📚 Topic 090: Integration by Substitution and Partial Fractions | 主题090: 换元积分法与部分分式分解
Integration by substitution and integration using partial fractions are two cornerstone techniques in the Edexcel A-Level pure mathematics syllabus. Together they equip you to handle a broad class of rational, radical, and composite integrals that cannot be tackled by standard power-rule alone. This revision guide walks you through the theory, the most common forms, and the practical steps required to score full marks on exam questions.
换元积分法与部分分式积分是爱德思 A-Level 纯数大纲中的两项核心技巧。将两者结合起来，便能处理大量仅靠幂法则无法解决的有理函数、根式函数以及复合函数积分。本篇复习指南将带你梳理相关理论、最常见的形式，以及考场上拿满分所需的实操步骤。
1. Understanding the Need for Substitution | 理解换元法的必要性
Many integrals contain an inner function and its derivative. This structure mirrors the chain rule for differentiation, and the substitution method reverses that rule.
许多被积函数都含有内层函数及其导数。这种结构与微分的链式法则相呼应，而换元法正是对链式法则的逆用。
For example, consider ∫ 2x·cos(x²) dx. The term 2x is the derivative of the inner function x².
例如，考虑 ∫ 2x·cos(x²) dx。其中 2x 正是内层函数 x² 的导数。
By setting u = x², we obtain du/dx = 2x, so du = 2x dx, which transforms the integral into ∫ cos u du – a direct integration.
令 u = x²，得到 du/dx = 2x，因此 du = 2x dx，原积分便化成了 ∫ cos u du，可以直接积出。
ivative appears, but a change of variable simplifies the expression, such as letting u = √x to remove a square root.
即便没有明显的导数形式出现，换元法也可通过变量代换简化表达式，例如令 u = √x 来消去根号。
2. The Method of Substitution: Step by Step | 换元法：逐步分解
tegrand. Look for a function whose derivative also appears, or for a core expression that is repeatedly used.
首先，找到一个合适的代换 u = g(x)，使被积函数得到简化。关注导数值也出现的位置，或者反复用到的核心表达式。
Next, compute du/dx and rearrange to express dx in terms of du: dx = du / g’(x). Ensure every x is replaced by an expression in u.
其次，计算 du/dx 并整理成用 du 表示 dx 的形式：dx = du / g’(x)。务必把所有 x 都替换成关于 u 的表达式。
Then perform the integration in u. This should now be a standard form: polynomial, trigonometric, exponential, or logarithmic.
然后对 u 进行积分。此时积分应当已经化为标准形式：多项式、三角、指数或对数。
Finally, substitute back g(x) for u to return the answer to the original variable. Do not forget the constant of integration, +C.
最后，用 g(x) 代回 u，将答案还原为原来的变量。千万不要遗漏积分常数 +C。
∫ f(g(x))g’(x) dx = ∫ f(u) du
3. Common Substitutions: Linear Functions | 常见换元：线性函数
a linear expression such as (ax + b) inside a power, root, or denominator, use u = ax + b.
当被积函数在幂、根号或分母中含有如 (ax + b) 这样的线性表达式时，设 u = ax + b。
For instance, ∫ (3x + 5)⁴ dx becomes ∫ u⁴ × (du/3) because du = 3 dx. The result is (1/15)(3x + 5)⁵ + C.
例如，∫ (3x + 5)⁴ dx 变为 ∫ u⁴ × (du/3)，因为 du = 3 dx。结果为 (1/15)(3x + 5)⁵ + C。
Similarly, ∫ 1/(5x − 2) dx is transformed using u = 5x − 2, giving (1/5) ∫ 1/u du = (1/5)ln|5x − 2| + C.
同样地，∫ 1/(5x − 2) dx 使用 u = 5x − 2 代换，得到 (1/5) ∫ 1/u du = (1/5)ln|5x − 2| + C。
This technique systematically handles any integral of the form ∫ (ax + b)ⁿ dx or ∫ 1/(ax + b) dx where n ≠ −1.
