Exercise 17C: Integration by Substitution | 练习17C：换元积分法

📚 Exercise 17C: Integration by Substitution | 练习17C：换元积分法

Integration by substitution, often called u-substitution, is one of the most powerful tools for evaluating integrals in the IB Mathematics Analysis and Approaches syllabus. This technique reverses the chain rule and allows us to simplify a complicated integrand by introducing a new variable. In Exercise 17C we focus on applying substitution to indefinite and definite integrals involving polynomials, trigonometric functions, exponentials, and logarithms. Mastering this method will give you the confidence to tackle a wide range of integration problems on Paper 1 and Paper 2.

换元积分法，通常称为 u 替代法，是 IB 数学分析与方法课程中求积分最强有力的工具之一。这项技巧反向运用了链式法则，通过引入新变量把一个复杂的被积函数简化。在练习 17C 中，我们专注于把换元法应用到涉及多项式、三角函数、指数函数和对数函数的不定积分与定积分上。熟练掌握这一方法将使你信心十足地应对 Paper 1 和 Paper 2 中大量的积分题目。

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1. The Substitution Rule | 换元法则

The substitution rule states that if u = g(x) is a differentiable function, then ∫ f(g(x))·g'(x) dx = ∫ f(u) du. In practice, we look for an inner function whose derivative also appears in the integrand (up to a constant factor). This transforms the integral into a basic form that can be evaluated directly.

换元法则指出，如果 u = g(x) 是一个可微函数，那么 ∫ f(g(x))·g'(x) dx = ∫ f(u) du。实际操作中，我们寻找一个内层函数，其导数（至多差一个常数倍数）也出现在被积函数中。这样就把原积分转化成可以直接积出的基本形式。

∫ f(g(x))·g'(x) dx = ∫ f(u) du, where u = g(x)

• Identify u = g(x).
• Compute du = g'(x) dx.
• Rewrite the entire integral in terms of u.
• Integrate with respect to u.
• Substitute back to express the result in x (for indefinite integrals).

关键步骤如下：选定 u = g(x)；计算 du = g'(x) dx；将被积函数全部用 u 表示；对 u 积分；最后将结果换回 x（不定积分）。

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2. Worked Example 1 – Simple Polynomial | 例题1 – 简单多项式

Evaluate ∫ 2x (x² + 1)³ dx.

计算 ∫ 2x (x² + 1)³ dx。

Let u = x² + 1. Then du = 2x dx. Notice that 2x dx appears exactly in the integral. Substituting gives ∫ u³ du. This is a basic power integral: ∫ u³ du = u⁴/4 + C. Returning to x, we obtain (x² + 1)⁴/4 + C.

令 u = x² + 1，则 du = 2x dx。观察到 2x dx 恰好出现在积分中。代入后得到 ∫ u³ du。这是基本的幂函数积分：∫ u³ du = u⁴/4 + C。换回 x，得到 (x² + 1)⁴/4 + C。

• Choose u = x² + 1
• du = 2x dx ⇒ dx = du/(2x)
• Substitute: ∫ 2x · u³ · (du/(2x)) = ∫ u³ du
• Integrate: u⁴/4 + C
• Back-substitute: (x² + 1)⁴/4 + C

步骤：选取 u = x² + 1；得到 du = 2x dx；代入化简为 ∫ u³ du；积出 u⁴/4 + C；代回得 (x² + 1)⁴/4 + C。

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3. Worked Example 2 – Trigonometric Function | 例题2 – 三角函数

Find ∫ sin(5x) dx.

求 ∫ sin(5x) dx。

Set u = 5x. Then du = 5 dx, so dx = du/5. The integral becomes ∫ sin u · (du/5) = (1/5) ∫ sin u du = –(1/5) cos u + C. Replacing u by 5x gives the final answer –(1/5) cos(5x) + C. This illustrates how substitution handles linear arguments of trigonometric functions.

