8.1 按部分整合:一次性解决

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  按部分合并:一次性解决

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  In the integration by parts technique that is presented in this concept, the integrand $\begin{align*}F(x)\end{align*}$  of the integral $\begin{align*}\int F(x)dx\end{align*}$  is broken up into two factors such that $\begin{align*}F(x)dx=f(x)g^\prime (x)dx\end{align*}$ , where $\begin{align*}f(x)\end{align*}$  and $\begin{align*}g^\prime (x)\end{align*}$  are differentiable (integrable) functions. A big part of using the technique is recognizing how to define $\begin{align*}f(x)\end{align*}$  and  $\begin{align*}g^\prime (x)\end{align*}$ so that it is easy to determine $\begin{align*}f^\prime (x)\end{align*}$  and $\begin{align*}g(x)\end{align*}$  that make the integral solution easy to obtain. To see how this works, try choosing as many $\begin{align*}f(x)\end{align*}$  and  $\begin{align*}g^\prime (x)\end{align*}$ combinations as you can for the following two examples of $\begin{align*}F(x)\end{align*}$ , then determine the corresponding $\begin{align*}f(x)\end{align*}$  and  $\begin{align*}g^\prime (x)\end{align*}$ for each combination:
  ::在按部分技术集成的概念中,集成 F(x)dx的整数F(x)分解成两个因素,例如F(x)dx=f(x)g(x)x(x)dx,其中f(x)和g(x)是可区别(不可分)的功能。使用该技术的很大一部分是确认如何定义 f(x)和g(x),以便易于确定使整体解决方案易于获得的f*(x)和g(x)。要了解如何使用,尽量选择F(x)和g(x)组合的多个f(x)和g*(x)组合,然后确定每种组合相应的f(x)和g*(x):

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  1. $\begin{align*}F(x)=x \sin x\end{align*}$
     ::F(x) =xsinx
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  2. $\begin{align*}F(x)=x^2 \sin x^2\end{align*}$
     ::F(x) =x2辛x2

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  Introduction to Integration by Parts
  ::按部分分列的合并介绍

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  Integration by parts is a technique to simplify the evaluation of an integral $\begin{align*}\int F(x)dx=\int f(x)g^\prime (x) dx\end{align*}$ . In this case, the function $\begin{align*}F(x)\end{align*}$  can be written as the product of two functions $\begin{align*}f(x)\end{align*}$ , and $\begin{align*}g^\prime (x)\end{align*}$ . This allows the orginal integral to be written as $\begin{align*}\int F(x) dx=\int f(x)g^\prime (x) dx=\int udv\end{align*}$ , where via substitution $\begin{align*}u=f(x)\end{align*}$ , and $\begin{align*}dv=g^\prime (x) dx\end{align*}$ .
  ::按部分整合是一种简化对 {F(x)dx}}}(x)g*(x)dxxxxx。 在这种情况下,函数 F(x) 可以写成为两个函数 f(x) 和 g}(x) 的产物。 这使得圆形内含可以写成 {F(x)dx}*f(x)g}(x)dx}}*udv, 通过替换 u=f(x) 和 dv=g*(x)dx。

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  This transformation is important because, based on the product rule of differentiation that you have already studied, we can solve for $\begin{align*}\int udv\end{align*}$  as follows:
  ::这种转变之所以重要,是因为根据你们已经研究过的差别化产品规则,我们可以解决以下udv问题:

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  $\begin{align*}\frac{d}{dx}[uv]=u \frac{dv}{dx}+v\frac{du}{dx}\end{align*}$ .
  ::ddx[前 =udvdx+vdudx。

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  If we integrate each side,
  ::如果我们把双方融为一体

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  $$\begin{align*}uv &= \int u \frac{dv}{dx}dx+\int v \frac{du}{dx}dx\\ &= \int udv+ \int vdu\end{align*}$$
  ::uvudvdxdxvdudxdxudvvdu

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  Solving for  $\begin{align*}\int udv\end{align*}$ yields
  ::解决 udv 产量

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  $\begin{align*}\int udv=uv-\int v du\end{align*}$ .
  ::.