这一技巧可以系统地处理所有形如 ∫ (ax + b)ⁿ dx 或 ∫ 1/(ax + b) dx（n ≠ −1）的积分。
4. Substitution with Inverse Trigonometric Functions | 反三角函数换元
Integrals that resemble the derivatives of arcsin or arctan are prime candidates for trigonometric or matching substitutions.
形如 arcsin 或 arctan 导数形式的积分尤其适合三角替换或匹配替换。
For ∫ 1/√(a² − x²) dx, the substitution x = a sin u leads to ∫ du = u + C = arcsin(x/a) + C.
对于 ∫ 1/√(a² − x²) dx，令 x = a sin u 可以得到 ∫ du = u + C = arcsin(x/a) + C。
If the integral has the form ∫ 1/(a² + x²) dx, use x = a tan u so that dx = a sec²u du and the denominator becomes a²sec²u, giving (1/a)arctan(x/a) + C.
如果积分形如 ∫ 1/(a² + x²) dx，使用 x = a tan u，于是 dx = a sec²u du，分母变为 a²sec²u，最终得到 (1/a)arctan(x/a) + C。
These substitutions appear regularly in Edexcel exam papers, often with a completing-the-square step to bring the integral into a recognisable form.
这些换元在爱德思试卷中频繁出现，通常需要先进行配平方，将积分化成可辨识的形式。
5. Introduction to Partial Fractions | 部分分式简介
Partial fraction decomposition breaks a complex rational function into a sum of simpler fractions, each of which can be integrated directly.
部分分式分解将一个复杂的有理函数拆分为简单分式的和，其中每一项都可以直接积分。
This method is used when the integrand is a rational function P(x)/Q(x) and the degree of the numerator is less than that of the denominator (proper fraction).
该方法适用于被积函数为有理函数 P(x)/Q(x)，且分子的次数低于分母的次数（真分式）时。
If the fraction is improper, perform polynomial long division first to obtain a polynomial plus a proper rational remainder.
如果分式为假分式，先进行多项式长除法，得到一个多项式与一个真分式余式。
The main goal is to express the fraction as sums of terms like A/(ax + b), B/(ax + b)², or (Cx + D)/(ax² + bx + c).
主要目标是将分式表示为形如 A/(ax + b)、B/(ax + b)² 或 (Cx + D)/(ax² + bx + c) 的项的和。
6. Decomposing Proper Fractions: Distinct Linear Factors | 分解真分式：不同线性因子
When the denominator factorises into distinct linear factors (ax + b)(cx + d)…, we write the fraction as A/(ax + b) + B/(cx + d) + …
当分母可分解为互异的一次因子 (ax + b)(cx + d)… 时，我们将分式写成 A/(ax + b) + B/(cx + d) + … 的形式。
For example, to decompose 1/(x² − 1), first note x² − 1 = (x − 1)(x + 1). Set 1/(x² − 1) = A/(x − 1) + B/(x + 1).
例如，分解 1/(x² − 1) 时，先注意到 x² − 1 = (x − 1)(x + 1)。设 1/(x² − 1) = A/(x − 1) + B/(x + 1)。
Multiply through by the denominator: 1 = A(x + 1) + B(x − 1). Solve by equating coefficients or substituting convenient values of x.
两边同乘分母：1 = A(x + 1) + B(x − 1)。通过比较系数或代入特殊 x 值求解。
When x = 1, 1 = 2A ⇒ A = 1/2; when x = −1, 1 = −2B ⇒ B = −1/2. Hence the decomposition is ½/(x − 1) − ½/(x + 1).
当 x = 1 时，1 = 2A ⇒ A = 1/2；当 x = −1 时，1 = −2B ⇒ B = −1/2。因此分解式为 ½/(x − 1) − ½/(x + 1)。
This technique immediately yields logarithmic integrals: ∫ 1/(x² − 1) dx = (1/2)ln|x − 1| − (1/2)ln|x + 1| + C.
