📚 Integration Techniques: Finding Antiderivatives from Derivatives | 由导函数求原函数的积分技巧
Integration is the reverse operation of differentiation. Given a function’s derivative, our task is to recover the original function—a process called antidifferentiation or indefinite integration. This article presents the essential techniques for doing this accurately in an exam setting.
积分是微分的逆运算。给定一个函数的导函数，我们的任务是要还原出原来的函数——这一过程称为反导数法或不定积分。本文介绍在考试中准确完成该操作的核心技巧。
1. The Fundamental Link: Derivatives and Antiderivatives | 基本联系：导数与反导数
If \(F'(x) = f(x)\), then \(F(x)\) is an antiderivative of \(f(x)\). Since the derivative of a constant is zero, any antiderivative must include an arbitrary constant \(C\).
若 \(F'(x) = f(x)\)，则 \(F(x)\) 是 \(f(x)\) 的一个反导数。由于常数的导数为零，任何反导数都必须包含任意常数 \(C\)。
For example, if \(f(x) = 2x\), then \(F(x) = x^2 + C\). The constant \(C\) represents a vertical shift of the curve, and it is determined only when an initial condition is given.
例如，若 \(f(x) = 2x\)，则 \(F(x) = x^2 + C\)。常数 \(C\) 表示曲线的垂直平移，只有在给定初始条件时才能确定。
2. Basic Integration Formulas | 基本积分公式
Memorize the reverse of standard derivatives. For \(n \neq -1\):
记住标准导数的逆运算。对于 \(n \neq -1\)：
∫ xⁿ dx = xⁿ⁺¹ / (n+1) + C
Also know the special cases: ∫ 1/x dx = ln|x| + C, ∫ eˣ dx = eˣ + C, ∫ aˣ dx = aˣ / ln a + C.
还需掌握特殊情形：∫ 1/x dx = ln|x| + C，∫ eˣ dx = eˣ + C，∫ aˣ dx = aˣ / ln a + C。
For trigonometric functions, the sign patterns matter: ∫ sin x dx = -cos x + C, ∫ cos x dx = sin x + C, ∫ sec² x dx = tan x + C.
对于三角函数，符号模式很关键：∫ sin x dx = -cos x + C，∫ cos x dx = sin x + C，∫ sec² x dx = tan x + C。
3. Linearity of Integration | 积分的线性法则
The integral of a sum is the sum of integrals, and constant factors can be pulled out:
和的积分等于积分的和，常数因子可以提出：
∫ [a f(x) + b g(x)] dx = a ∫ f(x) dx + b ∫ g(x) dx
This allows us to integrate term-by-term. For example, ∫ (3x² + 4) dx = x³ + 4x + C.
这允许我们逐项积分。例如，∫ (3x² + 4) dx = x³ + 4x + C。
Be careful with products and quotients: there is no “product rule” for integrals. Each factor must be handled via specific techniques.
注意乘积与商没有“积的积分法则”。每个因子必须通过特定技巧处理。
4. Recognizing Reverse Chain Rule | 识别逆向链式法则
When the integrand looks like \(f'(x) \cdot g(f(x))\), the antiderivative often involves a composite function.
当被积函数看起来像 \(f'(x) \cdot g(f(x))\) 时，其反导数通常涉及复合函数。
A simple pattern is ∫ 2x eˣ² dx = eˣ² + C, because the derivative of eˣ² is 2x eˣ².
一个简单模式是 ∫ 2x eˣ² dx = eˣ² + C，因为 eˣ² 的导数为 2x eˣ²。
In general, if \(\frac{d}{dx}g(x) = h(x)\), then \(\int h(x) dx = g(x) + C\). Always try to “see” the derivative of an inner function multiplied by the outer derivative.
一般地，若 \(\frac{d}{dx}g(x) = h(x)\)，则 \(\int h(x) dx = g(x) + C\)。始终尝试“看出”内层函数的导数与外层导数的乘积。
5. Substitution (u-Substitution) | 换元法（u 代换）
When the integrand contains a nested function and its derivative (up to a constant), set \(u =\) inner function.