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  The above equation is the formula for integration by parts. We know $\begin{align*}u=f(x)\end{align*}$ , and if $\begin{align*}\int dv=v= \int g^\prime (x) dx\end{align*}$ is easy enough to determine, then the integral  $\begin{align*}\int v du\end{align*}$ may be easier to integrate. With the proper choice of  $\begin{align*}u\end{align*}$  and $\begin{align*}dv\end{align*}$ ,  use of the formula for integration by parts can transform the original integral into some.
  ::以上方程是按部件整合的公式。 我们知道 u=f( x), 如果 =dv=vg* (x)dx 足够容易确定, 那么集成的 {vdu 可能比较容易整合。 如果正确选择 u 和 dv, 使用按部件整合的公式可以将原来的集成转换为部分 。

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  The process for integrating by parts is as follows:
  ::按部分分类的整合过程如下:

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  • Evaluate the “difficult” integral and split it into parts $\begin{align*}\int F(x)dx=\int f(x)g^\prime (x) dx\end{align*}$ .
    ::评估“困难”的内在部分,并将其分成部分F(x)dx*f(x)g*(x)dx。
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  • Transform $\begin{align*}\int f(x) g^\prime (x) dx\end{align*}$  to the form  $\begin{align*}\int udv=uv-\int vdu\end{align*}$ with the right choice for  $\begin{align*}u=f(x)\end{align*}$ and $\begin{align*}dv=g^\prime (x) dx\end{align*}$  that makes  $\begin{align*}\int vdu\end{align*}$ easier to evaluate than $\begin{align*}\int udv\end{align*}$ .
    ::f(x)g(x)dx 至窗体 udv=uvvdu, 且对 u=f(x) 和 dv=g=g}(x)dx 进行正确选择, 使得 vdu 的评审比 udv 更容易 。

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  To make solving the integral easier, 
  ::为了更容易解决整体问题

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  • Choose  $\begin{align*}dv\end{align*}$  to be the more complicated portion of the integrand that fits a basic integration formula. Choose  $\begin{align*}u\end{align*}$  to be the remaining term in the integrand.
    ::选择 dv 是符合基本集成公式的整数中较复杂的部分。 选择 u 是整数中的剩余词 。
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  • Choose  $\begin{align*}u\end{align*}$  to be the portion of the integrand whose derivative is simpler than $\begin{align*}u\end{align*}$ . Choose  $\begin{align*}dv\end{align*}$  to be the remaining term.
    ::选择 u 成为 integrand 的一部分, 其衍生值比 u 简单。 选择 dv 作为剩余任期 。

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  Remember, the goal of the integration by parts is to start with an integral in the form  $\begin{align*}\int udv\end{align*}$  that is hard to integrate directly and change it to an integral  $\begin{align*}\int vdu\end{align*}$  that looks easier to evaluate. In the best case, the integral  $\begin{align*}\int vdu\end{align*}$ is evaluated, so that the solution to the original integral is completed in one pass using integration by parts.
  ::记住,各部分集成的目的是从一个以 udv 形式组成的整体开始,它很难直接整合,并改变为一个看起来更容易评估的集成 vdu 。 在最好的情况下,对集成 vdu 进行了评估,这样,原始集成的解决方案就可以通过一个分集完成。

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  Let's apply the information from above and evaluate $\begin{align*}\int x \sin xdx\end{align*}$ .
  ::让我们应用上面的信息 来评估xsinxdx。

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  Use integration by parts: $\begin{align*}\int udv=uv-\int vdu\end{align*}$ .
  ::按部件分列的使用情况: udv=uvvdu。

