分部整合

积分是积分的计算，在数学中用于查找许多有用的量，例如面积、体积、位移等，这些量是由于无法单独测量的小数据的集合而发生的。积分用“∫”表示。

积分符号

集成的概念用于解决以下两种类型的问题：

• 给定导数后，求问题函数，
• 在某些特定约束下找到由函数图包围的区域。

这两个问题导致了称为“积分”的概念的发展，该概念由以下两种积分组成：

• 定积分：包括上限和下限的积分称为定积分。定积分也称为黎曼积分。一个明确的积分表示为：

$\int_{b}^{a} f(x)dx$

• 不定积分：不定积分没有上限和下限。它表示为：

∫f(x)dx = F(x) + C

这里 C 是一个常数，f(x) 是一个函数 这被称为被积函数。

有五种类型的集成方法可用：

1. 替代整合
2. 按部分规则集成
3. 部分分数积分
4. 使用三角恒等式积分
5. 某些特定函数的集成

在这里，我们只讨论部分集成。

分部整合

分部集成是一种集成方法，常用于集成两个功能的产品。该技术用于通过将积分简化为标准形式来查找积分。以下公式用于按部分执行积分：

$\int u(\frac{dv}{dx})dx = uv - \int v(\frac{du}{dx})dx$

在哪里：

u 是 x 的第一个函数：u(x)

v 是 x 的第二个函数：v(x)

分部积分公式：

$\int_{a}^{b} u(\frac{dv}{dx})dx = [uv]_{a}^{b} - \int_{a}^{b} v(\frac{du}{dx})dx$

在哪里：

a 是下限

b 是上限

按零件公式推导积分：

Suppose u and w are any two differentiable functions of a single variable y. 

As we already know how to differentiate a product, 

y = uw

then, by the product rule of differentiation

dy/dx = d⁄dx(uw) = u d/dx(w) + w d/dx(u)

Rearranging this, we get

u d/dx(w) = d⁄dx(uw) – w d/dx(u)

or

uw’ = d⁄dx(wu) – wu’

On integrating both sides w.r.t x, we get

∫(uw’)dx = ∫d/dx(uw) dx – ∫(wu’)dx

The first term on the right simplifies, since 

we are simply integrating what has been differentiated.

Therefore,

∫(uw’)dx = uw – ∫(u’w) dx

Now put

w’ = v

w = ∫v  

Substituting the above values, we get

∫(uv)dx = u(∫vdx) – ∫(u’∫vdx)dx

部分积分规则：

1. 选择第一个函数为u ，第二个函数为dv/dx 。
2. 区分你：你'
3. 积分 dv/dx： (∫(dv/dx)dx = v)
4. 将 u, u', v 和 ∫(dv/dx)dx 代入： u(∫vdx) −∫(u' ∫vdx) dx
5.  简化和解决

ILATE 规则

在按部分积分中，通常，我们使用优先顺序来决定第一个函数，使用ILATE规则（反三角函数、对数函数、代数函数、三角函数、指数函数）。该规则规定，在执行积分时，应将函数假定为第一个函数。因此，根据所涉及的函数类型，通过将（ILATE 的）左项视为第一个函数，将（ILATE 的）第二项视为第二个函数，假设函数按从左到右的顺序排列。所以， ILATE 是我们选择第一个函数的顺序。

以下是函数u 的以下步骤，其中

I = Inverse trigonometric functions

such as sin^-1 (x), cos^-1 (x), tan^-1 (x)

L = Logarithmic functions

such as ln(x), log(x)

A = Algebraic functions

such as x², x³

T =  Trigonometric functions

such as sin(x), cos(x), tan(x)

E = Exponential functions

such as e^x, 3^x

示例问题

问题 1. 计算 ∫2x cos(x)dx

解决方案：

Given that, ∫2x cos(x)dx

According to the rule of integration by parts, 

Let us consider,

u = 2x and dv/dx = cos(x) 

Then

du/dx = 2  and  v = ∫cos(x) =  sin(x)

Now, using the formula for integration by parts;

∫u(dv/dx)dx = uv – ∫v(du/dx)dx

We get,

∫2x cox(x)dx = 2x sin(x) – ∫sinx.2dx

∫2x cos(x)dx = 2x sin(x) + 2cos(x) + c

where c is a constant.

问题 2. 计算∫3x ln|x| dx

解决方案：

Given, ∫3xln|x| dx 

According to rule of integration by parts, 

Let us consider,

u = ln|x|  and dv/dx = 3x

Then

du/dx = 1/x  and  v = ∫3xdx = 3x²/2

Now, using the formula for integration by parts,

∫u(dv/dx)dx = uv – ∫v(du/dx)dx

∫3x ln|x|dx = ln|x|.3x²/2 – ∫3x²/2(1/x)dx

= 3x²/2(ln|x|) – ∫3x/2 dx

= 3x²/2(ln|x|) – 3x²/4 + C

where C is a constant.

