Indefinite Integrals: Basic Formulas and Calculation Techniques | 不定积分：基本公式与计算技巧

📚 Indefinite Integrals: Basic Formulas and Calculation Techniques | 不定积分：基本公式与计算技巧

Indefinite integration is the reverse process of differentiation. Given a function f(x), an indefinite integral, denoted as ∫f(x)dx, is a family of functions F(x) such that F'(x) = f(x). Since differentiation of a constant is zero, the general solution always includes an arbitrary constant C, called the constant of integration.

不定积分是微分的逆运算。对于函数 f(x)，不定积分记为 ∫f(x)dx，是满足 F'(x) = f(x) 的一族函数 F(x)。由于常数微分后为零，通解中总包含一个任意常数 C，称为积分常数。

For example, because d/dx (x²) = 2x, we write ∫2x dx = x² + C. The symbol ∫ is the integral sign, and dx indicates the variable of integration.

例如，由于 d/dx (x²) = 2x，我们写作 ∫2x dx = x² + C。符号 ∫ 是积分号，dx 表示积分的变量。

1. The Power Rule | 幂函数积分公式

The most fundamental formula is the power rule for integration. For any real number n ≠ -1,

最基本的公式是幂函数积分公式。对任意实数 n ≠ -1，

∫ xⁿ dx = xⁿ⁺¹ / (n+1) + C

This rule works for positive, negative, and fractional exponents, as long as n ≠ -1. For example, ∫ x³ dx = x⁴/4 + C, and ∫ √x dx = ∫ x^(1/2) dx = (2/3)x^(3/2) + C.

该公式适用于正、负以及分数指数，只要 n ≠ -1。例如，∫ x³ dx = x⁴/4 + C，而 ∫ √x dx = ∫ x^(1/2) dx = (2/3)x^(3/2) + C。

When integrating a polynomial, apply the rule term by term: ∫ (3x² + 2x – 5) dx = x³ + x² – 5x + C.

积分多项式时，逐项应用公式：∫ (3x² + 2x – 5) dx = x³ + x² – 5x + C。

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

When n = -1, the power rule fails because the denominator becomes zero. The correct formula involves the natural logarithm:

当 n = -1 时，幂函数公式失效，因为分母变为零。正确的公式涉及自然对数：

∫ x⁻¹ dx = ∫ (1/x) dx = ln|x| + C

The absolute value is essential to allow x to be negative. For example, ∫ (2/x) dx = 2 ln|x| + C. This formula is frequently tested in A-level exams.

绝对值符号必不可少，它使 x 可以为负数。例如，∫ (2/x) dx = 2 ln|x| + C。这个公式在 A-level 考试中经常出现。

3. Exponential and Logarithmic Integrals | 指数与对数函数积分

For exponential functions, the key formula is:

对于指数函数，关键公式是：

∫ eˣ dx = eˣ + C, and ∫ aˣ dx = aˣ / ln(a) + C (a > 0, a ≠ 1)

In particular, ∫ e^(kx) dx = (1/k)e^(kx) + C for a constant k. For example, ∫ e^(3x) dx = (1/3)e^(3x) + C. Similarly, ∫ 2ˣ dx = 2ˣ / ln(2) + C.

特别地，∫ e^(kx) dx = (1/k)e^(kx) + C，其中 k 为常数。例如，∫ e^(3x) dx = (1/3)e^(3x) + C。类似地，∫ 2ˣ dx = 2ˣ / ln(2) + C。

The integral of ln(x) itself is a classic result derived by integration by parts: ∫ ln x dx = x ln x – x + C, which we will revisit later.

ln(x) 本身的积分是利用分部积分法得到的经典结果：∫ ln x dx = x ln x – x + C，我们稍后会再次讨论。

4. Trigonometric Integrals | 三角函数积分

The basic trigonometric integrals are derived directly from standard derivatives:

三角函数的基本积分直接来自标准导数公式：

∫ cos x dx = sin x + C, ∫ sin x dx = -cos x + C, ∫ sec² x dx = tan x + C

Additional important formulas include ∫ sec x tan x dx = sec x + C and ∫ csc x cot x dx = -csc x + C. For linear arguments like ax + b, divide by the coefficient a: ∫ cos(2x) dx = (1/2) sin(2x) + C.

