PDF Joiner (4) – Question 188: Integration by Substitution | PDF练习题(4)第188题：代换积分法

📚 PDF Joiner (4) – Question 188: Integration by Substitution | PDF练习题(4)第188题：代换积分法

In this detailed walkthrough, we solve Question 188 from the Edexcel A-Level Mathematics revision collection PDF Joiner (4). The problem requires integration by substitution, one of the core skills tested in Pure Mathematics Paper 1 and Paper 2. We will break down every step, explain the underlying principles and highlight the exam techniques that can help you secure full marks on similar questions.

在这篇详尽的解题指南中，我们将解答Edexcel A-Level数学复习题集《PDF Joiner (4)》中的第188题。该题要求使用代换积分法，这是纯数试卷1和试卷2中考查的核心技能之一。我们将逐步拆解，讲解背后的原理，并强调能帮助你在类似题目中拿到满分的考试技巧。

1. Problem Statement | 题目陈述

Question 188 (Edexcel style): Use the substitution u = 2x + 1 to evaluate the indefinite integral

∫ x √(2x + 1) dx

第188题（Edexcel风格）：使用代换 u = 2x + 1 计算不定积分

∫ x √(2x + 1) dx

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2. Understanding the Substitution Method | 理解代换积分法

Integration by substitution is essentially the reverse of the chain rule for differentiation. When we spot a composite function whose derivative appears (up to a constant multiple) in the integrand, we can set the inner function as a new variable u. This transforms a complicated integral into a simpler one in terms of u, which we then integrate before substituting back to the original variable.

代换积分法本质上是链式法则求导的逆过程。当我们发现被积函数中包含一个复合函数，且其导数（乘上一个常数）也出现在被积函数中时，就可以令内层函数为新变量 u。这样做的目的是将一个复杂的积分转化为关于 u 的简单积分，积出结果后，再代回原变量。

The general formula for indefinite integration by substitution is

∫ f(g(x)) g'(x) dx = ∫ f(u) du

In our question, the substitution is given explicitly, so we do not need to recognise the pattern — we simply follow the mechanical process.

不定积分代换法的一般公式是

∫ f(g(x)) g'(x) dx = ∫ f(u) du

在我们的题目中，代换式已直接给出，因此无需自己识别形式 — 只需遵循标准操作流程即可。

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3. Choosing the Right Substitution | 选择合适的代换

The problem supplies u = 2x + 1. This is the natural choice because the expression inside the square root, 2x+1, is the ‘inner function’, and its derivative 2 is essentially a constant multiplier linking dx and du. Using this substitution will eliminate the square root and turn the integrand into powers of u.

本题给出 u = 2x + 1。这是一个很自然的选择，因为平方根内的表达式 2x+1 就是“内层函数”，其导数 2 在关联 dx 与 du 时仅仅是一个常数倍。使用该代换将消去根号，使被积函数完全变成 u 的幂函数。

In many examination questions, the substitution is given, but sometimes you are expected to identify it yourself. A rule of thumb is to let u be the part of the integrand that makes the integral look complicated, such as the argument of a square root, a trigonometric function or an exponential, especially when its derivative sits nearby.

在很多试题中，代换式是直接给出的，但有时也需要你自己识别。一个经验法则是：令 u 为被积函数中让积分看起来复杂的部分，比如根号内、三角函数内或指数函数的指数部分，特别是当这些部分的导数就出现在附近的时候。

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4. Computing du and dx | 计算 du 与 dx

From u = 2x + 1 we differentiate with respect to x:

du/dx = 2

This can be rearranged to express dx in terms of du:

dx = du / 2

由 u = 2x + 1 对 x 求导得

du/dx = 2

整理后得到用 du 表示 dx 的式子：

dx = du / 2

We also need to write the original variable x completely in terms of u. Solving u = 2x + 1 for x yields

x = (u − 1) / 2

This step is crucial because the integrand contains an x factor outside the square root. Leaving any x in the u-integral is a very common mistake.

我们还需要将原变量 x 完全用 u 表示。由 u = 2x + 1 解出 x 得

x = (u − 1) / 2

这一步至关重要，因为被积函数中在根号外面还有一个 x 因子。许多同学的常见错误就是忘记彻底替换，在关于 u 的积分里留下了 x。

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5. Rewriting the Integral | 重写积分式

Now substitute every piece into the original integral:

• Replace √(2x+1) with √u.
• Replace x with (u−1)/2.
• Replace dx with du/2.

