Integration of Power Functions xⁿ | 幂函数xⁿ的积分公式详解

📚 Integration of Power Functions xⁿ | 幂函数xⁿ的积分公式详解

In A-Level Mathematics, the ability to integrate power functions of the form xⁿ is one of the most fundamental skills you will develop. Whether you are working with positive exponents, negative exponents, or fractional exponents, the general formula provides a consistent and reliable method for finding antiderivatives.

在A-Level数学中，掌握形如xⁿ的幂函数的积分是你将培养的最基本技能之一。无论你处理的是正整数指数、负整数指数还是分数指数，通用公式都提供了一种一致且可靠的方法来求解原函数。

This guide will take you through every aspect of integrating power functions — from the core formula and its special cases, to algebraic manipulation, definite integrals, common pitfalls, and exam-style worked examples.

本指南将带你全面了解幂函数积分的各个方面——从核心公式及其特殊情况，到代数变形、定积分、常见错误以及考试风格的例题解答。

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1. The General Formula | 通用公式

The integration of a power function xⁿ follows a simple and elegant rule. For any real number n, provided that n ≠ −1, the indefinite integral is:

幂函数xⁿ的积分遵循一条简洁而优雅的规则。对于任何实数n，只要n ≠ −1，不定积分为：

∫ xⁿ dx = xⁿ⁺¹ ÷ (n+1) + C, n ≠ −1

Here, C represents the constant of integration. This rule is sometimes written in fraction form as x^(n+1)/(n+1) + C.

这里的C代表积分常数。这条规则有时也用分数形式写为x^(n+1)/(n+1) + C。

The derivation follows directly from the rule for differentiating power functions. Since the derivative of xⁿ⁺¹ is (n+1)xⁿ, dividing by (n+1) reverses the process of differentiation.

该推导直接来自幂函数的微分法则。因为xⁿ⁺¹的导数是(n+1)xⁿ，所以除以(n+1)就逆转了微分的过程。

Verification | 验证：

To verify that the formula is correct, we differentiate the result:

为了验证公式的正确性，我们对结果求导：

d/dx [ xⁿ⁺¹/(n+1) + C ] = (n+1)xⁿ/(n+1) = xⁿ

This confirms that our integration formula is indeed the reverse of differentiation.

这证实了我们的积分公式确实是微分的逆运算。

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2. The Special Case n = −1 | n = −1的特殊情况

The general formula fails when n = −1, because the denominator (n+1) becomes zero, and we cannot divide by zero. This is not merely a technical obstacle — it represents a fundamentally different situation that requires a separate rule.

当n = −1时，通用公式失效，因为分母(n+1)变为零，我们不能除以零。这不仅仅是一个技术上的障碍——它代表了一种根本不同的情形，需要单独的处理规则。

For the case n = −1, the function is x⁻¹, which is 1/x. The integral of this function involves the natural logarithm:

对于n = −1的情形，函数是x⁻¹，即1/x。这个函数的积分涉及自然对数：

∫ x⁻¹ dx = ∫ (1/x) dx = ln|x| + C

The absolute value bars around x are essential. Since the natural logarithm is only defined for positive arguments, the absolute value ensures that the formula works for negative values of x as well.

x周围的绝对值符号是必不可少的。由于自然对数只对正数有定义，绝对值确保公式对负值的x同样成立。

Why does this happen? | 为什么会这样？

When we differentiate ln|x|, we obtain 1/x. This can be shown using the chain rule and the fact that the derivative of ln(x) is 1/x for x > 0. For x < 0, we can write x = −u where u > 0, and ln|−u| = ln(u), giving the same derivative 1/x in both cases.

当我们对ln|x|求导时，得到1/x。这可以通过链式法则以及ln(x)的导数为1/x（当x > 0时）来证明。对于x < 0，我们可以令x = −u，其中u > 0，而ln|−u| = ln(u)，两种情况下的导数都为1/x。

In many exam problems, particularly those involving definite integrals, you may drop the absolute value if the interval of integration lies entirely within positive x values. However, for indefinite integrals, always include the absolute value.

在许多考试问题中，特别是涉及定积分的问题，如果积分区间完全在正的x值范围内，你可以省略绝对值。然而，对于不定积分，务必加上绝对值。

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3. Working Through the General Formula | 深入理解通用公式

Let us apply the general formula to several concrete examples to illustrate how it works in practice.

