Course: 8.4 按部分分数合并

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按部分分数分列的整合情况
Often a difficult rational integrand can be evaluated by breaking it up into multiple, simpler components that are each easier to evaluate. The concept is analogous to taking a fraction like $\begin{aligned} \frac{7}{12} \end{aligned}$ and breaking into the two- term representation of $\begin{aligned} \frac{1}{3} + \frac{1}{4} \end{aligned}$ , because each term might be easier to use. Note, however, that $\begin{aligned} \frac{1}{3} + \frac{1}{4} \end{aligned}$ is not the only representation that works; the two-term representation $\begin{aligned} \frac{A}{6} + \frac{B}{2} \end{aligned}$ could be chosen. How would you find appropriate values for $\begin{aligned} A \end{aligned}$ and $\begin{aligned} B \end{aligned}$ such that $\begin{aligned} \frac{7}{6 \cdot 2} = \frac{A}{6} + \frac{B}{2} \end{aligned}$ ? This is the basic concept of partial fraction approach.
::通常,一个困难的理性指数可以通过将其分成多个更容易评估的简单组成部分来评估。这个概念类似于采取712这样的分数,并打破13+14的两期代表制,因为每个术语可能比较容易使用。但请注意,13+14并不是唯一可行的代表制;可以选择两期代表制A6+B2。你如何为A和B找到合适的值,如76+2=A6+B2?这是部分代表制的基本概念。

Integration by Partial Fractions
::按部分分数分列的整合情况

The Integration by Partial Fractions technique involves integrating a rational function $\begin{aligned} \frac{f (x)}{g (x)} \end{aligned}$ by using two steps:
::“按部分分数整合”技术涉及采用以下两个步骤整合一个合理的函数f(x)g(x):

decompose the rational function into a sum of two or more simpler rational functions, and
::将合理功能分解成两个或两个以上简单合理功能的总和,

integrate the sum of partial fractions.
::整合部分分数的总和。

For example, what is $\begin{aligned} \int \frac{x + 4}{x^{2} + x - 2} d x \end{aligned}$ ?
::例如,什么是x+4x2+x-2dx?

The integrand can be decomposed into
::整数可以分解成

$\begin{aligned} \frac{f (x)}{g (x)} = \frac{x + 4}{x^{2} + x - 2} = \frac{3}{5 (x - 1)} - \frac{3}{2 (x + 2)} \end{aligned}$ .
:xx)g(x)=x+4x2+x-235(x-1)-32(x+2)。

The two terms on the right are called partial fraction . Note that the denominators of the partial fractions are the factors of $\begin{aligned} g (x) \end{aligned}$ .
::权利的两个术语称为部分分数,请注意部分分数的分母是g(x)系数。

By decomposing it into two partial fractions, the integral becomes manageable:
::通过将其分解成两个部分,整体部分变得可以管理:

$\begin{aligned} \int \frac{x + 4}{x^{2} + x - 2} d x & = \int (\frac{3}{5 (x - 1)} - \frac{3}{2 (x + 2)}) d x \\ = \frac{3}{5} \int (\frac{1}{x - 1}) d x - \frac{3}{2} \int (\frac{1}{x + 2}) d x \\ = \frac{3}{5} \ln | x - 1 | - \frac{3}{2} \ln | x + 1 | + C \end{aligned}$

::X+4x2+x-2dx(35(x-1)-32(x+2))dx=35*(1x-1)dx-32*(1x-1)dx-32*(1x+2)dx=35lnx-1}(32(x-1)x+1)c

For the above technique to work, the partial fractions must be determined. The general conditions for decomposing a rational function $\begin{aligned} \frac{f (x)}{g (x)} \end{aligned}$ are given below:
::为使上述技术发挥作用,必须确定部分分数。

The degree of $\begin{aligned} f (x) \end{aligned}$ must be less than the degree of $\begin{aligned} g (x) \end{aligned}$ .
::f(x) 度必须小于 g(x) 度。