现在将所有部分代入原积分：

• 将 √(2x+1) 替换为 √u；
• 将 x 替换为 (u−1)/2；
• 将 dx 替换为 du/2。

The integral becomes

∫ [(u−1)/2] · √u · (du/2)

Multiplying the constants 1/2 and 1/2 gives 1/4, so we have

∫ x √(2x+1) dx = ¼ ∫ (u−1) √u du

积分变为

∫ [(u−1)/2] · √u · (du/2)

常数 1/2 与 1/2 相乘得 1/4，因此得到

∫ x √(2x+1) dx = ¼ ∫ (u−1) √u du

At this stage all the x-dependence has been eliminated. The integral is now entirely in the variable u.

到了这一步，所有对 x 的依赖都已被消去，积分完全处在变量 u 之下。

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6. Simplifying the Integrand | 简化被积函数

Rewrite the square root as a fractional exponent: √u = u^(1/2). Then expand the product:

(u−1) · u^(1/2) = u · u^(1/2) − 1 · u^(1/2) = u^(3/2) − u^(1/2)

将平方根写成分数指数的形式：√u = u^(1/2)。然后展开乘积：

(u−1) · u^(1/2) = u · u^(1/2) − 1 · u^(1/2) = u^(3/2) − u^(1/2)

Therefore the integral in terms of u simplifies to

¼ ∫ (u^(3/2) − u^(1/2)) du

这是一个简洁的形式，能够直接积分。

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7. Performing the Integration | 执行积分运算

We now integrate term by term using the power rule for integration: ∫ uⁿ du = uⁿ⁺¹/(n+1) + C, where the exponent n is not equal to −1.

现在我们使用幂函数的积分法则逐项积分：∫ uⁿ du = uⁿ⁺¹/(n+1) + C，其中指数 n 不等于 −1。

For the first term, n = 3/2, so n+1 = 5/2 and division by n+1 gives a factor of 2/5:

∫ u^(3/2) du = (2/5) u^(5/2)

For the second term, n = 1/2, so n+1 = 3/2 and we get a factor of 2/3:

∫ u^(1/2) du = (2/3) u^(3/2)

Putting the constant ¼ outside and including the constant of integration C yields

¼ [ (2/5) u^(5/2) − (2/3) u^(3/2) ] + C

对第一项，指数 n = 3/2，则 n+1 = 5/2，除以 n+1 得到系数 2/5：

∫ u^(3/2) du = (2/5) u^(5/2)

对第二项，指数 n = 1/2，则 n+1 = 3/2，除以 n+1 得到系数 2/3：

∫ u^(1/2) du = (2/3) u^(3/2)

将括号外的 ¼ 乘入并补上积分常数 C 得

¼ [ (2/5) u^(5/2) − (2/3) u^(3/2) ] + C

Multiply through by ¼:

= (1/10) u^(5/2) − (1/6) u^(3/2) + C

将 ¼ 乘进去：

= (1/10) u^(5/2) − (1/6) u^(3/2) + C

These two terms are the antiderivative in u. The expression is complete and ready for back-substitution.

这就是关于 u 的不定积分原函数，随时可以代回原变量。

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8. Back-Substituting to x | 将 x 代回

Recall that u = 2x + 1. Substituting this back into our result gives the final answer in terms of x:

∫ x √(2x+1) dx = (1/10)(2x+1)^(5/2) − (1/6)(2x+1)^(3/2) + C

我们记得 u = 2x + 1。将其代回结果，就得到关于 x 的最终答案：

∫ x √(2x+1) dx = (1/10)(2x+1)^(5/2) − (1/6)(2x+1)^(3/2) + C

The answer can be left in this form, but it is often good practice to tidy it up. Factorise the highest common factor, which is (1/30)(2x+1)^(3/2):

= (1/30)(2x+1)^(3/2) [3(2x+1) − 5] + C

Simplify the bracket:

= (1/30)(2x+1)^(3/2) (6x + 3 − 5) + C = (1/30)(2x+1)^(3/2)(6x − 2) + C

Factor a 2 from the last bracket to give a neater expression:

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