其他重要公式包括 ∫ sec x tan x dx = sec x + C 和 ∫ csc x cot x dx = -csc x + C。对于 ax + b 形式的线性内项，需除以系数 a：∫ cos(2x) dx = (1/2) sin(2x) + C。

5. Integration by Substitution | 换元积分法

Substitution, or u-substitution, reverses the chain rule. We choose a new variable u = g(x), then replace dx with du = g'(x)dx. This simplifies the integrand into a known form.

换元积分法（又称 u 代换法）是链式法则的逆过程。我们令新变量 u = g(x)，然后用 du = g'(x)dx 替换 dx，从而将被积函数化为已知形式。

Example: Find ∫ 2x e^(x²) dx. Let u = x², then du = 2x dx. The integral becomes ∫ eᵘ du = eᵘ + C = e^(x²) + C.

例：求 ∫ 2x e^(x²) dx。令 u = x²，则 du = 2x dx。积分变为 ∫ eᵘ du = eᵘ + C = e^(x²) + C。

Choose u as the inner function of a composite, or as the denominator in a rational function.
Compute du and ensure every x is expressed in terms of u.
Integrate with respect to u, then substitute back.

选择 u 为复合函数的内层，或分母中的某个部分。
计算 du，确保所有 x 都用 u 表示。
对 u 积分，最后代回原变量。

For definite integrals, remember to change the limits when using substitution. However, for indefinite integrals, simply substitute back and add C.

对于定积分，换元时记得更改上下限。但对于不定积分，只需代回原变量并加上常数 C。

6. Integration by Parts | 分部积分法

Integration by parts is the reverse of the product rule. The formula is:

分部积分法是乘积法则的逆过程，其公式为：

∫ u dv = uv – ∫ v du

We choose u and dv from the integrand. Typically, pick u as a function that becomes simpler when differentiated (like x, x², ln x), and dv as a function that can be easily integrated.

我们从被积函数中选择 u 和 dv。通常选择 u 为微分后变得简单的函数（如 x、x²、ln x），而 dv 为容易积分的部分。

Example: Find ∫ x eˣ dx. Let u = x, dv = eˣ dx. Then du = dx, v = eˣ. The formula gives:

例：求 ∫ x eˣ dx。令 u = x，dv = eˣ dx，则 du = dx，v = eˣ。代入公式得：

∫ x eˣ dx = x eˣ – ∫ eˣ dx = x eˣ – eˣ + C = eˣ(x – 1) + C

For integrals like ∫ x² sin x dx, apply integration by parts twice. A common trick for ∫ ln x dx is to write it as ∫ (1)ln x dx, with u = ln x, dv = dx.

对于 ∫ x² sin x dx，需要连续两次使用分部积分法。求 ∫ ln x dx 的常用技巧是写成 ∫ (1)ln x dx，令 u = ln x，dv = dx。

7. Partial Fractions | 部分分式法

Rational functions of the form P(x)/Q(x) can often be integrated after decomposing into partial fractions. This is especially useful when Q(x) factorises into distinct or repeated linear factors.

形如 P(x)/Q(x) 的有理函数，通常可以通过分解为部分分式后积分。当 Q(x) 能分解为一次因式（含重因式）时尤其有效。

Example: Find ∫ 1/(x² – 1) dx. Since x² – 1 = (x-1)(x+1), we write:

例：求 ∫ 1/(x² – 1) dx。因为 x² – 1 = (x-1)(x+1)，我们设：

1/(x²-1) = A/(x-1) + B/(x+1)

Solving gives A = 1/2, B = -1/2. Thus the integral becomes (1/2) ln|x-1| – (1/2) ln|x+1| + C = (1/2) ln|(x-1)/(x+1)| + C.

解得 A = 1/2，B = -1/2。因此积分为 (1/2) ln|x-1| – (1/2) ln|x+1| + C = (1/2) ln|(x-1)/(x+1)| + C。

For repeated factors like (x+1)², include terms A/(x+1) + B/(x+1)². For irreducible quadratics, use linear numerators like Ax + B.