If so, the rational function is called proper.
::如果是这样,理性功能被称为适当功能。

If not, divide $\begin{aligned} f (x) \end{aligned}$ by $\begin{aligned} g (x) \end{aligned}$ (use long division) and work with the remainder term.
::如果不是,则将f(x)除以g(x)(使用长的分界线),与剩余任期一起工作。

The factors of $\begin{aligned} g (x) \end{aligned}$ must be determined in the form $\begin{aligned} (a x + b)^{n} \end{aligned}$ or $\begin{aligned} (a x^{2} + b x + c)^{n} \end{aligned}$ , where $\begin{aligned} n \geq 1 \end{aligned}$ and $\begin{aligned} (a x^{2} + b x + c) \end{aligned}$ is an irreducible quadratic , i.e., has no real zeros.
::g(x) 系数必须以表( 轴+b)n 或 ( 轴2+bx+c)n 确定,nn, n1 和 (轴2+bx+c) 是不可复制的二次方块, 即没有实际零。

If so, follow the guide below to find the partial fractions.
::如果是,按以下指南查找部分分数。

If not, partial fractions cannot be determined and this technique cannot be used.
::如果没有,则无法确定部分分数,因此无法使用这一技术。

If the denominator $\begin{aligned} g (x) \end{aligned}$ of the rational function $\begin{aligned} \frac{f (x)}{g (x)} \end{aligned}$ has been fully factored according to the second bullet above, then the following guide can be used to determine the terms of the sum of the partial fraction representation:
::如果理性函数f(x)g(x)分母g(x)的g(x)已根据以上第二个圆点充分计算在内,则以下指南可用于确定部分部分表示数之和的条件:

Forms of partial fractions composing $\begin{aligned} \frac{f (x)}{g (x)} \end{aligned}$
::F(x)g(x)

When $\begin{aligned} g (x) \end{aligned}$ can be factored with factors of the form $\begin{aligned} (a x + b)^{n} \end{aligned}$ or $\begin{aligned} (a x^{2} + b x + c)^{n} \end{aligned}$ , the partial fraction (PF) expansion of $\begin{aligned} \frac{f (x)}{g (x)} \end{aligned}$ looks like:
::当 g(x) 可以与窗体( 轴+b) n 或 ( 轴+bx+c) n 的因数乘以时, f( x) g( x) 的部分部分扩展( PF) 看起来像 :

$\begin{aligned} \frac{f (x)}{g (x)} = \sum PFs for all (a x + b)^{n} factors + \sum PFs for all (a x^{2} + b x + c)^{n} factors \end{aligned}$

::f(x)g(x) +(x) +(x) +(x) +(b)n 系数 +(x) n 系数 +(x) +(b)+(c)n

For each distinct factor of $\begin{aligned} g (x) \end{aligned}$ of the form $\begin{aligned} (a x + b)^{n} \end{aligned}$ , with $\begin{aligned} n \geq 1 \end{aligned}$ , its contribution to the partial fraction sum is:
::对于表(ax+b)n的克(x)的每一种不同系数,加上n1,其部分分数总和的分摊额是:

$\begin{aligned} \sum_{i = 1}^{n} \frac{A_{i}}{(a x + b)^{i}} = \frac{A_{1}}{a x + b} + \frac{A_{2}}{(a x + b)^{2}} + \dots + \frac{A_{n}}{(a x + b)^{n}}, \end{aligned}$

::i=1nAi( 轴+b)i=A1ax+b+A2( 轴+b)2AN( 轴+b)n,

where each $\begin{aligned} A_{i} \end{aligned}$ is a real number.
::每个Ai都是真实数字的地方

For each distinct factor of $\begin{aligned} g (x) \end{aligned}$ of the form $\begin{aligned} (a x^{2} + b x + c)^{n} \end{aligned}$ , with $\begin{aligned} n \geq 1 \end{aligned}$ , its contribution to the partial fraction sum is:
::对于窗体(xxxxxxxxxxxxxxxxxxxn)n 的每一种不同的因数,加上n1,其部分分数总和的分摊情况如下:

