Mastering Integration by Substitution: Solving Exercise 23D.3 | 掌握代换积分法：解练习23D.3

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📚 Mastering Integration by Substitution: Solving Exercise 23D.3 | 掌握代换积分法：解练习23D.3

, presents an indefinite integral that requires a careful application of substitution: ∫ x√2x+1 dx. This integral features a product of a linear term and a composite square‑root function, making it an ideal candidate for the method of u‑substitution. In this article we break down the solution step by step, explore the reasoning behind every choice, and highlight the most common traps students fall into. Whether you are preparing for an internal exam or the final IB assessment, understanding this exercise deeply will sharpen your calculus skills.

在IB数学分析与方法HL课程中，掌握积分技巧对取得成功至关重要。练习23D.3选自广泛使用的Haese Mathematics教材，给出了一道需要谨慎运用代换法的不定积分：∫ x√2x+1 dx。这道积分包含一个一次项与一个复合平方根函数的乘积，非常适合采用u代换法。本文将逐步拆解求解过程，探讨每一步背后的逻辑，并指出学生最常落入的陷阱。无论你是在准备校内考试还是IB终考，深刻理解这道练习都将提升你的微积分能力。

1. The Task at Hand | 题目呈现

Exercise 23D.3 asks us to find the indefinite integral ∫ x√2x+1 dx. At first glance, the integrand looks like a product of two simple functions, but the presence of the square root applied to a linear expression inside makes it non‑trivial. Direct integration using the power rule is not possible because the derivative of the inside function (2x+1) is not present as a standalone factor. The integral demands a technique that can handle composite functions—substitution.

练习23D.3要求我们求不定积分 ∫ x√2x+1 dx。乍看之下，被积函数像是两个简单函数的乘积，但由于平方根作用于一个内部线性表达式，它并非平凡。直接运用幂法则进行积分是不可能的，因为内部函数(2x+1)的导数并未作为独立因子出现。这道积分需要一种能够处理复合函数的方法——代换法。

2. Why Substitution? | 为何使用代换法？

Integration by substitution reverses the chain rule. When we see a composition such as √2x+1, we suspect that differentiating an expression like (2x+1)^3/2 n terms of u and obtain an integrand purely in u. This transforms a messy integral into a sum of powers that are straightforward to integrate.

代换积分法是链式法则的逆运算。当我们看到形如√2x+1的复合结构时，会联想到对(2x+1)^3/2这样的表达式求导会产生内部导数的因子。在这道练习中，额外的x因子提示我们，令u = 2x+1后，可以将x用u表示，从而得到一个完全关于u的被积函数。这把一个复杂的积分转化为若干幂函数的和，积分起来十分直接。

3. Selecting the Substitution | 选择代换变量

(u - 1)/2. Choosing u as the entire linear expression inside the square root is strategic—it simplifies the radical to √u = u^1/2 and allows us to replace both x and dx cleanly.

令 u = 2x+1，则 du/dx = 2，因此 dx = du/2。由于原积分含有x，我们还需用u表示x：由u = 2x+1可得x = (u - 1)/2。选择u为平方根内的整个线性表达式是有策略的——它将根式简化为√u = u^1/2，并使我们能够简洁地替换掉x和dx。

4. Rewriting the Integral | 重写积分式

Substitute every piece into ∫ x√2x+1 dx:

∫ x√2x+1 dx = ∫ [(u - 1)/2] · √u · (du/2)

This simplifies to (1/4) ∫ (u - 1) · u^1/2 du. The constant 1/4 can be factored outside, leaving an integral that is a sum of two power functions once expanded.

将每一部分代入 ∫ x√2x+1 dx：

∫ x√2x+1 dx = ∫ [(u - 1)/2] · √u · (du/2)

这化简为 (1/4) ∫ (u - 1) · u^1/2 du。常数1/4可以提取到积分号外，括号展开后剩下的是两个幂函数之和的积分。

5. Expanding and Simplifying | 展开与化简

Expand (u - 1)u^1/2 = u · u^1/2 - 1 · u^1/2 = u^3/2 - u^1/2. Now the integral becomes:

(1/4) ∫ (u^3/2 - u^1/2) du

This is exactly the kind of integrand we love—simple powers of u. No more square root of a linear expression, no more extra x factor. The substitution has done its job.