对于 (x+1)² 这样的重因式，需包含 A/(x+1) + B/(x+1)² 两项。对于不可因式分解的二次式，分子需设为 Ax + B。

8. Algebraic Manipulation | 代数变形技巧

Some integrals require algebraic simplification before applying basic formulas. Recognising perfect squares and completing the square can transform integrals into standard arctangent or arcsine forms.

有些积分需要先进行代数变形才能套用基本公式。识别完全平方或配方，可以将积分化为标准的反正切或反正弦形式。

Useful formulas include:

常用公式包括：

∫ 1/(x² + a²) dx = (1/a) arctan(x/a) + C

∫ 1/√(a² – x²) dx = arcsin(x/a) + C

For example, ∫ 1/(x² + 4x + 5) dx = ∫ 1/((x+2)² + 1) dx = arctan(x+2) + C.

例如，∫ 1/(x² + 4x + 5) dx = ∫ 1/((x+2)² + 1) dx = arctan(x+2) + C。

Another useful trick is to expand products: ∫ (x+1)(x-2) dx = ∫ (x² – x – 2) dx = x³/3 – x²/2 – 2x + C.

另一个技巧是展开乘积：∫ (x+1)(x-2) dx = ∫ (x² – x – 2) dx = x³/3 – x²/2 – 2x + C。

9. Recognising Derivatives | 识别导数结构

Many exam questions are designed so that a part of the integrand is the derivative of the rest. If you spot a function g(x) whose derivative g'(x) also appears in the integrand, the integral is often ln|g(x)| or a power of g(x).

许多考试题目设计成被积函数中某一部分恰好是另一部分的导数。如果你发现某个函数 g(x) 的导数 g'(x) 也出现在被积函数中，积分结果往往与 ln|g(x)| 或 g(x) 的幂有关。

Example: ∫ (2x + 3)/(x² + 3x + 1) dx. Notice that d/dx (x² + 3x + 1) = 2x + 3, so the integral is simply ln|x² + 3x + 1| + C.

例：∫ (2x + 3)/(x² + 3x + 1) dx。注意到 d/dx (x² + 3x + 1) = 2x + 3，所以积分直接为 ln|x² + 3x + 1| + C。

Similarly, ∫ tan x dx = ∫ (sin x / cos x) dx, and since d/dx (cos x) = -sin x, we get -ln|cos x| + C.

类似地，∫ tan x dx = ∫ (sin x / cos x) dx，因为 d/dx (cos x) = -sin x，所以结果为 -ln|cos x| + C。

Always scan the integrand for derivative patterns before attempting more complex methods.

在尝试复杂方法之前，务必先观察被积函数中是否存在导数模式。

10. Summary and Exam Tips | 总结与考试技巧

Mastering indefinite integrals requires memorising the basic formulas and recognising which technique applies. Below is a quick reference table for the most common integrals.

掌握不定积分需要牢记基本公式并识别适用的技巧。下表中列出了最常见的积分公式，供快速查阅。

f(x)	∫ f(x) dx
xⁿ (n ≠ -1)	xⁿ⁺¹/(n+1) + C
1/x	ln\|x\| + C
eˣ	eˣ + C
sin x	-cos x + C
cos x	sin x + C
sec² x	tan x + C
1/(x² + a²)	(1/a) arctan(x/a) + C
1/√(a² – x²)	arcsin(x/a) + C

Always include + C for indefinite integrals. Check your answer by differentiating it – the derivative should equal the original integrand. During exams, if a direct substitution looks messy, try algebraic manipulation first.

不定积分中务必加上 + C。通过求导可以检验答案——导数应等于原被积函数。考试时，如果代换法显得繁琐，先试试代数变形。

As you practise, you will build intuition for which technique to use. The key is consistent practice and familiarity with the structure of common integrals.

通过不断练习，你会逐步培养出选择正确方法的直觉。关键在于持续练习，并熟悉常见积分的结构。

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Indefinite Integrals: Basic Formulas and Calculation Techniques | 不定积分：基本公式与计算技巧