📚 Indefinite Integrals: Basic Formulas and Calculation Techniques | 不定积分：基本公式与计算技巧
(x), an indefinite integral, denoted as ∫f(x)dx, is a family of functions F(x) such that F' eral 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。
= 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
ative, and fractional exponents, as long as n ≠ -1. For example, ∫ x³ dx = x⁴/4 + C, and ∫ √x dx = ∫ x1/2 dx = (2/3)x3/2 + 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 = ∫ x1/2 dx = (2/3)x3/2 + C。
grating 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
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, ∫ ekx dx = (1/k)ekx + C for a constant k. For example, ∫ e3x dx = (1/3)e3x + C. Similarly, ∫ 2ˣ dx = 2ˣ / ln(2) + C.
特别地，∫ ekx dx = (1/k)ekx + C，其中 k 为常数。例如，∫ e3x dx = (1/3)e3x + C。类似地，∫ 2ˣ dx = 2ˣ / ln(2) + C。
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
near 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。
Remember that ∫ tan x dx = -ln|cos x| + C = ln|sec x| + C, which is often required. You can verify it by differentiating ln|sec x|.
请记住 ∫ tan x dx = -ln|cos x| + C = ln|sec x| + C，这个公式经常用到。你可以通过对 ln|sec x| 求导来验证它。
5. Integration by Substitution | 换元积分法
n, 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 ex² dx. Let u = x², then du = 2x dx. The integral becomes ∫ eᵘ du = eᵘ + C = ex² + C.
例：求 ∫ 2x ex² dx。令 u = x²，则 du = 2x dx。积分变为 ∫ eᵘ du = eᵘ + C = ex² + 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
wice. 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。
l 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。
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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