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  $$\begin{align*}& \text{Choose:} && u=x \quad \ dv=\sin xdx\\ & \text{Form:} && du=dx \quad v=-\cos x\\ & \text{Evaluate:} && \int x \sin x dx=-x \cos x-\int (-\cos x)dx \quad \ldots \text{The new integral is simpler! CONTINUE!}\\ & && \qquad \qquad \quad \ =-x \cos x+\sin x+C \end{align*}$$
  ::选择 : u= x dv=sinxdxForm: du=dxvcosxEvaluate:\\\ xinxxxxxxxxxxxx}}{( - cosx) dxx}}... 新集成更简单 ! CONTINUE! @ xcosx+sinx+C

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  The above selections for $\begin{align*}u\end{align*}$  and $\begin{align*}dv\end{align*}$  lead to a successful evaluation of the original integral in one pass. But what if we had made a different choice as follows:
  ::u和dv的上述选择导致成功评价原始整体部分。 但如果我们作出以下不同选择:

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  $$\begin{align*}& \text{Choose:} && u=\sin x \qquad dv=xdx \\ & \text{Form:} && du=\cos xdx \quad v=\frac{x^2}{2} \\ & \text{Evaluate:} && \int x \sin x dx = ( \sin x ) \left ( \frac{x^2}{2} \right ) - \int \frac{x^2}{2} \cos x dx \quad \ldots \text{The new integral is more complicated! STOP!}\end{align*}$$
  ::选择 : u=sinxdv=xdxForm: du=cosxdxv=x22 valuate:\\\ xsinxdx=( sinx)(x22)\\ x22)\\ x22cosxxx... 新的集成更复杂! 停止 !

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  As you can see, the new integral is worse than what we started with! This tells us that we have made the wrong choice and we must change (in this case switch) our choices of  $\begin{align*}u\end{align*}$ and $\begin{align*}dv\end{align*}$ .
  ::如你所见, 新的整体体比我们开始的要糟糕! 这说明我们做了错误的选择, 我们必须改变(在此情况下, 换换)我们的选择 u 和 dv 。

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  A more specific guide for making a choice for $\begin{align*}u\end{align*}$ , called the LIATE rule, is shown below. While not 100% accurate, it does provide a guide.
  ::下面是更具体地选择u的指南,称为 " 期限规则 " 。它虽然不完全准确,但提供了指南。

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  LIATE Rule*: Guide for Selecting $\begin{align*}u\end{align*}$
  ::时效规则 *:选择u的指南

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  Whichever function comes first in the following list should be $\begin{align*}u\end{align*}$ :
  ::下列列表中无论哪个函数首当其冲,该函数应为 u:

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  $\begin{align*}L =\end{align*}$  Logarithmic (e.g., $\begin{align*}\ln x, \log_b x\end{align*}$ )
  ::L=对数(例如, Inx,logbx)

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  $\begin{align*}I =\end{align*}$  Inverse trigonometric (e.g., $\begin{align*}\arctan x, \text{arcsec } x\end{align*}$ )
  ::I=反三角测量(例如弧坦克斯、弧塞克x)

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  $\begin{align*}A =\end{align*}$  Algebraic (e.g.,  $\begin{align*}x^2, 3x^{50}\end{align*}$ )
  ::A=代数(例如x2,3x50)

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  $\begin{align*}T =\end{align*}$  Trigonometric (e.g., $\begin{align*}\sin x, \tan x\end{align*}$ )
  ::T=三角测量(例如,罪恶、唐氏)

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  $\begin{align*}E =\end{align*}$  Exponential (e.g., $\begin{align*}e^x , 19^x\end{align*}$ )
  ::E=指数(例如,ex,19x)

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  The function which is to be $\begin{align*}dv\end{align*}$  comes later in the list, that is, functions lower on the list have easier anti- than the functions above them.
  ::dv 的函数在列表中后面产生,也就是说,列表中下方的函数比上方的函数更容易反弹。

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  * Kasube, Herbert E. (1983). “A Technique for Integration by Parts”. The American Mathematical Monthly 90 (3): 210–211.
  ::* Kasube, Herbert E.(1983年), " 部分融合技术 " ,《美国数学月刊》90(3):210-211。

