📚 Substitution Method for ∫ f(ax² + b)x dx | 形如∫ f(ax² + b)x dx 的换元积分法
his article explains the method in detail, with worked examples and exam-focused insights.
在积分学中，换元法（也称 u-代换）是最强有力的技巧之一。在 IB 数学中，一种特别常见的模式是形如 ∫ f(ax² + b) x dx 的积分，其中的 x 因子使我们可以通过引入变量 u = ax² + b 来简化运算。本文详细解释该方法，并提供典型例题与考试要点。
1. Recognising the Pattern | 识别模式
The integrand contains a function of a quadratic expression ax² + b multiplied by x (or a constant multiple of x). Because the derivative of ax² + b is 2ax, the presence of x dx is the key clue for u-substitution.
被积函数包含一个关于二次式 ax² + b 的函数乘以 x（或 x 的常数倍）。因为 ax² + b 的导数是 2ax，所以 x dx 的出现是采用 u-代换的关键线索。
For instance, ∫ x sin(x² + 4) dx, ∫ x √3x² − 1 dx and ∫ x eˣ² dx all share this structure.
例如，∫ x sin(x² + 4) dx、∫ x √3x² − 1 dx 和 ∫ x eˣ² dx 都具有这种结构。
2. The Substitution u = ax² + b | 换元 u = ax² + b
= 1/(2a) du. This transforms the original integral into a much simpler one in terms of u.
设 u = ax² + b，可得 du/dx = 2ax，因此 du = 2ax dx。重新整理得到 x dx = 1/(2a) du。这样就将原积分化为关于 u 的简单积分。
∫ f(ax² + b) x dx = 1/(2a) ∫ f(u) du
The constant factor 1/(2a) can be taken outside the integral, and the problem reduces to integrating f(u) alone.
常数因子 1/(2a) 可以提到积分号外，问题归结为对 f(u) 积分。
3. Step-by-Step Procedure | 逐步步骤
Step 1: Identify the inner quadratic function inside the composition. Look for an expression of the form ax² + b inside another function, multiplied by x.
步骤 1：识别内层二次函数。找出嵌套在另一函数中的形如 ax² + b 的表达式，并确认其与 x 相乘。
Step 2: Set u = ax² + b and differentiate to find du/dx = 2ax, so du = 2ax dx.
步骤 2：令 u = ax² + b，求导得 du/dx = 2ax，从而 du = 2ax dx。
Step 3: Isolate the x dx term: x dx = du/(2a).
步骤 3：分离 x dx 项：x dx = du/(2a)。
Step 4: Substitute both the function and the x dx term into the integral.
步骤 4：将函数以及 x dx 项同时代入积分式。
Step 5: Integrate with respect to u. Remember to include the constant factor 1/(2a).
步骤 5：对 u 积分，记得保留常数因子 1/(2a)。
Step 6: or indefinite integrals, substitute back u = ax² + b to return to the original variable x.
步骤 6：对于不定积分，将 u = ax² + b 代回，还原为原变量 x。
4. Worked Example 1: Polynomial Function | 例 1：多项式函数
Evaluate ∫ x (x² + 3)⁴ dx.
计算 ∫ x (x² + 3)⁴ dx。
Let u = x² + 3. Then du/dx = 2x ⇒ du = 2x dx, so x dx = du/2.
令 u = x² + 3。则 du/dx = 2x ⇒ du = 2x dx，因此 x dx = du/2。
The integral becomes ½ ∫ u⁴ du = ½ × u⁵/5 + C = (1/10) u⁵ + C.
积分变为 ½ ∫ u⁴ du = ½ × u⁵/5 + C = (1/10) u⁵ + C。
Substituting back gives the final answer: (1/10)(x² + 3)⁵ + C.
代回后得到最终答案：(1/10)(x² + 3)⁵ + C。
5. Worked Example 2: Trigonometric Function | 例 2：三角函数
Find ∫ x cos(x² + 1) dx.
求 ∫ x cos(x² + 1) dx。
+ 1, so du = 2x dx and x dx = du/2. The integral simplifies to ½ ∫ cos u du = ½ sin u + C.
设 u = x² + 1，则 du = 2x dx，x dx = du/2。积分简化为 ½ ∫ cos u du = ½ sin u + C。
Replacing u gives the result: ½ sin(x² + 1) + C.
替换 u 得到结果：½ sin(x² + 1) + C。
6. Worked Example 3: Exponential Function | 例 3：指数函数
Evaluate ∫ x eˣ² dx.