展开 (u - 1)u^1/2 = u · u^1/2 - 1 · u^1/2 = u^3/2 - u^1/2。此时积分变为：

(1/4) ∫ (u^3/2 - u^1/2) du

这正是我们最喜欢的被积函数类型——单纯的u的幂次。不再有线性表达式的平方根，也不再有额外的x因子。代换已经完成了它的使命。

6. Integrating Term by Term | 逐项积分

Apply the power rule for integration: ∫ uⁿ du = uⁿ⁺¹ id for n ≠ -1. For the first term, n = 3/2, so n+1 = 5/2, and division by 5/2 gives (2/5)u^5/2. For the second term, n = 1/2 gives n+1 = 3/2, yielding (2/3)u^3/2. Including the outer factor 1/4, we obtain:

(1/4)[ (2/5)u^5/2 - (2/3)u^3/2 ] + C

应用积分的幂法则：∫ uⁿ du = uⁿ⁺¹ / (n+1) + C，n ≠ -1时成立。第一项n = 3/2，故n+1 = 5/2，除以5/2得到(2/5)u^5/2。第二项n = 1/2，n+1 = 3/2，得到(2/3)u^3/2。纳入外部因子1/4，我们得到：

(1/4)[ (2/5)u^5/2 - (2/3)u^3/2 ] + C

7. Back Substitution | 回代

Now replace u with the original expression 2x+1 and simplify the constants:

∫ x√2x+1 dx = (1/4)·(2/5)(2x+1)^5/2 - (1/4)·(2/3)(2x+1)^3/2 + C

Multiplying the constants gives (2/20) = 1/10 and (2/12) = 1/6. Hence the final answer is:

(1/10)(2x+1)^5/2 - (1/6)(2x+1)^3/2 + C

This is the antiderivative in its simplest form. You can leave it as two separate terms or factor out (2x+1)^3/2 if you prefer, but the IB typically accepts the unsimplified sum.

现在将u替换回原表达式2x+1，并化简常数：

∫ x√2x+1 dx = (1/4)·(2/5)(2x+1)^5/2 - (1/4)·(2/3)(2x+1)^3/2 + C

常数相乘得(2/20) = 1/10 和 (2/12) = 1/6。因此最终答案为：

(1/10)(2x+1)^5/2 - (1/6)(2x+1)^3/2 + C

这是最简形式的反导数。你可以保留两项分开，也可以提取公因子(2x+1)^3/2，IB考试通常接受未合并的和式。

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8. Verification by Differentiation | 微分验证

To build confidence, differentiate the result and check whether we recover x√2x+1. Let F(x) = (1/10)(2x+1)^5/2 - (1/6)(2x+1)^3/2. Using the chain rule:

F'(x) = (1/10)·(5/2)(2x+1)^3/2·2 - (1/6)·(3/2)(2x+1)^1/2·2

Simplify: (1/10)·5 = 1/2, times 2 gives 1, so the first term becomes (2x+1)^3/2. The second term: (1/6)·(3/2) = 1/4, times 2 gives 1/2, so it becomes (1/2)(2x+1)^1/2. Thus F'(x) = (2x+1)^3/2 - (1/2)(2x+1)^1/2. Factor (2x+1)^1/2 to get √2x+1[(2x+1) - 1/2] = √2x+1(2x + 1/2) = √2x+1·(1/2)(4x+1). Wait, that does not immediately look like x√2x+1. Let's re‑evaluate: actually (2x+1) - 1/2 = 2x + 1/2. Multiply by √2x+1 gives (2x + 1/2)√2x+1. That seems different from x√2x+1. I must have made an algebraic slip. Let's do it carefully without factoring: F'(x) = (5/10)*2*(2x+1)^3/2 - (3/12)*2*(2x+1)^1/2 = (1/2)*2*(2x+1)^3/2 - (1/4)*2*(2x+1)^1/2 = (2x+1)^3/2 - (1/2)(2x+1)^1/2. Now write the first term as (2x+1)·(2x+1)^1/2 = (2x+1)√2x+1. Then F'(x) = √2x+1[ (2x+1) - 1/2 ] = √2x+1(2x + 1/2) = (2x + 1/2)√2x+1. This is not x√2x+1. It appears I made an error earlier. Indeed, re‑check the substitution: I had (1/4) ∫ (u^3/2 - u^1/2) du. Integration gave (1/4)[ (2/5)u^5/2 - (2/3)u^3/2 ] = (1/10)u^5/2 - (1/6)u^3/2. Then differentiate: d/du of (1/10)u^5/2 = (1/10)*(5/2)u^3/2 = (1/4)u^3/2, times du/dx = 2 gives (1/2)u^3/2. For the second term: d/du of (1/6)u^3/2 = (1/6)*(3/2)u^1/2 = (1/4)u^1/2, times 2 gives (1/2)u^1/2. So F'(x) = (1/2)(2x+1)^3/2 - (1/2)(2x+1)^1/2. That's different! Let's recalc: I mistakenly applied the chain rule earlier. Actually F(x) = (1/10)u^5/2 - (1/6)u^3/2 where u=2x+1. dF/dx = (1/10)*(5/2)u^3/2*2 - (1/6)*(3/2)u^1/2*2 = (1/4)*2 u^3/2 - (1/4)*2 u^1/2 = (1/2)u^3/2 - (1/2)u^1/2 = (1/2)(2x+1)^3/2 - (1/2)(2x+1)^1/2. Factor (1/2)(2x+1)^1/2[ (2x+1) - 1 ] = (1/2)√2x+1(2x) = x√2x+1. Yes, perfect. My earlier numbers were off because I incorrectly multiplied constants. So the verification works. This is a great lesson: always differentiate to check your integration; if it doesn't match, trace back your steps. I'll present the correct verification in the article, showing the precise algebra.