让我们通过几个具体例子来应用通用公式，说明它在实践中如何运作。

Example 1: Positive Integer Exponent | 例1：正整数指数

Find ∫ x⁵ dx.

求∫ x⁵ dx。

Here n = 5. Applying the formula:

这里n = 5。应用公式：

∫ x⁵ dx = x⁶/6 + C

Example 2: Negative Exponent | 例2：负指数

Find ∫ x⁻² dx.

求∫ x⁻² dx。

Here n = −2. Since n ≠ −1, we can apply the general formula:

这里n = −2。由于n ≠ −1，我们可以应用通用公式：

∫ x⁻² dx = x⁻¹/(−1) + C = −x⁻¹ + C = −1/x + C

Example 3: Fractional Exponent | 例3：分数指数

Find ∫ √x dx.

求∫ √x dx。

First, rewrite the square root as a power: √x = x^(1/2). Then n = 1/2:

首先，将平方根改写为幂的形式：√x = x^(1/2)。然后n = 1/2：

∫ x^(1/2) dx = x^(3/2) ÷ (3/2) + C = (2/3)x^(3/2) + C

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4. Simplifying Before Integrating | 先化简再积分

In many examination questions, the integrand (the function being integrated) is not presented in the simple form xⁿ. Instead, you must first perform algebraic manipulation to express the function as a sum of power functions before applying the integration rule term by term.

在许多考试题中，被积函数（要求积分的函数）并非以简单的xⁿ形式呈现。相反，你必须先进行代数运算，将函数表达为若干幂函数的和，然后再逐项应用积分规则。

Key techniques include | 关键技巧包括：

• Expanding brackets — e.g. multiply out (x+1)(x−2) before integrating.
• 展开括号——例如在积分前先计算(x+1)(x−2)的乘积。
• Splitting fractions — e.g. write (x³+2x)/x as x² + 2.
• 拆分分数——例如将(x³+2x)/x写成x² + 2。
• Converting roots to fractional powers — e.g. ∛x = x^(1/3).
• 将根式转换为分数次幂——例如∛x = x^(1/3)。
• Rewriting reciprocal powers — e.g. 1/x² = x⁻².
• 改写倒数幂——例如1/x² = x⁻²。

Worked Example | 例题：

Find ∫ (x³ + 2x² − 5)/x² dx.

求∫ (x³ + 2x² − 5)/x² dx。

Step 1: Split the fraction into separate terms:

步骤1：将分数拆分为若干独立的项：

∫ (x³/x² + 2x²/x² − 5/x²) dx = ∫ (x + 2 − 5x⁻²) dx

Step 2: Integrate each term separately:

步骤2：分别对每一项积分：

∫ x dx + ∫ 2 dx − ∫ 5x⁻² dx = x²/2 + 2x + 5x⁻¹ + C

Note: When integrating −5x⁻², we get −5 × (x⁻¹/−1) = 5x⁻¹. Always check the sign carefully.

注意：对−5x⁻²积分时，得到−5 × (x⁻¹/−1) = 5x⁻¹。务必仔细检查符号。

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5. Definite Integrals | 定积分

When evaluating a definite integral, we apply the same rules to find the antiderivative, but then substitute the upper and lower limits and subtract. The constant of integration C cancels out and does not appear in the final answer.

当计算定积分时，我们应用相同的规则求原函数，然后将上限和下限代入并相减。积分常数C会相互抵消，不会出现在最终答案中。

The evaluation is performed using the notation:

计算使用如下记号：

∫ₐᵇ f(x) dx = [F(x)]ₐᵇ = F(b) − F(a)

where F(x) is the antiderivative of f(x).

其中F(x)是f(x)的原函数。

Worked Example | 例题：

Evaluate ∫₁³ x² dx.

计算∫₁³ x² dx。

Step 1: Find the antiderivative:

步骤1：求原函数：

∫ x² dx = x³/3

Step 2: Substitute the limits:

步骤2：代入上下限：

[x³/3]₁³ = (3³/3) − (1³/3) = 27/3 − 1/3 = 26/3

Step 3: The final answer is 26/3, which is approximately 8.67 square units.