设 u = 5x，则 du = 5 dx，因此 dx = du/5。积分变为 ∫ sin u · (du/5) = (1/5) ∫ sin u du = –(1/5) cos u + C。将 u 换回 5x 得到最终答案 –(1/5) cos(5x) + C。这个例子展示了换元法如何处理三角函数的线性参数。

• u = 5x
• du = 5 dx → dx = du/5
• ∫ sin u · (du/5) = (1/5)(–cos u) + C
• = –(1/5) cos(5x) + C

步骤：u = 5x；得到 dx = du/5；积分得 –(1/5) cos u + C；代回为 –(1/5) cos(5x) + C。

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4. Worked Example 3 – Exponential Function | 例题3 – 指数函数

Evaluate ∫ x e^x² dx.

计算 ∫ x e^x² dx。

Here the inner function is x². Let u = x², so du = 2x dx. The integrand contains x dx, which is (1/2) du. Substituting: ∫ e^u · (1/2) du = (1/2) e^u + C = (1/2) e^x² + C. This is a classic example where the derivative of the exponent is closely related to the remaining factor.

这里内层函数是 x²。令 u = x²，则 du = 2x dx。被积函数含有 x dx，正好是 (1/2) du。代入得：∫ e^u · (1/2) du = (1/2) e^u + C = (1/2) e^x² + C。这是一个经典例子，指数部分的导数与剩余因子密切相关。

• u = x²
• du = 2x dx → x dx = du/2
• ∫ e^u · (du/2) = (1/2) e^u + C
• = (1/2) e^x² + C

步骤：设 u = x²；则 x dx = du/2；积出 (1/2) e^u + C；代回得 (1/2) e^x² + C。

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5. Worked Example 4 – Logarithmic Integration | 例题4 – 对数积分

Compute ∫ (ln x)/x dx.

计算 ∫ (ln x)/x dx。

A natural choice is u = ln x, because du = (1/x) dx. The integral becomes ∫ u du = u²/2 + C = (ln x)²/2 + C. This problem highlights how substitution can turn a seemingly tricky integrand into a simple power function.

很自然的选择是 u = ln x，因为 du = (1/x) dx。积分变成 ∫ u du = u²/2 + C = (ln x)²/2 + C。这道题突显了换元法如何把看似棘手的被积函数化为简单的幂函数。

• u = ln x
• du = (1/x) dx
• ∫ (ln x)·(1/x) dx = ∫ u du
• = u²/2 + C = (ln x)²/2 + C

步骤：令 u = ln x；则 du = (1/x) dx；积分变为 ∫ u du；积出 u²/2 + C；代回得 (ln x)²/2 + C。

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6. Definite Integrals by Substitution | 定积分的换元法

When evaluating a definite integral ∫_a^b f(x) dx using substitution, we have two options: change the limits to match the new variable u, or revert to x after integrating and then apply the original limits. The first method is often cleaner: once u = g(x) is chosen, compute u(a) and u(b) and use them as the new limits. There is no need to back-substitute.

用换元法计算定积分 ∫_a^b f(x) dx 时，有两种选择：把上下限换成新变量 u 的对应值，或者积分后换回 x 再代入原上下限。第一种方法通常更简洁：选定 u = g(x) 后，计算 u(a) 和 u(b) 并作为新的上下限，无需回代。

Example: Evaluate ∫₀² 2x (x² + 1)³ dx.

例子：计算 ∫₀² 2x (x² + 1)³ dx。

Let u = x² + 1. When x = 0, u = 1; when x = 2, u = 5. Also du = 2x dx. The integral becomes ∫₁⁵ u³ du = [u⁴/4]₁⁵ = (625/4) – (1/4) = 624/4 = 156.

