📚 Integration by Parts: Formula and Classic Examples | IB数学：分部积分法公式与经典例题
Integration by parts is one of the most important integration techniques in the IB Mathematics: Analysis and Approaches Higher Level course. It transforms a difficult integral into a simpler one by reversing the product rule of differentiation.
分部积分法是IB数学：分析与方法（Analysis and Approaches）高级课程中最重要的一种积分技巧。它通过逆用微分的乘积法则，将较难的积分转化为更简单的积分。
1. Deriving the Formula | 公式推导
of differentiation: if u and v are differentiable functions of x, then (uv)′ = u′v + uv′.
我们从微分的乘积法则出发：若u和v都是关于x的可微函数，则(uv)′ = u′v + uv′。
o x gives uv = ∫u′v dx + ∫uv′ dx. Rearranging, we obtain the integration by parts formula:
对等式两边关于x积分，得到uv = ∫u′v dx + ∫uv′ dx。整理后即可得到分部积分公式：
∫ u dv = uv − ∫ v du
Here u and v are usually written as functions of x, and du = u′ dx, dv = v′ dx.
这里u和v通常都是x的函数，且du = u′ dx，dv = v′ dx。
2. Choosing u and dv: The LIATE Rule | 选择u与dv：LIATE法则
, I = inverse trigonometric, A = algebraic/polynomial, T = trigonometric, E = exponential.
成功使用分部积分法的关键在于选择哪个因子作为u。LIATE法则提供了一个有用的优先顺序：L = 对数函数，I = 反三角函数，A = 代数/多项式函数，T = 三角函数，E = 指数函数。
Choose u from the leftmost category present in the integrand, and let the remaining factor be dv. This choice usually makes du simpler than u.
优先从被积函数中选取最靠左的类别作为u，其余部分作为dv。这样的选择通常会让du比u更简单。
| Order | 顺序 | Function Type | 函数类型 | Example | 例子 |
| 1 | Logarithmic | 对数 | ln x |
| 2 | Inverse trigonometric | 反三角 | arctan x |
| 3 | Algebraic | 代数 | x, x² |
| 4 | Trigonometric | 三角 | sin x, cos x |
| 5 | Exponential | 指数 | eˣ |
3. Classic Example: ∫ x eˣ dx | 经典例题：∫ x eˣ dx
Let u = x and dv = eˣ dx. Then du = dx and v = eˣ.
取u = x，dv = eˣ dx，则du = dx，v = eˣ。
Applying the formula: ∫ x eˣ dx = x eˣ − ∫ eˣ dx.
代入公式：∫ x eˣ dx = x eˣ − ∫ eˣ dx。
Since ∫ eˣ dx = eˣ + C, the result is:
因为∫ eˣ dx = eˣ + C，所以结果为：
∫ x eˣ dx = x eˣ − eˣ + C = (x − 1) eˣ + C
This example shows the LIATE rule at work: the algebraic x is chosen as u, leaving the exponential as dv.
这个例子体现了LIATE法则的运用：代数因子x被选为u，而指数函数作为dv。
4. Classic Example: ∫ x cos x dx | 经典例题：∫ x cos x dx
For this integral, we choose u = x and dv = cos x dx. Then du = dx and v = sin x.
对于这个积分，我们选择u = x，dv = cos x dx，则du = dx，v = sin x。
Using the formula: ∫ x cos x dx = x sin x − ∫ sin x dx.
使用公式：∫ x cos x dx = x sin x − ∫ sin x dx。
Now ∫ sin x dx = −cos x + C, so:
而∫ sin x dx = −cos x + C，因此：
∫ x cos x dx = x sin x + cos x + C
Notice that the remaining integral is simpler than the original because the polynomial factor x has been removed.
注意剩余积分比原积分更简单，因为多项式因子x已经被消去了。
5. Classic Example: ∫ ln x dx | 经典例题：∫ ln x dx
ance, this integral seems to have only one factor. The trick is to write ln x as 1 × ln x.
乍一看，这个积分似乎只有一个因子。技巧在于将ln x写成1 × ln x。
Let u = ln x and dv = 1 dx. Then du = (1/x) dx and v = x.
取u = ln x，dv = 1 dx，则du = (1/x) dx，v = x。
Applying the formula: ∫ ln x dx = x ln x − ∫ x × (1/x) dx = x ln x − ∫ 1 dx.
代入公式：∫ ln x dx = x ln x − ∫ x × (1/x) dx = x ln x − ∫ 1 dx。
Therefore:
因此：
∫ ln x dx = x ln x − x + C
This example demonstrates why logarithmic functions are given the highest priority in LIATE.
这个例子说明对数函数在LIATE中被赋予最高优先级的原因。
6. Classic Example: ∫ eˣ sin x dx | 经典例题：∫ eˣ sin x dx
Here both eˣ and sin x are common functions. Let u = eˣ and dv = sin x dx. Then du = eˣ dx and v = −cos x.
这里eˣ和sin x都是常见函数。我们取u = eˣ，dv = sin x dx，则du = eˣ dx，v = −cos x。
Thus ∫ eˣ sin x dx = −eˣ cos x + ∫ eˣ cos x dx.