$\begin{aligned} \sum_{i = 1}^{n} \frac{A_{i} x + B_{i}}{(a x^{2} + b x + c)^{i}} = \frac{A_{1} x + B_{1}}{a x^{2} + b x + c} + \frac{A_{2} x + B_{2}}{(a x^{2} + b x + c)^{2}} + \dots + \frac{A_{n} x + B_{n}}{(a x^{2} + b x + c)^{n}}, \end{aligned}$

::i=1nAix+Bi(轴2+bx+c)i=A1x+B1ax2+bx+c+A2x+B2(轴2+bx+c)2Ax+Bn(轴2+bx+c)n,

where each $\begin{aligned} A_{i} \end{aligned}$ and $\begin{aligned} B_{i} \end{aligned}$ is a real number.
::在那里,每个爱与爱是真实的数字。

Let's find the partial fraction decomposition of $\begin{aligned} \frac{2 x - 19}{x^{2} + x - 6} \end{aligned}$ .
::让我们找到 2x-19x2+x-6 的局部分解分解。

Note that the function is a proper fraction .
::请注意,该函数是一个适当的分数。

We begin by factoring the denominator as $\begin{aligned} x^{2} + x - 6 = (x + 3) (x - 2) \end{aligned}$ .
::我们首先考虑分母为 x2+x-6=(x+3)(x-2)。

Then write the partial fraction decomposition as:
::然后将部分分解分解写为:

$\begin{aligned} \frac{2 x - 19}{x^{2} + x - 6} = \frac{2 x - 19}{(x + 3) (x - 2)} = \frac{A}{x + 3} + \frac{B}{x - 2} \end{aligned}$ .
::2x-19x2+x-6=2x-19(x+3)(x-2)=Ax+3+Bx-2。

Our goal at this point is to find the values of $\begin{aligned} A \end{aligned}$ and $\begin{aligned} B \end{aligned}$ . To do so, multiply both sides of the equation by the factored denominator $\begin{aligned} (x + 3) (x - 2) \end{aligned}$ to produce the basic equation
::我们目前的目标是找到A和B的值。为此,将方程的两侧乘以乘数分母(x+3(x-2),得出基本方程。

$\begin{aligned} 2 x - 19 = A (x - 2) + B (x + 3) \end{aligned}$
::2x-19=A(x-2)+B(x+3)

This equation is true for all values of $\begin{aligned} x \end{aligned}$ . The most convenient values are the ones that make a factor equal to zero, namely, $\begin{aligned} x = 2 \end{aligned}$ and $\begin{aligned} x = - 3 \end{aligned}$ . Substituting $\begin{aligned} x = 2 \end{aligned}$ ,
::x的所有值都是如此。最方便的值是使一个系数等于零的值,即 x=2 和 x3。替代 x=2, 替代 x=2。

$\begin{aligned} 2 (2) - 19 & = A (2 - 2) + B (2 + 3) \\ 2 (2) - 19 & = 0 + 5 B \\ - 3 & = B \end{aligned}$

::2(2)-19=A(2-2)2+B(2+3)2(2)-19=0+5B-3=B

Similarly, substituting for $\begin{aligned} x = - 3 \end{aligned}$ into the basic equation we get
::同样地,用x3来取代我们得到的基本方程

$\begin{aligned} 2 (- 3) - 19 & = A (- 3 - 2) + B (- 3 + 3) \\ - 25 & = - 5 A \\ 5 & = A \end{aligned}$

::2(-3)-19=A(-3-2)+B(-3+3)-25=5A5=A

We have found the values of $\begin{aligned} A \end{aligned}$ and $\begin{aligned} B \end{aligned}$ . Therefore, the partial fraction decomposition is $\begin{aligned} \frac{2 x - 19}{x^{2} + x - 6} = \frac{5}{x + 3} - \frac{3}{x - 2} \end{aligned}$ .
::我们发现了A和B的值。因此,部分分解为 2x-19x2+x-6=5x+3-3x-2。