令 u = x² + 1。当 x = 0 时 u = 1；当 x = 2 时 u = 5。同时 du = 2x dx。积分变为 ∫₁⁵ u³ du = [u⁴/4]₁⁵ = (625/4) – (1/4) = 624/4 = 156。

• u = x² + 1 → du = 2x dx
• New limits: from u = 1 to u = 5
• ∫₁⁵ u³ du = [u⁴/4]₁⁵
• = 156

步骤：换元后上下限变为 1 和 5；积分 u³；代入计算得 156。

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7. Exercise 17C Problem 1 – Square Root Substitution | 练习17C 问题1 – 根式换元

Evaluate ∫ √(2x + 1) dx.

计算 ∫ √(2x + 1) dx。

Let u = 2x + 1. Then du = 2 dx, so dx = du/2. The integrand becomes √u = u^1/2. Thus ∫ u^1/2 · (du/2) = (1/2) ∫ u^1/2 du = (1/2) · (2/3) u^3/2 + C = (1/3) (2x + 1)^3/2 + C. Remember to adjust for the constant factor from dx.

令 u = 2x + 1，则 du = 2 dx，所以 dx = du/2。被积函数变成 √u = u^1/2。于是 ∫ u^1/2 · (du/2) = (1/2) ∫ u^1/2 du = (1/2) · (2/3) u^3/2 + C = (1/3) (2x + 1)^3/2 + C。务必调整 dx 带来的常数倍数。

• u = 2x + 1
• du = 2 dx ⇒ dx = du/2
• ∫ √u · du/2 = (1/2) × (2/3) u^3/2 + C
• = (1/3) (2x + 1)^3/2 + C

步骤：换元后简化成幂函数积分；消去常数后得到最终结果。

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8. Exercise 17C Problem 2 – Trigonometric Powers | 练习17C 问题2 – 三角函数的幂

Find ∫ cos³ x sin x dx.

求 ∫ cos³ x sin x dx。

Here the derivative of cos x is –sin x, which is almost present. Let u = cos x. Then du = –sin x dx, so –du = sin x dx. The integral becomes ∫ u³ · (–du) = –∫ u³ du = –u⁴/4 + C = –(cos⁴ x)/4 + C. This technique works for any odd power of sine or cosine paired with the other function.

这里 cos x 的导数是 –sin x，几乎出现在积分中。令 u = cos x，则 du = –sin x dx，于是 –du = sin x dx。积分变成 ∫ u³ · (–du) = –∫ u³ du = –u⁴/4 + C = –(cos⁴ x)/4 + C。这一技巧适用于任何与另一个函数配对的正弦或余弦的奇次幂。

• u = cos x
• du = –sin x dx ⇒ sin x dx = –du
• ∫ u³ (–du) = –∫ u³ du
• = –u⁴/4 + C = –(cos⁴ x)/4 + C

步骤：巧妙利用负号，使 sin x dx 转化为 –du；积分后得到负的余弦四次方。

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9. Exercise 17C Problem 3 – Rational Function with Limits | 练习17C 问题3 – 带有理函数的定积分

Evaluate ∫₀¹ x/(1 + x²) dx using substitution.

用换元法计算 ∫₀¹ x/(1 + x²) dx。

Let u = 1 + x². Then du = 2x dx, so x dx = du/2. Change the limits: when x = 0, u = 1; when x = 1, u = 2. The integral becomes ∫₁² (1/u) · (du/2) = (1/2) ∫₁² (1/u) du = (1/2) [ln|u|]₁² = (1/2)(ln 2 – ln 1) = (1/2) ln 2. Since ln 1 = 0, the result simplifies nicely.