当被积函数包含复合函数及其导数（相差常数倍）时，令 \(u\) 等于内层函数。
Example: ∫ 2x(x²+1)⁵ dx. Let \(u = x² + 1\), then \(du = 2x dx\). The integral becomes ∫ u⁵ du = u⁶/6 + C = (x²+1)⁶/6 + C.
例：∫ 2x(x²+1)⁵ dx。令 \(u = x² + 1\)，则 \(du = 2x dx\)。积分变为 ∫ u⁵ du = u⁶/6 + C = (x²+1)⁶/6 + C。
For definite integrals, change limits as well: when \(x=a\) then \(u=u(a)\), and when \(x=b\) then \(u=u(b)\). This avoids substituting back.
对于定积分，还要换限：当 \(x=a\) 时 \(u=u(a)\)，当 \(x=b\) 时 \(u=u(b)\)。这样可避免代回。
6. Integration by Parts | 分部积分法
This technique comes from the product rule of differentiation. The formula is:
该技巧源于微分乘积法则，公式为：
∫ u dv = u v – ∫ v du
Choose \(u\) as the part that simplifies when differentiated (often \(x^n\), \(\ln x\)), and \(dv\) as the part that integrates easily (often \(e^x\), \(\sin x\)).
选择 \(u\) 为微分后变简单的部分（通常是 \(x^n\)、\(\ln x\)），而 \(dv\) 为容易积分的部分（通常是 \(e^x\)、\(\sin x\)）。
Example: ∫ x eˣ dx. Let \(u = x\), \(dv = eˣ dx\); then \(du = dx\), \(v = eˣ\). So the integral is \(x eˣ – ∫ eˣ dx = x eˣ – eˣ + C\).
例：∫ x eˣ dx。令 \(u = x\)，\(dv = eˣ dx\)，则 \(du = dx\)，\(v = eˣ\)。故积分为 \(x eˣ – ∫ eˣ dx = x eˣ – eˣ + C\)。
For products like \(x^n \ln x\), choose \(u = \ln x\) because its derivative \(1/x\) simplifies the new integral.
对于 \(x^n \ln x\) 这类乘积，选择 \(u = \ln x\)，因为其导数 \(1/x\) 会让新积分简化。
7. Trigonometric Integrals | 三角积分
Use identities to rewrite integrands into manageable forms. Common identities: \(\sin² x + \cos² x = 1\), \(\tan² x + 1 = \sec² x\), and double-angle formulas.
使用恒等式将被积函数改写为可处理的形式。常用恒等式：\(\sin² x + \cos² x = 1\)，\(\tan² x + 1 = \sec² x\)，以及倍角公式。
For \(\int \sin² x dx\), use the identity \(\sin² x = (1 – \cos 2x)/2\):
对于 \(\int \sin² x dx\)，使用恒等式 \(\sin² x = (1 – \cos 2x)/2\)：
∫ sin² x dx = x/2 – (sin 2x)/4 + C
Products of sines and cosines like \(\int \sin x \cos x dx\) can be solved by substitution \(u = \sin x\) or using \(\sin 2x = 2\sin x \cos x\).
\(\int \sin x \cos x dx\) 这类正弦余弦乘积可用 \(u = \sin x\) 代换或利用 \(\sin 2x = 2\sin x \cos x\) 求解。
8. Partial Fractions | 部分分式法
Rational functions \(\frac{P(x)}{Q(x)}\) can often be decomposed into simpler fractions before integration.
有理函数 \(\frac{P(x)}{Q(x)}\) 通常可先分解为较简单的分式再积分。
If \(Q(x) = (ax+b)(cx+d)\), write:
若 \(Q(x) = (ax+b)(cx+d)\)，可写为：
P(x)/Q(x) = A/(ax+b) + B/(cx+d)
Solve for \(A\) and \(B\) by multiplying through and matching coefficients, then integrate each term using \(\int \frac{1}{ax+b} dx = \frac{1}{a} \ln|ax+b| + C\).