问题 3. 评估∫10x ²

解决方案：

Given that, ∫10x²

According to rule of integration by parts, 

Let us consider,

u = 10x² and dv/dx = 1

Then

du/dx = 20x  and  v = ∫(1)dx = x 

Now, using the formula for integration by parts;

∫u(dv/dx)dx = uv – ∫v(du/dx)dx

∫10x² dx = 10x²x – ∫x(20x)dx

∫10x² dx = 10x³ – ∫20x²dx

∫10x² dx = 10x³ – 20/3x³ + C

∫10x² dx = 10x³/3 + C

where C is a constant.

问题 4. 计算 ∫x ² e ^4x dx

解决方案：

Given that, ∫x² e^4x dx

According to rule of integration by parts, 

Let us consider,

u = x² and dv/dx = e^4x

Then

du/dx = 2x  and  v = ∫e^4x dx = 1/4(e^4x)

Then, using the formula for integration by parts,

∫u(dv/dx)dx = uv – ∫v(du/dx)dx

∫x²e^4x dx = (x²)1/4(e^4x) – ∫1/4(e^4x)(2x)dx

= 1/4(x²)(e^4x) – ∫2/4(x)(e^4x)dx 

= 1/4(x²)(e^4x) – ∫1/2(x)(e^4x)dx 

The resulting integral is still a product of 

two functions i.e. 1/2(x) and e^4x. 

So, We use the formula for integration again.

This time we take

u = 1/2(x)  and  dv/dx = e^4x

So

du/dx = 1/2  and  v = ∫e^4xdx = 1/4(e^4x)

Now, on putting these values in above equation,

∫x²e^4x dx = 1/4(x²)(e^4x) – ∫2/4(x)(e^4x)dx

= 1/4(x²e^4x) – {2/4(x).1/4(e^4x) – ∫(1/4(e^4x).2/4 dx}

= 1/4(x²e^4x) – 1/8(xe^3x) + 1/8(e^4x) + C

where C is a constant.

问题 5. 计算 ∫ 2e ^x sin(x) dx

解决方案：

Given that, ∫ 2e^xsin(x) dx

According to rule of integration by parts, 

Let us consider,

u = 2e^x  and  dv/dx = sin(x)

Then

du/dx = 2e^x  and v = ∫sin(x)dx = -cos(x)

Now, using the formula for integration by parts,

∫u(dv/dx)dx = uv – ∫v(du/dx)dx

∫ 2e^xsin(x) dx = (2e^x)(-cos(x)) – ∫(-cos(x))(2e^x)dx

= -(2e^x)cos(x) + ∫ 2e^x cos(x)dx

The resulting integral is yet a product of two

functions i.e. 2e^x and cos(x). 

So, we use the formula for integration again.

This time we take

u = 2e^x  and  dv/dx = cos(x)

So

du/dx = 2e^x  and  v = ∫cos(x)dx = sin(x)

Now, on putting these values in the above equation, we get

∫ 2e^xsin(x) dx = -cos(x)(2e^x) + ∫ 2e^x cos(x)dx

= -(2e^x)cos(x) + {2e^xsin(x) – ∫sin(x)(2e^x)dx}

= -2e^xcos(x) + 2e^xsin(x) – ∫2e^xsin(x)dx

Now, notice that the integral ∫e^xsin(x)dx in the result

is same as the integral in the given question.

So, let us assume this as I.

Therefore,

I = ∫ 2e^xsin(x) dx

So,

I = 2e^xsin(x) – 2e^xcos(x) – I

2I = 2e^xsin(x) – 2e^xcos(x)

2I = 2(e^xsin(x) – e^xcos(x))

I = (e^xsin(x) – e^xcos(x))

So,

∫ 2e^xsin(x) dx = e^xsin(x) – e^xcos(x) + C

where C is a constant.

问题 6. 计算 ∫2tan ^-1 x dx

解决方案：

Given, ∫2tan^-1x dx 

According to the rule of integration by parts, 

Let us consider,

u = tan^-1x  and  dv/dx = 2

Then

du/dx = 1/(1 + x²)  and  v = ∫(2)dx = 2x 

Now, using the formula for integration by parts,

∫u(dv/dx)dx = uv – ∫v(du/dx)dx

∫2tan^-1x dx = tan^-1x(2x) – ∫2x(1/(1+x²))dx 

let, p = (1 + x²)

So, dp/dx = 2x 

⇒ x = 1/2

Now, substituting the above values, we get

∫tan^-1x dx = tan^-1x(2x) – ∫(1/p)dx

= 2xtan^-1x – ∫(1/p)dx

= 2xtan^-1x – lnp

= 2xtan^-1x – ln(1 + x²) 

= 2xtan^-1x – ln√(1 + x²) + C

where C is a constant.