ + C
You can verify this instantly by differentiating: d/dx[(1/a) eᵃˣ⁺ᵇ] = (1/a) × a eᵃˣ⁺ᵇ = eᵃˣ ating the answer is a fast and reliable habit, and it takes only a few seconds in an exam.
你可以立刻通过求导验证：d/dx[(1/a) eᵃˣ⁺ᵇ] = (1/a) × a eᵃˣ⁺ᵇ = eᵃˣ⁺ᵇ。把答案求导来检验积分，是一个快速而可靠的习惯，在考试中只需几秒钟。
evel Mathematics does not require you to integrate general exponential functions such as aˣ, though the result ∫ aˣ dx = aˣ / ln a + C follows from writing aˣ = eˣ ˡⁿ ᵃ. If it appears at all, it will normally be guided.
注意，A-Level 数学不要求你积分一般形式的指数函数如 aˣ，尽管由 aˣ = eˣ ˡⁿ ᵃ 可以推出 ∫ aˣ dx = aˣ / ln a + C。如果它真的出现，通常会有提示引导。
5. The Reciprocal Function and Logarithms | 倒数函数与对数
0, the integral of 1/x is a natural logarithm. Because the integrand is undefined at zero and ln x is undefined for negative x, the modulus signs are essential.
幂函数法则无法处理的 n = −1 情形，恰恰就是 1/x 的积分。因为当 x > 0 时 d/dx(ln x) = 1/x，所以 1/x 的积分是自然对数。由于被积函数在零点无定义、而 ln x 对负数无定义，绝对值符号是必不可少的。
∫ (1/x) dx = ln|x| + C
As with exponential functions, the linear version introduces a factor of 1/a:
与指数函数一样，线性版本会引入一个 1/a 因子：
∫ 1/(ax + b) dx = (1/a) ln|ax + b| + C
applies with n = −3, giving ∫ x⁻³ dx = −x⁻²/2 + C = −1/(2x²) + C. Only a denominator of exactly degree one, with no extra power of x, produces a logarithm.
一个非常常见的考试陷阱是分母为 x 的幂次的分式，例如 1/x³。这并不产生对数。因为 1/x³ = x⁻³，应用幂函数法则时 n = −3，得到 ∫ x⁻³ dx = −x⁻²/2 + C = −1/(2x²) + C。只有分母恰好为一次、且没有额外的 x 幂次时，才会产生对数。
The same idea extends to fractions such as f'(x)/f(x), where the numerator is the derivative of the denominator. In that situation ∫ f'(x)/f(x) dx = ln|f(x)| + C. For instance, ∫ 2x/(x² + 5) dx = ln|x² + 5| + C. This pattern is worth recognising because it saves a full substitution.
同样的思路可推广到形如 f'(x)/f(x) 的分式，即分子是分母的导数。此时 ∫ f'(x)/f(x) dx = ln|f(x)| + C。例如 ∫ 2x/(x² + 5) dx = ln|x² + 5| + C。这个模式值得识别，因为它可以省去一次完整的换元。
6. Integrals of Trigonometric Functions | 三角函数积分
cause d/dx(sin x) = cos x but d/dx(cos x) = −sin x, integrating cosine is straightforward while integrating sine produces a minus sign.
标准积分表中的三角函数条目可以直接从相应的导数读出，但要记住正弦项会变号。因为 d/dx(sin x) = cos x，而 d/dx(cos x) = −sin x，所以积分余弦很直接，而积分正弦则会产生一个负号。
sp; ∫ cos x dx = sin x + C ∫ sec² x dx = tan x + C
C and ∫ cosec² x dx = −cot x + C appear in some specifications and are useful to be aware of, although the core Edexcel A-Level requirement is sin, cos and sec².
第三条结果来自 d/dx(tan x) = sec² x，这是你应当掌握的标准导数。相关结果如 ∫ sec x tan x dx = sec x + C 和 ∫ cosec² x dx = −cot x + C 在某些考纲中出现，了解它们很有用，不过 Edexcel A-Level 的核心要求是 sin、cos 与 sec²。
With linear arguments, the pattern continues unchanged:
当自变量为线性式时，规律保持不变：
- ∫ 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
Angles in these formulas are always in radians for the calculus results to hold in this simple form. Edexcel calculus questions use radians by default, and marks are regularly lost by students who switch their calculator into degree mode.
在这些公式中，角度必须始终使用弧度，微积分结果才能保持这种简洁形式。Edexcel 的微积分题目默认使用弧度，而把计算器切到角度模式是学生经常失分的原因。
7. Linear Forms: The 1/a Factor | 线性形式：1/a 因子
Almost every standard integral has a companion version in which x is replaced by ax + b. The rule is uniform and easy to remember: differentiate the inner function to get a, and divide the result by a. The table below collects the forms you need for Edexcel A-Level Mathematics.
几乎每一个标准积分都有一个对应版本：把 x 替换为 ax + b。规律统一且易于记忆：对内层函数求导得到 a，然后把结果除以 a。下表汇总了 Edexcel A-Level 数学所需的各种形式。
| Integrand | 被积函数 | Integral | 积分结果 |
| (ax + b)ⁿ, n ≠ −1 | (ax + b)ⁿ⁺¹ / [a(n + 1)] + C |
| eᵃˣ⁺ᵇ | (1/a) eᵃˣ⁺ᵇ + C |
| 1/(ax + b) | (1/a) ln|ax + b| + C |
| sin(ax + b) | −(1/a) cos(ax + b) + C |
| cos(ax + b) | (1/a) sin(ax + b) + C |
| sec²(ax + b) | (1/a) tan(ax + b) + C |
ease the power to 5, divide by 5, and then divide by the coefficient 3: the answer is (3x + 1)⁵ / 15 + C. Differentiating this mentally confirms the result in one step.