于是∫ eˣ sin x dx = −eˣ cos x + ∫ eˣ cos x dx。
Apply integration by parts again to the new integral: let u = eˣ, dv = cos x dx. Then du = eˣ dx, v = sin x.
对新积分再次应用分部积分：取u = eˣ，dv = cos x dx，则du = eˣ dx，v = sin x。
This gives ∫ eˣ cos x dx = eˣ sin x − ∫ eˣ sin x dx.
得到∫ eˣ cos x dx = eˣ sin x − ∫ eˣ sin x dx。
Let I = ∫ eˣ sin x dx. Then I = −eˣ cos x + eˣ sin x − I.
设I = ∫ eˣ sin x dx，则I = −eˣ cos x + eˣ sin x − I。
Add I to both sides: 2I = eˣ(sin x − cos x), hence:
两边加上I：2I = eˣ(sin x − cos x)，因此：
∫ eˣ sin x dx = ½ eˣ(sin x − cos x) + C
This is a classic "cyclic" example in which the original integral appears again.
这是一个经典的“循环”型例题，原积分在运算过程中再次出现。
7. Repeated Integration by Parts: ∫ x² eˣ dx | 重复使用分部积分：∫ x² eˣ dx
When the algebraic power is greater than 1, integration by parts must be applied more than once.
当代数幂次大于1时，分部积分法需要多次应用。
First choose u = x² and dv = eˣ dx. Then du = 2x dx and v = eˣ.
首先取u = x²，dv = eˣ dx，则du = 2x dx，v = eˣ。
Therefore ∫ x² eˣ dx = x² eˣ − 2∫ x eˣ dx.
因此∫ x² eˣ dx = x² eˣ − 2∫ x eˣ dx。
From Example 3, ∫ x eˣ dx = (x − 1) eˣ + C. Substituting back:
由例题3知∫ x eˣ dx = (x − 1) eˣ + C。代回：
∫ x² eˣ dx = x² eˣ − 2(x − 1)eˣ + C = (x² − 2x + 2) eˣ + C.
∫ x² eˣ dx = x² eˣ − 2(x − 1)eˣ + C = (x² − 2x + 2) eˣ + C。
∫ x² eˣ dx = (x² − 2x + 2) eˣ + C
Each application of integration by parts reduces the degree of the polynomial by one.
每一次分部积分都会使多项式的次数降低一次。
8. Definite Integrals: ∫₀¹ x eˣ dx | 定积分：∫₀¹ x eˣ dx
or definite integrals, apply the formula with the endpoints: ∫ₐᵇ u dv = [uv]ₐᵇ − ∫ₐᵇ v du.
对于定积分，代入上下限使用公式：∫ₐᵇ u dv = [uv]ₐᵇ − ∫ₐᵇ v du。
Using u = x, dv = eˣ dx as before, we have:
和之前一样取u = x，dv = eˣ dx，则：
∫₀¹ x eˣ dx = [x eˣ]₀¹ − ∫₀¹ eˣ dx.
∫₀¹ x eˣ dx = [x eˣ]₀¹ − ∫₀¹ eˣ dx。
Evaluate the boundary term: [x eˣ]₀¹ = 1×e¹ − 0×e⁰ = e.
计算边界项：[x eˣ]₀¹ = 1×e¹ − 0×e⁰ = e。
Evaluate the remaining integral: ∫₀¹ eˣ dx = e − 1.
计算剩余积分：∫₀¹ eˣ dx = e − 1。
Therefore ∫₀¹ x eˣ dx = e − (e − 1) = 1.
因此∫₀¹ x eˣ dx = e − (e − 1) = 1。
∫₀¹ x eˣ dx = 1
Always keep the limits attached to the boundary term when using a definite integral.
使用定积分时，一定要把上下限代入边界项中。
9. Common Mistakes and Tips | 常见错误与技巧
One common mistake is forgetting the constant of integration C when finding antiderivatives. Always add + C unless the problem is a definite integral.
常见错误之一是求不定积分时忘记积分常数C。除非是定积分，否则一定要加上+C。
ctly. If the resulting ∫v du is more complicated than the original, try swapping u and dv.
另一个错误是u选择不当。如果化简后的∫v du比原积分更复杂，可以尝试交换u和dv。
For cyclic integrals like ∫ eˣ ot panic when the original integral reappears. Collect like terms and solve algebraically.
对于类似∫ eˣ sin x dx这样循环出现的积分，不必惊慌。合并同类项并代数求解即可。
Always double-check signs, especially when integrating trigonometric functions such as sin x or cos x.
始终检查符号，尤其是在积分sin x或cos x这样的三角函数时。
10. Summary and Practice | 总结与练习
ns. The core formula is ∫u dv = uv − ∫v du, and the LIATE rule helps choose u effectively.
分部积分法为函数乘积的积分提供了强大的工具。核心公式是∫u dv = uv − ∫v du，而LIATE法则有助于有效选择u。
mials × exponentials, polynomials × trigonometric functions, logarithms, and cyclic cases.
掌握这种方法需要练习多项式×指数、多项式×三角函数、对数以及循环型积分。
In IB exams, show every step clearly and simplify your final answer. This will earn full marks for method and accuracy.
在IB考试中，请清晰地展示每一步并化简最终答案，这样可以在方法和准确性上获得满分。
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