步骤3：最终答案为26/3，约等于8.67平方单位。

When the lower limit is negative, special care must be taken with signs. For example, evaluating ∫₋₂² x³ dx:

当下限为负数时，必须特别注意符号。例如，计算∫₋₂² x³ dx：

[x⁴/4]₋₂² = (2⁴/4) − ((−2)⁴/4) = 16/4 − 16/4 = 0

The result is zero because x³ is an odd function, symmetric about the origin, so the area below the axis on the negative side exactly cancels the area above the axis on the positive side.

结果为零，因为x³是奇函数，关于原点对称，所以负半轴下方的面积恰好抵消了正半轴上方的面积。

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6. Common Mistakes and How to Avoid Them | 常见错误及避免方法

Students frequently make several avoidable errors when integrating power functions. Being aware of these pitfalls is key to scoring well in exams.

学生在积分幂函数时经常犯几个可以避免的错误。意识到这些陷阱是在考试中获得高分的关键。

Mistake | 错误	Incorrect | 错误做法	Correct | 正确做法
Forgetting the constant C | 忘记常数C	∫ x³ dx = x⁴/4	∫ x³ dx = x⁴/4 + C
Dividing by n instead of n+1 | 除以n而不是n+1	∫ x² dx = x³/2	∫ x² dx = x³/3 + C
Using the general rule for n = −1 | 对n = −1使用通用规则	∫ x⁻¹ dx = x⁰/0	∫ x⁻¹ dx = ln|x| + C
Omitting absolute value in ln | 在ln中省略绝对值	∫ 1/x dx = ln x + C	∫ 1/x dx = ln|x| + C
Incorrectly handling constants | 常数处理不当	∫ 3x² dx = 3x³/3	∫ 3x² dx = x³ + C

Another common error is attempting to integrate a product such as x² × x³ as if each factor could be integrated separately. In this case, simplify first: x² × x³ = x⁵, and then integrate to get x⁶/6 + C.

另一个常见错误是试图将x² × x³这样的乘积当作每个因子可以分别积分来处理。在这种情况下，先化简：x² × x³ = x⁵，然后积分得到x⁶/6 + C。

Remember: you cannot integrate a product by integrating each factor separately and multiplying the results. Always simplify the product into a single power first.

记住：你不能先将每个因子分别积分再将结果相乘来得到乘积的积分。务必先将乘积化简为单一的幂。

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7. Integration with Linear Arguments | 线性内层函数的积分

When the power function has a linear argument, such as (ax + b)ⁿ, the integration rule is extended with a factor of 1/a in front:

当幂函数具有线性内层函数时，例如(ax + b)ⁿ，积分规则前面需要乘以1/a的因子：

∫ (ax + b)ⁿ dx = (ax + b)ⁿ⁺¹ ÷ [a(n+1)] + C, n ≠ −1

Worked Example | 例题：

Find ∫ (2x + 3)⁴ dx.

求∫ (2x + 3)⁴ dx。

Here a = 2, b = 3, n = 4. Applying the rule:

这里a = 2，b = 3，n = 4。应用规则：

∫ (2x + 3)⁴ dx = (2x + 3)⁵ ÷ (2 × 5) + C = (2x + 3)⁵/10 + C

For the reciprocal case n = −1:

对于n = −1的倒数情形：

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

For example, ∫ 1/(3x − 2) dx = (1/3) ln|3x − 2| + C.

例如，∫ 1/(3x − 2) dx = (1/3) ln|3x − 2| + C。

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8. Exam-Style Worked Problems | 考试风格例题精解

Let us work through some realistic examination problems step by step.

让我们一步一步地解答一些真实的考试风格问题。

Problem A | 题目A：

Find ∫ (2x + 1)(x − 4) dx.

求∫ (2x + 1)(x − 4) dx。

Solution | 解答：

Step 1: Expand the brackets:

步骤1：展开括号：

(2x + 1)(x − 4) = 2x² − 8x + x − 4 = 2x² − 7x − 4

Step 2: Integrate term by term:

步骤2：逐项积分：

∫ (2x² − 7x − 4) dx = 2x³/3 − 7x²/2 − 4x + C

Always remember the constant of integration.

始终记得加上积分常数。

Problem B | 题目B：

Evaluate ∫₀⁴ (3√x − x) dx.