一个例子可以说明其自然之处。求 ∫ (3x + 1)⁴ dx 时，把指数升到 5、除以 5，再除以系数 3：答案为 (3x + 1)⁵ / 15 + C。心算求导一次即可验证结果。
8. Definite Integrals and the Constant | 定积分与积分常数
A definite integral evaluates the antiderivative between two limits, and the constant C cancels out in the subtraction. This is why the constant is normally written only for indefinite integrals, and why omitting it in a definite integral costs nothing.
定积分是在两个上下限之间对被积函数的原函数求值，而常数 C 在相减时会被抵消。这就是为什么通常只在不定积分中写出常数，也是为什么定积分中不写常数并不会被扣分。
∫ₐᵇ f(x) dx = F(b) − F(a)
racket with the limits before substituting - it makes your working easy to follow and reduces the risk of arithmetic slips.
例如，∫₁³ (2x + 1) dx = [x² + x]₁³ = (9 + 3) − (1 + 1) = 10。在代入之前先写出带上下限的方括号——这能让解题过程清晰易读，并降低计算失误的风险。
elow the axis contribute negative values, which is why questions sometimes ask you to find the total area by splitting the integral at each root and taking absolute values.
定积分还具有几何意义：其值等于曲线 y = f(x) 与 x 轴之间从 x = a 到 x = b 的有向面积。位于 x 轴下方的区域贡献负值，这就是为什么有些题目要求你在每个零点处拆分积分并取绝对值，以求出总面积。
For indefinite integrals, always add '+ C' h as y = 4 when x = 0, substitute it to find the specific value of C and state it clearly.
对于不定积分，务必加上 '+ C'。如果答案是 x³/3 而没有常数，即使其他步骤完全正确，也确实会被扣分。如果题目给出了初始条件，例如当 x = 0 时 y = 4，就把它代入求出 C 的具体值并清楚写出。
9. Using Standard Integrals with Substitution and Parts | 在换元法与分部积分中使用标准积分
amiliar integral into one or more standard integrals by changing the variable, usually writing u = ax + b or u = the inner function.
标准积分是更高级技巧的基石。在换元积分法中，核心目的正是通过改变变量（通常设 u = ax + b 或 u 为内层函数），把一个陌生的积分转化成一个或多个标准积分。
d back-substitution gives (x² + 1)⁶/6 + C. The power rule did all the work; the substitution simply made the power rule applicable.
例如，求 ∫ 2x(x² + 1)⁵ dx 时，令 u = x² + 1，则 du = 2x dx。积分变为 ∫ u⁵ du = u⁶/6 + C，回代得到 (x² + 1)⁶/6 + C。幂函数法则完成了全部工作，换元只是让幂函数法则变得可用。
s a second integral that must itself be standard or simpler. Choosing u and dv wisely usually means the remaining integral is one you can write down immediately from the standard table.
在分部积分法中，公式 ∫ u (dv/dx) dx = uv − ∫ v (du/dx) dx 会留下第二个积分，而这个积分本身必须是标准积分或更简单。明智地选取 u 与 dv，通常意味着剩下的积分可以直接从标准积分表写出。
Partial fractions combine the same idea with logarithms. Once a rational expression such as (3x + 5)/[(x − 1)(x + 2)] is split into A/(x − 1) + B/(x + 2), each piece integrates to a logarithm of a linear function, using the 1/(ax + b) entry from Section 7.
部分分式则把同样的思想与对数结合起来。一旦把形如 (3x + 5)/[(x − 1)(x + 2)] 的有理式拆成 A/(x − 1) + B/(x + 2)，每一部分都可以利用第 7 节的 1/(ax + b) 条目，积成一个线性函数的对数。
10. Standard Integrals in Differential Equations | 标准积分在微分方程中的应用
Many A-Level differential equations are solved by separating the variables and then integrating both sides using standard results. The skill being tested is as much about rearranging as it is about integrating.
许多 A-Level 微分方程通过分离变量、再用标准积分对两边同时积分来求解。这里考查的能力，既在于积分本身，也同样在于移项变形。
= ky, a model of exponential growth or decay. Separating gives ∫ (1/y) dy = ∫ k dx, so ln|y| = kx + c. Exponentiating both sides yields y = A eᵏˣ, where A = ec is an arbitrary positive constant determined by an initial condition.
考虑 dy/dx = ky，这是指数增长或衰减的模型。分离变量得到 ∫ (1/y) dy = ∫ k dx，于是 ln|y| = kx + c。两边取指数得到 y = A eᵏˣ，其中 A = ec 是由初始条件确定的任意正常数。
Notice how three standard results were used in that single solution: the logarithm for 1/y, the constant rule for k, and the exponential integral in reverse to undo the log. Fluency with the standard table turns an intimidating differential equation into three lines of routine work.
注意这一个解中就用到了三条标准结果：1/y 对应对数，常数 k 对应常数积分，而指数积分则被反向用来消去对数。对标准积分表的熟练，能把一道令人生畏的微分方程变成三行常规计算。
final answer should be stated as an equation for y in terms of t rather than left as an implicit relationship.
在建模题中，记得结合情境解释常数的意义。如果 y 表示以千为单位的人口数，且当 t = 0 时 y = 4，那么 A = 4，最终答案应写成 y 关于 t 的方程，而不是停留在隐式关系上。
11. Common Mistakes and Exam Technique | 常见错误与应试技巧
Most lost marks in integration questions come from a small number of recurring errors rather than from conceptual confusion. Knowing them in
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