为了建立信心，对结果求导并检验是否还原为 x√2x+1。令 F(x) = (1/10)(2x+1)^5/2 - (1/6)(2x+1)^3/2。运用链式法则：

F'(x) = (1/10)·(5/2)(2x+1)^3/2·2 - (1/6)·(3/2)(2x+1)^1/2·2

化简：(1/10)·(5/2)·2 = (1/2)，因此第一项成为 (1/2)(2x+1)^3/2；第二项：(1/6)·(3/2)·2 = (1/2)，因此第二项为 (1/2)(2x+1)^1/2。于是 F'(x) = (1/2)(2x+1)^3/2 - (1/2)(2x+1)^1/2。提取因子 (1/2)(2x+1)^1/2 得 (1/2)√2x+1[(2x+1) - 1] = (1/2)√2x+1·2x = x√2x+1。完全吻合。验证成功。这个步骤强调了检查的重要性：万一对不上，就要回溯计算。

9. Common Mistakes | 常见错误

Students often stumble on this exercise for a few recurring reasons:

Forgetting to change dx: dx = du without accounting for the derivative of u. The correct differential is dx = du/2.
Not expressing x in terms of u: tuting u = 2x+1 but leaving x as x. The integral must be entirely in u before integrating.
Misapplying the power rule: Adding 1 to 3/2 gives 5/2, but dividing by 5/2 requires multiplying by its reciprocal (2/5), a step often mishandled.
Constant factor errors: Losing the 1/4 factor from the substitution or incorrectly combining coefficients at the end.
Missing the +C: An indefinite integral always requires the constant of integration.

学生在解这道练习时常因几个反复出现的原因而失分：

忘记变换 dx：写成了 dx = du，却没有考虑 u 的导数。正确的微分是 dx = du/2。
未将 x 用 u 表示：代换 u = 2x+1 后却仍保留 x。积分必须在积分前完全化为 u 的函数。
幂法则运用错误：对 3/2 加 1 得到 5/2，但除以 5/2 需要乘以其倒数 (2/5)，这一步经常出错。
常数因子错误：丢失代换得来的 1/4 因子，或最后合并系数时算错。
遗漏 +C：不定积分永远需要积分常数。

10. Extension: Definite Integral Version | 拓展：定积分情形

If Exercise 23D.3 had included bounds, say ∫₀¹ x√2x+1 u = 2x+1, x = 0 gives u = 1, x = 1 gives u = 3. The definite integral becomes (1/4) ∫₁³ (u^3/2 - u^1/2) du. Evaluating this yields numbers without needing to return to x. Always remember: change the limits when doing a definite substitution to avoid back‑substitution errors.

如果练习23D.3带有积分限，例如 ∫₀¹ x√2x+1 dx，我们仍可采取相同代换，但同时要变换上下限。当 u = 2x+1 时，x = 0 对应 u = 1，x = 1 对应 u = 3。定积分就变成 (1/4) ∫₁³ (u^3/2 - u^1/2) du。计算得到数值，无需回到 x。请务必记住：计算定积分代换时，改变积分限可以避免回代带来的错误。

11. Why This Skill Matters | 为何此技能重要

IB exam papers frequently include integrals that require substitution to unlock. Often these are hidden inside area problems, volume of revolution calculations, or differential equations. Mastering Exercise 23D.3 prepares you to recognise patterns where an inner function’s derivative is not perfectly present but can be manufactured by expressing the remaining variable in terms of u. This pattern—linear inside a radical multiplied by a linear term—is a classic that appears in many guises.

IB试卷中经常出现需要借助代换才能求解的积分。它们常常隐藏在面积问题、旋转体体积或微分方程中。掌握练习23D.3能帮助你识别这样一种模式：内部函数的导数并未完全出现，但可以通过把剩余变量也用 u 表示来构造。这种模式——根号内含线性式乘以一个线性项——是一个经典的变形，在多种题型中都会现身。

12. Final Tips and Practice | 最终提示与练习建议

To cement your understanding, try variations such as ∫ x√3x+4 dx or ∫ (2x+1)√x-3 dx. Always start by identifying the inner function, set u, compute du, and express everything else in terms of u. Write each step clearly, and never skip the verification. If you struggle, revisit the chain rule for differentiation: substitution is simply the reverse. With deliberate practice, you will handle any IB integration problem confidently.

为巩固理解，请尝试变体练习，如 ∫ x√3x+4 dx 或 ∫ (2x+1)√x-3 dx。始终先识别内部函数，设定 u，计算 du，并将其他一切用 u 表达。清晰地写出每一步，绝不要跳过验证。若感到困难，重温微分的链式法则：代换不过是其逆过程。刻意练习后，你将能自信地处理任何 IB 积分题。

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