计算∫₀⁴ (3√x − x) dx。

Solution | 解答：

Step 1: Rewrite √x as x^(1/2):

步骤1：将√x改写为x^(1/2)：

∫₀⁴ (3x^(1/2) − x) dx

Step 2: Find the antiderivative:

步骤2：求原函数：

[3 × (2/3)x^(3/2) − x²/2]₀⁴ = [2x^(3/2) − x²/2]₀⁴

Step 3: Substitute the limits:

步骤3：代入上下限：

(2 × 4^(3/2) − 16/2) − (0 − 0) = (2 × 8 − 8) = 16 − 8 = 8

The answer is 8.

答案为8。

Problem C | 题目C：

Find the equation of the curve y = f(x) given that dy/dx = 6x² − 4x + 3 and the curve passes through the point (1, 5).

已知dy/dx = 6x² − 4x + 3，且曲线经过点(1, 5)，求曲线y = f(x)的方程。

Solution | 解答：

Step 1: Integrate to find f(x):

步骤1：积分求f(x)：

f(x) = ∫ (6x² − 4x + 3) dx = 2x³ − 2x² + 3x + C

Step 2: Use the given point to find C:

步骤2：利用给定点求C：

5 = 2(1)³ − 2(1)² + 3(1) + C = 2 − 2 + 3 + C = 3 + C

C = 2

Step 3: The equation of the curve is f(x) = 2x³ − 2x² + 3x + 2.

步骤3：曲线方程为f(x) = 2x³ − 2x² + 3x + 2。

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9. Tables of Integrals — Quick Reference | 积分表——快速参考

The following table summarises the most important power function integrals you need to memorise for your examination.

下表总结了考试中你需要记住的最重要的幂函数积分。

Function | 函数	Integral | 积分	Condition | 条件
xⁿ	xⁿ⁺¹/(n+1) + C	n ≠ −1
1/x = x⁻¹	ln|x| + C	x ≠ 0
√x = x^(1/2)	(2/3)x^(3/2) + C	x ≥ 0
1/√x = x^(−1/2)	2√x + C	x > 0
(ax + b)ⁿ	(ax + b)ⁿ⁺¹/[a(n+1)] + C	n ≠ −1
1/(ax + b)	(1/a)ln|ax + b| + C	ax + b ≠ 0

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10. Discussion: Why the Formula Works | 深度讨论：公式为何成立

The correctness of the power rule for integration is grounded in the Fundamental Theorem of Calculus. Integration is, by definition, the inverse operation of differentiation. When we differentiate xⁿ⁺¹/(n+1), we recover xⁿ exactly. This inverse relationship is symmetric: just as differentiation reduces the exponent by one and multiplies by the original exponent, integration increases the exponent by one and divides by the new exponent.

幂函数积分规则的正确性建立在微积分基本定理的基础之上。根据定义，积分是微分的逆运算。当我们对xⁿ⁺¹/(n+1)求导时，恰好恢复xⁿ。这种逆关系是对称的：正如微分将指数减一并乘以原指数，积分将指数加一并除以新指数。

One might intuitively ask why the rule breaks down at n = −1. The reason becomes clear when we examine the behaviour of the formula as n approaches −1. The denominator n + 1 approaches zero, causing the result to grow without bound. Meanwhile, xⁿ⁺¹ approaches x⁰ = 1. The logarithmic function ln|x| emerges naturally as the limiting case that bridges this gap, and it is a consequence of the fact that the natural logarithm is defined as the area under the hyperbola y = 1/t from 1 to x.

你可能会直觉地问，为什么这条规则在n = −1时会失效。当我们考察公式当n趋近于−1时的行为时，原因就变得清晰了。分母n + 1趋近于零，导致结果无界增长。与此同时，xⁿ⁺¹趋近于x⁰ = 1。对数函数ln|x|自然地作为连接这一缺口的极限情况出现，这是自然对数被定义为双曲线y = 1/t从1到x的面积的必然结果。

This understanding is not required to perform calculations, but it deepens your mathematical intuition and helps you remember the special case. For A-Level purposes, the key message is simple: n = −1 is special; every other exponent follows the general power rule.

这种理解并非解题所必需，但它能加深你的数学直觉，帮助你记住特殊情况。从A-Level考试的角度看，关键信息很简洁：n = −1是特殊的；其他所有指数都遵循通用幂函数积分规则。

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