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  Examples
  ::实例

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  Example 1
  ::例1

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  Earlier, you were asked to try choosing as many $\begin{align*}f(x)\end{align*}$  and $\begin{align*}g^\prime (x)\end{align*}$ factor combinations as you can for $\begin{align*}F(x)=x \sin x\end{align*}$  and  $\begin{align*}F(x)=x^2 \sin x^2\end{align*}$ . Then you were asked to determine the corresponding $\begin{align*}f(x)\end{align*}$  and $\begin{align*}g^\prime (x)\end{align*}$ :
  ::早些时候,您被要求尽量选择 F (x) 和 g 和 g 和 g 的因子组合, F (x) =xsinx 和 F (x) =x2sinx2. 然后您被要求确定相应的 f (x) 和 g =(x) :

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  $\begin{align*}F(x)=x \sin x \Rightarrow\end{align*}$   
  ::F(x) =xsinx

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  1. $\begin{align*}f(x)=x, g^\prime (x)=\sin x; f^\prime (x)=1, g(x)=-\cos x\end{align*}$
     :x) =x, g}(x) = sinx; f}(x) = 1, g(x) cosx
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  2. $\begin{align*}f(x)=\sin x, g^\prime (x)=x; f^\prime (x)=\cos x, g(x)=\frac{x^2}{2}\end{align*}$
     ::f( x) =sinx, g}( x) =x; f}( x) =cosx, g( x) =x22

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  Of the 2 options, the option (a) selection of $\begin{align*}f(x)\end{align*}$  and $\begin{align*}g^\prime(x)\end{align*}$  results in the simplest $\begin{align*}f(x)\end{align*}$  and  $\begin{align*}g^\prime (x)\end{align*}$ combination. This is what we try to achieve!
  ::在2个选项中,选项 (a) 从(x) 和 g(x) 中选择, 结果是简单stf(x) 和 g(x) 组合。 这就是我们试图实现的目标 !

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  $\begin{align*}F(x)=x^2 \sin x^2 \Rightarrow\end{align*}$
  ::F(x) =x2sinx2

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  1. $\begin{align*}f(x)=x^2, g^{\prime} (x)=\sin x^2 \Rightarrow f^{\prime} (x)=2x, g(x)=-\cos x^2\end{align*}$  
     ::f_(x)=2x2,g_(x)=sinx2\\\\ f_(x)=2x,g(x)\cosx2
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  2. $\begin{align*}f(x)=\sin x^2, g^\prime (x)=x^2 \Rightarrow f^\prime (x)=2x \cos x^2, g(x)=\frac{x^3}{3}\end{align*}$
     ::f( x) =sinx2, g_( x) =x2, g_( x) =x2, f_( x) = 2xcosx2, g( x) =x33
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  3. $\begin{align*}f(x)=x, g^\prime (x)=x \sin x^2 \Rightarrow f^\prime (x)=1, g(x)=-\frac{\cos x^2}{2}\end{align*}$  
     ::f_( x) = 1, g( x) =x, g_( x) =xsinxx2 @ f_( x) =1, g( x) cosx22
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  4. $\begin{align*}f(x)=x \sin x^2, g^\prime (x)=x \Rightarrow f^\prime (x)=2x^2 \cos x^2, g(x)=\frac{x^2}{2}\end{align*}$
     ::f( x) =xsinx2, g}( x) =xx_xx2, g( x) =x_f}( x) =2x2cosx2, g( x) =x22

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  Of the 4 options, the option (c) selection of $\begin{align*}f(x)\end{align*}$  and $\begin{align*}g^\prime (x)\end{align*}$  results in the simplest  $\begin{align*}f(x)\end{align*}$ and  $\begin{align*}g^\prime (x)\end{align*}$ combination. This is what we try to achieve!
  ::在4个选项中,f(x)和g(x)的选项(c)选择F(x)和g(x)的结果是简单f(x)和g(x)的组合。这就是我们试图实现的目标!