这一技巧能立刻给出对数积分：∫ 1/(x² − 1) dx = (1/2)ln|x − 1| − (1/2)ln|x + 1| + C。
7. Handling Repeated Linear Factors | 处理重复线性因子
If the denominator contains a repeated factor such as (ax + b)², the decomposition must include terms up to that power: A/(ax + b) + B/(ax + b)².
若分母含有重复因子如 (ax + b)²，分解时就必须包含直到该次幂的项：A/(ax + b) + B/(ax + b)²。
Consider (2x + 1)/(x − 1)². We write it as A/(x − 1) + B/(x − 1)².
考虑 (2x + 1)/(x − 1)²，我们将其写成 A/(x − 1) + B/(x − 1)² 的形式。
Multiplying gives 2x + 1 = A(x − 1) + B. Equating coefficients: for x, 2 = A; constant, 1 = −A + B ⇒ B = 3.
两边相乘得 2x + 1 = A(x − 1) + B。比较系数：x 项，2 = A；常数项，1 = −A + B ⇒ B = 3。
Thus the integral ∫ (2x + 1)/(x − 1)² dx becomes ∫ [2/(x − 1) + 3/(x − 1)²] dx = 2ln|x − 1| − 3/(x − 1) + C.
所以积分 ∫ (2x + 1)/(x − 1)² dx 变为 ∫ [2/(x − 1) + 3/(x − 1)²] dx = 2ln|x − 1| − 3/(x − 1) + C。
Always check for missing lower-degree terms; including each power up to the multiplicity guarantees that all possible constants are found.
务必检查是否漏掉了低次项；包含直到重复次数的每一幂次能够确保所有可能的常数都被求出。
8. Dealing with Irreducible Quadratic Factors | 处理不可约二次因子
When the denominator includes an irreducible quadratic factor like (x² + px + q) that has no real roots, the corresponding numerator must be linear: Cx + D.
当分母含有无实根的不可约二次因子如 (x² + px + q) 时，相应的分子必须是一次式：Cx + D。
For instance, decompose 1/(x(x² + 1)) as A/x + (Bx + C)/(x² + 1).
例如，将 1/(x(x² + 1)) 分解为 A/x + (Bx + C)/(x² + 1)。
Multiplying out gives 1 = A(x² + 1) + (Bx + C)x. Comparing coefficients yields A = 1, B = −1, C = 0.
展开得 1 = A(x² + 1) + (Bx + C)x。比较系数可得 A = 1，B = −1，C = 0。
The resulting integral ∫ [1/x − x/(x² + 1)] dx can be solved using standard log and arctan forms, possibly with a minor u-substitution for the quadratic term.
得到的积分 ∫ [1/x − x/(x² + 1)] dx 可以利用标准对数和 arctan 形式求解，其中二次项可能需要一次简单的 u 替换。
Thus ∫ x/(x² + 1) dx = (1/2)ln(x² + 1) + C, completing the solution.
因此 ∫ x/(x² + 1) dx = (1/2)ln(x² + 1) + C，整个积分得以完成。
9. Integrating Using Partial Fractions | 用部分分式进行积分
Once the rational function is decomposed, each term leads to a familiar integral type: logarithmic, inverse tangent, or power-rule for denominators with (ax + b)ⁿ.
一旦有理函数被分解，每一项都会化归为熟悉的积分类型：对数、反正切，或分母为 (ax + b)ⁿ 的幂法则形式。
Term of the form A/(ax + b) integrates to (A/a) ln|ax + b|. A repeated linear term B/(ax + b)ⁿ (n > 1) gives −B/[a(n − 1)(ax + b)&supn;⁻¹].
形如 A/(ax + b) 的项积分为 (A/a) ln|ax + b|。重复线性项 B/(ax + b)ⁿ （n > 1）给出 −B/[a(n − 1)(ax + b)&supn;⁻¹]。
For irreducible quadratic factor (Cx + D)/(ax² + bx + c), split the numerator to match the derivative of the denominator, leading to a log plus an arctan integral.