Now, let's use the method of partial fractions to evaluate $\begin{aligned} \int \frac{x + 1}{(x + 2)^{2}} d x \end{aligned}$ .
::现在,让我们使用部分分数的方法来评价 x+1(x+2)2dx。

Note that the integrand is a proper fraction. According to the guide above, take the form:
::请注意,整数是一个适当的分数。根据上文的指南,采取下列形式:

$\begin{aligned} \frac{x + 1}{(x + 2)^{2}} = \frac{A}{(x + 2)} + \frac{B}{(x + 2)^{2}} \end{aligned}$ .
::x+1(x+2)2=A(x+2)+B(x+2)2。

Multiplying both sides of the equation by $\begin{aligned} (x + 2)^{2} \end{aligned}$ yields
::用(x+2)2 乘以方方程两侧的产量2

$\begin{aligned} x + 1 & = A (x + 2) + B \\ x + 1 & = A x + (2 A + B) \end{aligned}$

::x+1=A(x+2)+Bx+1=Ax+2(2A+B)

Equating the coefficients of like terms from both sides,
::将双方相同条件的系数进行等同计算,

$\begin{aligned} 1 & = A \\ 1 & = 2 A + B \end{aligned}$

::1=A1=2A+B

Thus
::因此,

$\begin{aligned} 1 & = A \\ - 1 & = B \end{aligned}$

::1=A-1=B

Therefore the partial fraction decomposition is
::因此,部分分部分分解是

$\begin{aligned} \frac{x + 1}{(x + 2)^{2}} = \frac{1}{(x + 2)} - \frac{1}{(x + 2)^{2}} \end{aligned}$ .
::x+1(x+2)2=1(x+2)-1(x+2)-1(x+2)-2。

The integral will become
::整体体将成为

$\begin{aligned} \int \frac{x + 1}{(x + 2)^{2}} d x & = \int (\frac{1}{x + 2} - \frac{1}{(x + 2)^{2}}) d x \\ = \int (\frac{1}{x + 2}) d x - \int \frac{1}{(x + 2)^{2}} d x \\ = \ln | x + 2 | + \frac{1}{x + 2} + C \dots Substitution of u = x + 2 was used in the send integral. \end{aligned}$

::X+1(x+2)2dx(1x+2)-2-1(x+2)2)dx(1x+2)dx(1x+2)x1(x+2)2dx=lnx+2x1x+2+C...在发送元件中使用了 u=x+2的替代。

Examples
::实例

Example 1
::例1

Earlier, you were asked how to find appropriate values for $\begin{aligned} A \end{aligned}$ and $\begin{aligned} B \end{aligned}$ for $\begin{aligned} \frac{7}{12} = \frac{7}{6 \cdot 2} = \frac{A}{6} + \frac{B}{2} \end{aligned}$ . This is one of the basic concepts of partial fraction approach.
::早些时候,有人问您如何找到712=762=A6+B2的A和B的适当值。这是部分分数法的基本概念之一。

The equation above requires that $\begin{aligned} 7 = 2 A + 6 B \end{aligned}$ . Because there are two variables, and no other constraining relationship, either variable can be set to a value to determine the second. If $\begin{aligned} B = 1 \end{aligned}$ , then $\begin{aligned} A = \frac{1}{2} \end{aligned}$ , so that the relationship is $\begin{aligned} \frac{7}{12} = \frac{\frac{1}{2}}{6} + \frac{1}{2} \end{aligned}$ .
::以上方程式需要 7 = 2A+6B。因为有两个变量, 没有其它制约关系, 任何变量都可以设定为一个值来确定第二个值。如果 B= 1, 那么 A= 12, 则关系为 712= 126+12 。

Example 2
::例2

Evaluate $\begin{aligned} \int \frac{3 x^{2} + 3 x + 1}{x^{3} + 2 x^{2} + x} d x \end{aligned}$ .
::3x2+3x1x3+2x2+xdx。