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  Example 2
  ::例2

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  Evaluate $\begin{align*}\int xe^x dx\end{align*}$ .
  ::评估 xexdx。

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  Again, we use the Integration by Parts formula: $\begin{align*}\int udv=uv-\int vdu\end{align*}$ .
  ::我们再次使用“按部分组合”的公式:udv=uvvdu。

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  $$\begin{align*}& \text{Choose:} && u=x \quad \ dv=e^xdx\\ & \text{Form:} && du=dx \quad v=e^x\\ & \text{Evaluate:} && \int xe^x dx=xe^x-\int e^x dx \quad \ldots \text{The new integral is simpler! CONTINUE!}\\ & && \qquad \qquad =xe^x-e^x+C\\ & && \qquad \qquad =(x-1)e^x+C \end{align*}$$
  ::选择 : u= x dv=exdxForm: du=dxv=exEvaluate:\\\ xexdx=xex=xex*exdx... 新组合更简单! CONTINUE!=xex-ex+C=(x- 1)ex+C

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  The above selections for $\begin{align*}u\end{align*}$  and $\begin{align*}dv\end{align*}$  lead to a successful evaluation of the original integral in one pass.  
  ::u和dv的上述选择导致在一张通行证中成功评价原组成部分。

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  Again, what if we had made a different choice as follows:
  ::倘若我们作出以下不同的选择:

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  $$\begin{align*}& \text{Choose:} && u=e^x \qquad dv=xdx\\ & \text{Form:} && du=e^xdx \quad v=\frac{x^2}{2}\\ & \text{Evaluate:} && \int x e^x dx=(e^x)\left(\frac{x^2}{2}\right)-\int \frac{x^2}{2}e^x dx \quad \ldots \text{The new integral is more complicated! STOP!}\end{align*}$$
  ::选择 : u= exdv= xdxForum: du= exdxv=x22 valuate:\\\ xexdx=( ex( x)( x22)\\ x22exdx)... 新的集成更复杂! 停止 !

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  As you can see, the new integral is worse than what we started with! This tells us that we have made the wrong choice and we must change (in this case switch) our choices of  $\begin{align*}u\end{align*}$ and $\begin{align*}dv\end{align*}$ .
  ::如你所见, 新的整体体比我们开始的要糟糕! 这说明我们做了错误的选择, 我们必须改变(在此情况下, 换换)我们的选择 u 和 dv 。

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  Example 3
  ::例3

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  Evaluate $\begin{align*}\int \ln x dx\end{align*}$ .
  ::评估 lnxxx。

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  Here, it appears we only have one term in the integrand, $\begin{align*}\ln x\end{align*}$ , but we can always assume that this term is multiplied by 1, i.e., we have $\begin{align*}\int \ln x \cdot 1 dx\end{align*}$ :
  ::这里,我们似乎只有一个术语 在前格朗, Inx, 但我们总是可以假设这个术语乘以 1, 也就是说,我们有 lnxx1dx:

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  $$\begin{align*}& \text{Choose:} && u=\ln x \quad \ dv=1 dx\\ & \text{Form:} && du=\frac{1}{x}dx \quad v=x\\ & \text{Evaluate:} && \int \ln x dx=\ln x \cdot x-\int x \frac{1}{x}dx \quad \ldots \text{The new integral is simpler! CONTINUE!}\\ & && \qquad \qquad =x \ln x-x+C\\ & && \qquad \qquad =x(\ln x-1)+C \end{align*}$$
  ::选择 : u= lnx dv= 1dxForm: du= 1xdxv= x valuate:\\ lnxdx= lnxxxxxxx\\\ x1xxxx... 新内装件更简单! CONTINUE!=xlnx-x+C=x( lnx- 1)+C

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  The above selections for $\begin{align*}u\end{align*}$  and $\begin{align*}dv\end{align*}$  lead to a successful evaluation of the original integral in one pass.
  ::u和dv的上述选择导致在一张通行证中成功评价原组成部分。