通过两边相乘并比较系数求解 \(A\) 和 \(B\)，然后用 \(\int \frac{1}{ax+b} dx = \frac{1}{a} \ln|ax+b| + C\) 逐项积分。
Repeated linear factors require terms up to the multiplicity; irreducible quadratics require a numerator of the form \(Ax+B\).
重复线性因子需要包含至相应幂次的项；不可约二次因子需要 \(Ax+B\) 形式的分子。
9. Dealing with Radicals | 处理根号
Radicals like \(\sqrt{ax+b}\) can be converted to fractional powers for direct integration:
\(\sqrt{ax+b}\) 这类根式可转化为分数指数幂直接积分：
∫ (ax+b)^{1/2} dx = (2/(3a)) (ax+b)^{3/2} + C
For expressions like \(\sqrt{x^2 + a^2}\), trigonometric substitution is often needed, but A-level questions usually avoid it or provide a formula.
对于 \(\sqrt{x^2 + a^2}\) 这类表达式，通常需要用三角代换，但 A-level 考试一般避免或提供公式。
Another common trick is to rewrite \(x \sqrt{x+1}\) as \((x+1-1)\sqrt{x+1}\) and integrate term by term.
另一个常用技巧是把 \(x \sqrt{x+1}\) 改写为 \((x+1-1)\sqrt{x+1}\)，然后逐项积分。
10. Definite Integrals: Applying the Fundamental Theorem | 定积分：应用微积分基本定理
Once an antiderivative \(F(x)\) is found, the definite integral from \(a\) to \(b\) is \(F(b) – F(a)\).
一旦找到反导数 \(F(x)\)，从 \(a\) 到 \(b\) 的定积分即为 \(F(b) – F(a)\)。
When using substitution in a definite integral, remember to update the limits. For example, \(\int_0^1 2x (x^2+1)^3 dx\), let \(u = x^2+1\) then \(u\) goes from 1 to 2.
在定积分中使用换元法时，记住更新上下限。例如 \(\int_0^1 2x (x^2+1)^3 dx\)，令 \(u = x^2+1\)，则 \(u\) 从 1 变到 2。
Even functions on symmetric intervals satisfy \(\int_{-a}^{a} f(x) dx = 2\int_0^a f(x) dx\) if \(f\) is even, and this often simplifies computation.
偶函数在对称区间上满足 \(\int_{-a}^{a} f(x) dx = 2\int_0^a f(x) dx\)，这常能简化计算。
11. Common Mistakes and Verification | 常见错误与验证
Always check your antiderivative by differentiating it. If \(F'(x)\) does not equal the original integrand, adjust the constant or the method.
始终通过求导验证你的反导数。如果 \(F'(x)\) 不等于原被积函数，请调整常数或方法。
Do not forget the \(+C\) for indefinite integrals. Missing the constant of integration is a common loss of marks.
不要忘记不定积分中的 \(+C\)。漏写积分常数是常见的失分点。
Watch the signs in integration by parts and substitution: a lost negative sign or an incorrect \(du\) can ruin the result.
注意分部积分和换元中的符号：漏负号或 \(du\) 错误会使结果出错。
Simplify before integrating: expand, cancel, or rewrite using identities to make the integrand simpler.
积分前先化简：展开、约分或利用恒等式改写，以使被积函数更简单。
12. Summary of Strategies | 策略总结
Facing an integral, follow this decision tree: try direct formulas, then reverse chain, then substitution, then parts, then partial fractions.
面对一个积分，按以下决策树进行：先试直接公式，再逆链式法则，然后换元，再分部积分，最后部分分式。
Practice is essential. The more patterns you recognize, the faster you can choose the right technique.
练习至关重要。你识别的模式越多，就能越快选择正确技巧。
Remember that every integration result can be verified by differentiation—make this your final safety check in any exam.
请记住，每个积分结果都可以通过微分来验证——在考试中请将其作为最后的安全检查。
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