Note that the integrand is a proper fraction.
::请注意,整数是适当的分数。

We begin by factoring the denominator as $\begin{aligned} x (x + 1)^{2} \end{aligned}$ . Then the partial fraction decomposition is
::我们首先将分母作为x(x+1)2的乘数乘以x(x+1)2。然后部分分解是

$\begin{aligned} \frac{3 x^{2} + 3 x + 1}{x^{3} + 2 x^{2} + x} = \frac{A}{x} + \frac{B}{x + 1} + \frac{C}{(x + 1)^{2}} \end{aligned}$ .
::3x2+3x+1x3+2x2+x=Ax+Bx+1+C(x+1)2。

Multiplying each side of the equation by $\begin{aligned} x (x + 1)^{2} \end{aligned}$ , we get the basic equation
::将方程的每侧乘以 x( x+1) 2, 我们得到基本方程

$\begin{aligned} 3 x^{2} + 3 x + 1 = A (x + 1)^{2} + B x (x + 1) + C x \end{aligned}$ .
::3x2+3x+1=A(x+1)2+Bx(x+1)+Cx。

This equation is true for all values of $\begin{aligned} x \end{aligned}$ . The most convenient values are the ones that make a factor equal to zero, namely, $\begin{aligned} x = - 1 \end{aligned}$ and $\begin{aligned} x = 0 \end{aligned}$ .
::x 的所有值都是如此。最方便的值是使系数等于零的值, 即 x1 和 x=0 。

Substituting $\begin{aligned} x = - 1 \end{aligned}$ ,
::替代 x1,

$\begin{aligned} 3 (- 1)^{2} + 3 (- 1) + 1 & = A (- 1 + 1)^{2} + B (- 1) (- 1 + 1) + C (- 1) 1 \\ 1 & = 0 + 0 - C \\ - 1 & = C \end{aligned}$

:三)-1)2+3(-1)3+3(-1)+3(-1)+1=A(-1+1)2+B(-1)(-1)+1+1)+C(-1)11=0+0-C-1=C

Substituting $\begin{aligned} x = 0 \end{aligned}$ ,
::替代x=0,

$\begin{aligned} 3 (0)^{2} + 3 (0) + 1 & = A (0 + 1)^{2} + B (0) (0 + 1) + C (0) 1 \\ 1 & = A + 0 + 0 \\ 1 & = A \end{aligned}$

::3(0)2+3(0)+1=A(0+1)2+B(0)(0)+1)+C(0)11=A+0+01=A

To find $\begin{aligned} B \end{aligned}$ , we can simply substitute any value of $\begin{aligned} x \end{aligned}$ along with the values of $\begin{aligned} A \end{aligned}$ and $\begin{aligned} C \end{aligned}$ obtained.
::要找到B,我们可以简单地将x的任何价值与A和C所获得的价值相替代。

Choose $\begin{aligned} x = 1 \end{aligned}$ :
::选择 x=1 :

$\begin{aligned} 3 (1)^{2} + 3 (1) + 1 & = A (1 + 1)^{2} + B (1) (1 + 1) + C (1) 1 \\ 7 & = 4 + 2 B - 1 \\ 2 & = B \end{aligned}$

::3(1)2+3(1)+11=A(1+1)2+B(1)(1+1)+C(1)17=4+2B-12=B

Now we have solved for $\begin{aligned} A, B, \end{aligned}$ and $\begin{aligned} C \end{aligned}$ so that we can use the partial fraction decomposition to integrate.
::现在我们已经解决了A,B,C, 这样我们就可以使用部分分解集成来整合。

$\begin{aligned} \int \frac{3 x^{2} + 3 x + 1}{x^{3} + 2 x^{2} + x} d x & = \int (\frac{1}{x} + \frac{2}{x + 1} - \frac{1}{(x + 1)^{2}}) d x \\ = \ln | x | + 2 \ln | x + 1 | + \frac{1}{x + 1} + C \end{aligned}$

::3x2+3x+1x3+2x2+2x2+xdxx( 1x+2x+1- 1( x+1)+2)dx=ln2lnx+1x1+1+C

Example 3
::例3

This problem is an example of an improper rational function. Evaluate the definite integral
::这一问题是不当合理功能的一个例子。