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  Example 4
  ::例4

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  Evaluate  $\begin{align*}\int x(x+7)^{\frac{3}{2}}dx\end{align*}$ .
  ::评价 x(x+7)32dx。

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  We use the Integration by Parts formula: $\begin{align*}\int udv=uv-\int vdu\end{align*}$ .
  ::我们使用部件组合公式:udv=uvvdu。

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  $$\begin{align*}& \text{Choose:} && u=x \quad \ dv=(x+7)^{\frac{3}{2}}dx\\ & \text{Form:} && du=dx \quad v=\frac{2}{5}(x+7)^{\frac{5}{2}}\\ & \text{Evaluate:} && \int x(x+7)^{\frac{3}{2}}dx=x \frac{2}{5}(x+7)^{\frac{5}{2}}-\int \frac{2}{5} (x+7)^{\frac{5}{2}} dx \quad \ldots \text{The new integral is simpler! CONTINUE!}\\ & && \qquad \qquad \qquad \ \ =x \frac{2}{5} (x+7)^{\frac{5}{2}}-\frac{2}{5} \frac{2}{7} (x+7)^{\frac{7}{2}}+C\\ & && \qquad \qquad \qquad \ \ =\frac{2}{5}x(x+7)^{\frac{5}{2}}-\frac{4}{35}(x+7)^{\frac{7}{2}}+C \end{align*}$$
  ::选择 : u=x dv= (x+7) 32dxForm: du=dxv=25(x+7) 52Evaluate:\\x(x+7) 32dx=x25(x+7) 5225(x+7) 52(x+7) 52dx... 新元件更简单! CONTINUE! =x25(x+7) 52- 2527(x7) 72+C=25x(x+7) 52- 435(x+7)72+C

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  The above selections for $\begin{align*}u\end{align*}$  and $\begin{align*}dv\end{align*}$  lead to a successful evaluation of the original integral in one pass.
  ::u和dv的上述选择导致在一张通行证中成功评价原组成部分。

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  Review
  ::回顾

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  Evaluate the following integrals using integration by parts. ( Remark:  In some, you may need to use  $\begin{align*}u\end{align*}$ -substitution along with integration by parts.)
  :备注:在有些情况下,您可能需要使用替代,同时使用部分集成。 )

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  1. $\begin{align*}\int 3 xe^x dx\end{align*}$
     ::3xexdx
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  2. $\begin{align*}\int (3-2x)e^{4x}dx\end{align*}$
     :3-2x) e4xxx
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  3. $\begin{align*}\int \ln (3x+2)dx\end{align*}$
     ::00( 3x+2) dx
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  4. $\begin{align*}\int 4x \cos 5x dx\end{align*}$
     ::4xcos5xxx
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  5. $\begin{align*}\int \theta \sin \theta d \theta\end{align*}$
     ::{\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}
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  6. $\begin{align*}\int 2x \ln (3x) dx\end{align*}$
     ::2xln( 3x) dx
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  7. $\begin{align*}\int x^4 \ln x dx\end{align*}$
     ::
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  8. $\begin{align*}\int x \sqrt{5x-2}dx\end{align*}$
     ::5x-2dx
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  9. $\begin{align*}\int\limits^e_1 \frac{\ln x}{x^3}dx\end{align*}$
     ::e1lnx23dx
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  10. $\begin{align*}\int \frac{x^3}{(x^2+2)^2}dx\end{align*}$
      ::x3 (x2+2) 2dx
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  11. $\begin{align*}\int\limits^3_1 \ln (x+1)dx\end{align*}$
      ::31ln( x+1) dx
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  12. $\begin{align*}\int \sin^{-1} \theta d \theta\end{align*}$
      ::-1d
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  13. $\begin{align*}\int \tan^{-1}xdx\end{align*}$
      ::- 1xdx
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  14. $\begin{align*}\int x^5 \cos (x^3)\end{align*}$
      ::x5cos( x3)
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  15. $\begin{align*}\int 3(x+2) \cos x \sin x dx\end{align*}$
      ::3 (x+2) oxsinxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

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