对于不可约二次因子 (Cx + D)/(ax² + bx + c)，拆分分子以匹配分母的导数，从而得到一个对数积分和一个 arctan 积分。
Always add the constant of integration at the end and, if the question involves definite integration, evaluate the limits after integrating.
始终在最后加上积分常数；如果是定积分题目，则在积分完成后代入上下限。
10. Combining Substitution and Partial Fractions | 结合换元法与部分分式
Some integrals require both techniques: first a substitution to turn the integrand into a rational function, then partial fractions to split it.
有些积分需要同时使用这两种技巧：首先通过换元将被积函数转化为有理函数，然后再用部分分式进行拆分。
A classic example is ∫ (eˣ)/(e²ˣ − 1) dx. Let u = eˣ, so du = eˣ dx, transforming the integral into ∫ 1/(u² − 1) du.
一个典型的例子是 ∫ (eˣ)/(e²ˣ − 1) dx。令 u = eˣ，则 du = eˣ dx，积分转化为 ∫ 1/(u² − 1) du。
Now use partial fractions on 1/(u² − 1) to get (1/2)[1/(u − 1) − 1/(u + 1)], integrate and substitute back to x.
现在对 1/(u² − 1) 使用部分分式，得到 (1/2)[1/(u − 1) − 1/(u + 1)]，积分后再换回 x。
tan(x/2) for rational functions of sin x and cos x, which then demands partial fractions.
另一种常见情形涉及三角替换，比如处理 sin x 和 cos x 的有理函数时使用 t = tan(x/2)，之后往往就需要部分分式。
Mastery of both techniques in tandem will allow you to tackle the most demanding Edexcel integration problems with confidence.
熟练掌握这两项技巧的协同运用，将使你信心十足地应对爱德思考试中最棘手的积分问题。
11. Definite Integrals: Changing Limits | 定积分：变换上下限
When using substitution in a definite integral, you have two options: change the limits to u-values, or substitute back to x and use the original limits.
在定积分中使用换元法时，你有两种选择：将上下限换成 u 值，或者换回 x 后代入原来的上下限。
ts as soon as you choose u = g(x). Compute u at x = a and x = b to get new limits c and d.
一般来说，当选定了 u = g(x) 后，立即同时变换上下限更为稳妥。计算 x = a 和 x = b 时的 u 值，得到新限 c 和 d。
Then evaluate ∫cd f(u) du directly; there is no need to return to the original variable.
然后直接计算 ∫cd f(u) du，无需再回到原变量。
For example, ∫0π/2 2x cos(x²) dx. Let u = x²; limits become 0 and π²/4, giving ∫0π²/4 cos u du = sin(π²/4).
例如，∫0π/2 2x cos(x²) dx。令 u = x²，上下限变为 0 和 π²/4，得到 ∫0π²/4 cos u du = sin(π²/4)。
When partial fractions are involved, integrate first, then apply the limits to the final antiderivative in x, ensuring proper logarithmic simplifications.
涉及部分分式时，先求不定积分，再将上下限代入最终的 x 反导函数，并正确化简对数表达式。
12. Common Mistakes and Exam Tips | 常见错误与应试技巧
dx properly is the most frequent error. Always write dx = du / (du/dx) in its exact form.
最常见错误是忘了正确替换 dx。始终写出精确形式 dx = du / (du/dx)。
In partial fractions, candidates often miss necessary terms for repeated or quadratic factors; check the decomposition structure before solving.
在部分分式中，考生常漏掉重复因子或二次因子所需的项；先核对分解结构再求解。
When equating coefficients, use a table to keep track of each term. A table helps avoid sign errors and ensures all equations are accounted for.
比较系数时，使用表格来跟踪每一项。表格有助于避免符号错误，并确保所有方程都被列入。
| Coefficient of x² | A + B = 1 |
| Coefficient of x | 2A − C = 0 |
| Constant term | −3A
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