令 u = 1 + x²，则 du = 2x dx，所以 x dx = du/2。换上下限：x = 0 时 u = 1；x = 1 时 u = 2。积分变为 ∫₁² (1/u) · (du/2) = (1/2) ∫₁² (1/u) du = (1/2) [ln|u|]₁² = (1/2)(ln 2 – ln 1) = (1/2) ln 2。由于 ln 1 = 0，结果变得非常简洁。

• u = 1 + x², du = 2x dx
• Limits: u(0)=1, u(1)=2
• (1/2) ∫₁² du/u = (1/2) ln u |₁²
• = (1/2) ln 2

步骤：换元后积分变为 1/u 的形式；代入新的上下限得到 (1/2) ln 2。

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10. Exercise 17C Problem 4 – Recognizing f'(x)/f(x) | 练习17C 问题4 – 识别 f'(x)/f(x) 型

Integrate ∫ (e^x)/(1 + e^x) dx.

求 ∫ (e^x)/(1 + e^x) dx。

Notice that the numerator is exactly the derivative of the denominator. Set u = 1 + e^x, then du = e^x dx. The integral becomes ∫ (1/u) du = ln|u| + C = ln(1 + e^x) + C (since 1 + e^x > 0 for all real x). This pattern is extremely common in IB exam questions.

注意到分子恰好是分母的导数。设 u = 1 + e^x，则 du = e^x dx。积分变成 ∫ (1/u) du = ln|u| + C = ln(1 + e^x) + C（因为对所有实数 x 都有 1 + e^x > 0）。这种模式在 IB 考题中极为常见。

• u = 1 + e^x
• du = e^x dx
• ∫ du/u = ln|u| + C
• = ln(1 + e^x) + C

步骤：直接识别出分子是分母的导数；积分结果就是分母的自然对数。

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11. Common Pitfalls and Tips | 常见错误与提示

One frequent mistake is forgetting to change the limits when evaluating definite integrals. If you use the back-substitution method, you must apply the original limits to the antiderivative in x—not u. Another error is mishandling the constant factor: always check that du matches the exact differential present. If there is a missing constant, multiply the integral by its reciprocal. Additionally, avoid using substitution when the derivative of the chosen u is not present or cannot be introduced by a constant factor. Sometimes algebraic manipulation or trigonometric identities are needed first. Finally, for IB exams, remember to simplify logarithmic arguments and rationalise where appropriate.

一个常见错误是在计算定积分时忘记更换上下限。如果你采用回代法，则必须在关于 x 的原函数中使用原来的上下限，而不是 u 的上下限。另一个错误是处理常数因子不当：一定要检查 du 是否与积分中出现的微分精确匹配。如果差一个常数，要把该常数的倒数乘到积分上。另外，如果所选 u 的导数没有出现且无法通过常数因子引入，就不要强行使用换元。有时需要先进行代数变形或使用三角恒等式。最后，在 IB 考试中，记得化简对数参数并视情况进行有理化。

• Always update limits for definite integrals with u-substitution.
• Check that du is exactly present; adjust with constants if needed.
• If the derivative is missing, consider rewriting the integrand first.
• Back-substitution must return to the original variable before applying limits.

定积分换元务必更新上下限；检查 du 是否准确出现，必要时用常数调整；若导数缺失，考虑先改写被积函数；回代时必须在代入上下限之前换回原变量。

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12. Conclusion and Further Practice | 总结与进一步练习

Mastering integration by substitution is essential for success in IB Mathematics. The exercises in 17C provide a solid foundation, but true fluency comes from practicing a wide variety of functions: rational, trigonometric, exponential, logarithmic, and those requiring creative algebraic manipulation. As you work through additional problems, try to identify the substitution within seconds—this intuition will save valuable time during exams. Continue to the next exercise set to explore integration by parts, another key technique that pairs naturally with substitution.

掌握换元积分法对于 IB 数学取得成功至关重要。练习 17C 提供了坚实的基础，但真正的熟练来自于对各种函数的广泛练习：有理函数、三角函数、指数函数、对数函数以及那些需要创造性代数变形的题目。在练习更多问题时，尝试在几秒内识别出换元方式——这种直觉会在考试中为你节省宝贵的时间。继续进入下一个练习集，探索分部积分法，这是另一项与换元法自然配合的关键技巧。

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