计算 ∫ x eˣ² dx。
Let u = x² ⇒ du = 2x dx ⇒ x dx = du/2. The integral becomes ½ ∫ eᵘ du = ½ eᵘ + C. Substitute back: ½ eˣ² + C.
令 u = x² ⇒ du = 2x dx ⇒ x dx = du/2。积分变为 ½ ∫ eᵘ du = ½ eᵘ + C。代回得 ½ eˣ² + C。
7. Handling a Coefficient of x That Differs from 2a | 处理 x 系数与 2a 不一致的情况
Sometimes the term multiplied by the function is not exactly x, but a constant multiple like 3x. You can still use the same substitution; just adjust the constant factor.
有时与函数相乘的项不恰好是 x，而是常数倍比如 3x。你仍然可以使用同样的换元，只需要调整常数因子即可。
Example: ∫ 3x √2x² + 5 dx. Let u = 2x² + 5, then du = 4x dx ⇒ x dx = du/4. Therefore 3x dx = 3 × du/4 = ¾ du.
例：∫ 3x √2x² + 5 dx。令 u = 2x² + 5，则 du = 4x dx ⇒ x dx = du/4。因此 3x dx = 3 × du/4 = ¾ du。
The integral becomes (3/4) ∫ √u du = (3/4) × (2/3) u³⁄² + C = ½ u³⁄² + C = ½ (2x² + 5)³⁄² + C.
积分变为 (3/4) ∫ √u du = (3/4) × (2/3) u³⁄² + C = ½ u³⁄² + C = ½ (2x² + 5)³⁄² + C。
Always extract any constant multiplier from the x-term before completing the substitution.
务必在完成换元前将 x-项的任何常数乘子提取出来。
8. Substitution for Definite Integrals | 定积分的换元
When the integral has limits, you must change the limits according to the substitution to avoid back-substitution. This often saves time and reduces errors.
当积分带有上下限时，你必须根据换元改变积分限，从而避免回代。这通常能节省时间并减少错误。
¹ x (x² + 1)³ dx. Use u = x² + 1. When x = 0, u = 1; when x = 1, u = 2. Also, x dx = du/2.
例：计算 ∫₀¹ x (x² + 1)³ dx。用 u = x² + 1。当 x = 0 时，u = 1；当 x = 1 时，u = 2。同时，x dx = du/2。
integral becomes ½ ∫₁² u³ du = ½ [¼ u⁴]₁² = ½ (¼·16 − ¼·1) = ½ (4 − ¼) = ½ × 15/4 = 15/8.
定积分变为 ½ ∫₁² u³ du = ½ [¼ u⁴]₁² = ½ (¼·16 − ¼·1) = ½ (4 − ¼) = ½ × 15/4 = 15/8。
Always write the new limits clearly on the integral sign when changing variables in a definite integral.
在定积分中换元时，始终要在积分号上清楚地写出新的上下限。
9. Common Pitfalls and How to Avoid Them | 常见错误及避免方法
Forgetting the constant 1/(2a). but forget to divide by 2a when isolating x dx. Always rewrite x dx = du/(2a) explicitly.
忘记常数 1/(2a)。很多学生写出 du = 2x dx 但忘记在分离 x dx 时除以 2a。要始终明确写出 x dx = du/(2a)。
Incorrect back-substitution. After integrating with respect to u, you must replace u with ax² + b. Double-check that every u becomes the original expression.
回代错误。对 u 积分后，必须用 ax² + b 替换 u。仔细检查每一个 u 是否都变回了原表达式。
Not changing limits in definite integrals. If you keep the x-limits while the variable is u, the evaluation will be wrong. Convert the limits as soon as you substitute.
定积分不换限。如果变量是 u 却仍然保留 x 的积分限，计算结果将是错误的。一旦换元，立即转换积分限。
Misreading the form. the x factor, the substitution u = ax² + b does not simplify the integral in the same way.
误判形式。该方法仅在积分包含 x dx（或其常数倍）时有效。如果没有 x 因子，换元 u = ax² + b 并不能以同样方式简化积分。
10. Exam Tips for IB Students | IB 考试技巧
In IB papers, this type of integral often appears inside area or volume of revolution problems. Look for a composition with an inner quadratic and an outer function multiplied by x.
在 IB 试卷中，此类积分常出现在面积或旋转体体积问题中。注意寻找内层为二次式、外层为某函数并与 x 相乘的复合结构。
Always show the substitution step clearly: write "Let u = ...", give du/dx, and explicitly state the transformed integral. This earns
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