$\begin{aligned} \int_{1}^{2} \frac{x^{3} - 4 x^{2} - 3 x + 3}{x^{2} - 3 x} d x \end{aligned}$ .
::12x3 -4x2 - 3x+3x2 - 3xxxxx

This rational function is improper because its numerator has a degree that is higher than its denominator. The first step is to divide the denominator into the numerator by long division and obtain
::此合理函数不适当, 因为它的分子数具有高于其分母的分量。第一步是将分母分解成分子数, 并获得

$\begin{aligned} \frac{x^{3} - 4 x^{2} - 3 x + 3}{x^{2} - 3 x} = (x - 1) + \frac{- 6 x + 3}{x^{2} - 3 x} \end{aligned}$ .
::x3 - 4x2 - 3x+3x2 - 3x= (x - 1) 6x+3x2 - 3x

Now apply partial function decomposition only on the remainder,
::现在只对剩余部分应用部分函数分解,

$\begin{aligned} \frac{- 6 x + 3}{x^{2} - 3 x} = \frac{- 6 x + 3}{x (x - 3)} = \frac{A}{x} + \frac{B}{x - 3} \end{aligned}$ .
::-6x+3x2 -3x_3x_6x+3x(x-3)=Ax+Bx-3。

As we did in the previous examples, multiply both sides by $\begin{aligned} x (x - 3) \end{aligned}$ to obtain the basic equation $\begin{aligned} - 6 x + 3 = A (x - 3) + B x \end{aligned}$ .
::正如我们在前几个例子中所做的那样,将两边乘以x(x-3),以获得基本等式-6x+3=A(x-3)+Bx。

For $\begin{aligned} x = 0 \end{aligned}$ ,
::x=0 时

$\begin{aligned} 3 & = - 3 A + 0 \\ - 1 & = A \end{aligned}$

::33A+0-1=A

For $\begin{aligned} x = 3 \end{aligned}$ ,
::x=3,对x=3,对x=3,对x=3,对x=3,对x=3,对x=3,对x=3,对x=3,对x=3

$\begin{aligned} - 18 + 3 & = 0 + 3 B \\ - 15 & = 3 B \\ - 5 & = B \end{aligned}$

::- 18+3=0+3B-15=3B-5=B

Thus the integrand becomes
::这样一来,原的变成

$\begin{aligned} \frac{x^{3} - 4 x^{2} - 3 x + 3}{x^{2} - 3 x} = (x - 1) + \frac{- 6 x + 3}{x^{2} - 3 x} = x - 1 - \frac{1}{x} - \frac{5}{x - 3} \end{aligned}$ ,
::x3 - 4x2 - 3x+3x2 - 3x=(x-1)\\\\6x+3x2 - 3x=x-1 - 1x-5x-3)

And the integral can be evaluated as
::整体体可以被评估为

$\begin{aligned} \int_{1}^{2} \frac{x^{3} - 4 x^{2} - 3 x + 3}{x^{2} - 3 x} d x & = \int_{1}^{2} [x - 1 - \frac{1}{x} - \frac{5}{x - 3}] d x \\ = {[\frac{x^{2}}{2} - x - \ln | x | - 5 \ln | x - 3 |]}_{1}^{2} \\ = (\frac{4}{2} - 2 - \ln 2 - 5 \ln 1) - (\frac{1}{2} - 1 - \ln 1 - 5 \ln 2) \\ = 4 \ln 2 + \frac{1}{2} \end{aligned}$

::12x3-4x2-3x2-3x2-3x2-3xxxxx*12[x-1-1x-5x-3]dx=[x22-x-x-lnx 5xx-3]12=(42-2-ln___2-5ln___1)-(12-1-ln_1-5ln_2)=4ln_2+12

Review
::回顾

Evaluate the following integrals using partial fractions.
::使用部分分数评价以下整体部分。

$\begin{aligned} \int \frac{1}{x^{2} - 1} d x \end{aligned}$
::1x2 - 1dx

$\begin{aligned} \int \frac{x}{x^{2} - 2 x - 3} d x \end{aligned}$
::*x22-2-2x-3dx

$\begin{aligned} \int \frac{1}{x^{3} + x^{2} - 2 x} d x \end{aligned}$
::1x3+x2-2xxxx

$\begin{aligned} \int \frac{x^{3}}{x^{2} + 4} d x \end{aligned}$
::x3x2+4dx

$\begin{aligned} \int_{0}^{1} \frac{ϕ}{1 + ϕ} d ϕ \end{aligned}$
::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}我... {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}我... {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}

$\begin{aligned} \int_{1}^{5} \frac{x - 1}{x^{2} (x + 1)} d x \end{aligned}$
::15x- 1x2( x+1dx)

$\begin{aligned} \int \frac{10 x^{2} + 21 x - 9}{(x^{2} - 7 x + 12) (x + 5)} d x \end{aligned}$
::10x2+21x-9(x2-7x+12)(x+5)dx

$\begin{aligned} \int (\frac{3 e^{x}}{(e^{x} - 1) (e^{x} + 3)}) d x \end{aligned}$ Hint: Use $\begin{aligned} u \end{aligned}$ -substitution to convert to a rational function.
:3ex(ex-1)(ex+3)) dx hint: 使用 u- 替代来转换为合理函数。

$\begin{aligned} \int \frac{\cos θ}{\sin^{2} θ + 4 \sin θ - 5} d θ \end{aligned}$ Hint: Use $\begin{aligned} u \end{aligned}$ -substitution to convert to a rational function.
::=============================================================================================================================================== ============================================================================================== ==============================================================================================================================================================================================================================================================================

$\begin{aligned} \int \frac{3 e^{θ}}{e^{2 θ} - 1} d θ \end{aligned}$ Hint: Use $\begin{aligned} u \end{aligned}$ -substitution to convert to a rational function.
::======================================================================================================================= ============================================================================================================== ======================================================================================================================================================================================================================================================================================

Find the area under the curve $\begin{aligned} y = 1 \end{aligned}$ over the interval $\begin{aligned} [- \ln 3, \ln 4] \end{aligned}$ ( Hint : make a $\begin{aligned} u \end{aligned}$ -substitution to convert the integrand into a rational function.)
::在曲线 y= 1 的间隔处查找区域 [- ln3, ln4] (提示: 以 u 替代将整数转换为合理函数。)

Show that $\begin{aligned} \int \frac{1}{a^{2} - x^{2}} d x = \frac{1}{2 a} \ln | \frac{a + x}{a - x} | + C \end{aligned}$ .
::显示% 1a2- x2dx= 12alna+xa-xxQC。

Evaluate $\begin{aligned} \int \frac{2 x^{3} + 3}{(x^{2} - 4) (x^{2} + 3 x + 2)} d x \end{aligned}$
::评价2x3+3(x2-4)(x2+3x+2)dx

Evaluate $\begin{aligned} \int \frac{x^{2} + 6 x + 1}{3 x^{2} + 5 x - 2} d x \end{aligned}$
::====================================================================================================================================================================

Evaluate $\begin{aligned} \int \frac{d x}{\sinh x} d x \end{aligned}$ ; Hint: Let $\begin{aligned} u = e^{- x} \end{aligned}$ and use partial fractions.
::评估 dxsinh xdx; 提示 : Let u = e-x 并使用部分分数。

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::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键。

Section outline

General

按部分分数分列的整合情况

Integration by Partial Fractions ::按部分分数分列的整合情况

Forms of partial fractions composing f ( x ) g ( x ) ::F(x)g(x)

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Review ::回顾

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Integration by Partial Fractions
::按部分分数分列的整合情况

Forms of partial fractions composing $\begin{aligned} \frac{f (x)}{g (x)} \end{aligned}$
::F(x)g(x)

Examples
::实例

Example 1
::例1

Example 2
::例2

Example 3
::例3

Review
::回顾

Review (Answers)
::回顾(答复)