课程： 8.2 按部分合并:多通过

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按部分集成: 多个通道

In the previous concept, you learned that a “one-pass” application of the integration by parts technique could be adequate to solve an integral. For many integrals, more than one pass may be required before and integral can be evaluated. How many passes are required to evaluate the integral $\begin{align*}\int x^2 \sin 5xdx\end{align*}$ ?
::在前一个概念中,你了解到,“单程”应用部件集成技术可能足以解决一个整体问题。对于许多整体体而言,需要不止一张通行证才能评价一个整体问题。需要多少张通行证才能评价一个整体的x2sin5xdx?

Integration by Parts: Multiple Passes
::按部分集成: 多个通道

In the previous concept, the technique of Integration by parts was used to transform a “difficult” integral to an “easier” integral to evaluate by using the procedure below:
::在前一个概念中,采用按部分分类的整合技术,通过采用以下程序,改变“容易”的“困难”,而“困难”是评估“容易”的“障碍”的组成部分:

Evaluate the “difficult” integral $\begin{align*}\int F(x)dx = \int f(x)g^\prime(x) dx\end{align*}$ .
::评估“困难”的“困难”组成部分 @F(x)dxf(x)g_(x)dx。

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。

Guide:
::指南:

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。

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。

In the examples we looked at, one application of the integration by parts technique (sometimes in conjunction with $\begin{align*}u\end{align*}$ -substitution) was sufficient to find an analytical solution.
::在我们所研究的例子中,按部件技术整合的一个应用(有时结合替代技术)足以找到分析解决办法。

For example, the integral $\begin{align*}\int x\sin xdx\end{align*}$ was evaluated in one pass as follows:
::例如,在一张通行证中,对有机的 xsinxdx 进行了以下评价:

$\begin{align*}& \text{Use:} && \int udv = uv -\int vdu .\\ & \text{Choose:} && u=x \qquad \ dv= \sin xdx\\ & \text{Form:} && du=dx \qquad v=-\cos x \\ & \text{Evaluate:} && \int x\sin xdx = -x\cos x-\int (-\cos x) dx \quad \ldots \text{The new integral is simpler! CONTINUE!}\\ & && \qquad \qquad \quad \ =- x \cos x + \sin x +C \qquad \quad \ldots \text{One pass}\end{align*}$
::使用 : @ udv=uv@vddu. 选择: u=x dv=sinxdxForm: du=dxvcosxEvaluate:\\ xinxxxxxxxxxx(- cosx)dx... 新的组件更简单 ! @ xcosx+sinx+C... 1 pass

Some integrals, however, require more than one application (pass) of the integration by parts technique.
::然而,有些部件需要一种以上的部件技术集成应用(通行证)。

Take the integral $\begin{align*}\int (\ln x)^2 dx\end{align*}$ .
::取下积分 {( lnx) 2dx 。 <button class="play-button btn btn-success" style="float: right;" value="@s"> 播放段落 </button> Here, it appears we only have one term in the integrand, ( ln x ) 2 <script id="MathJax-Element-18" type="math/tex"> \begin{align*}(\ln x)^2\end{align*} , but we can always assume that this term is multiplied by 1, i.e., we have $\begin{align*}\int (\ln x)^2 \cdot 1dx\end{align*}$ :
::这里,我们似乎在整数中只有一个术语, (xx) 2, 但我们总是可以假设这个术语乘以 1, 也就是说, 我们有 (xx) 21dx :

$\begin{align*}& \text{Use:} && \int udv = uv - \int vdu. \qquad \qquad \qquad \quad \ldots \text{Pass 1}\\ & \text{Choose:} && u = (\ln x)^2 \qquad \ dv = 1dx\\ & \text{Form:} && du = 2 \frac{1}{x}\ln xdx \quad v=x \\ & \text{Evaluate:} && \int (\ln x)^2 dx =x(\ln x)^2 -\int 2x \frac{1}{x} \ln xdx \\ & && \qquad \qquad \quad =x(\ln x)^2 - 2 \int \ln xdx \quad \ldots \text{The new integral is simpler, but requires}\\ & && \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \text{more work! Try another iteration.}\end{align*}$
::使用: @ udv=uvvdu.... Pass 1Choose: u=( lnx) 2 dv=1dxForm: du= 21xlnxdxv=xEvaluate: @ (lnx) 2xx=x( lnx) 2}2x1xlnxdx=x( lnx) 2_\\\\\ lnxdx... 新的集成更简单, 但需要更多工作。

$\begin{align*}& \text{Use:} && \int udv= uv - \int vdu. \qquad \qquad \qquad \qquad \qquad \ldots \text{Pass 2}\\ & \text{Choose:} && u= \ln x \quad \ dv = 1dx\\ & \text{Form:} && du = \frac{1}{x} dx \quad v=x\\ & \text{Evaluate:} && \int (\ln x)^2 dx= x(\ln x)^2 -2 [x \ln x-\int\frac{1}{x} xdx] \\ & && \qquad \qquad \quad \ =x(\ln x)^2 -2 [x\ln x - \int dx] \quad \ldots \text{The new integral is simpler and can be evaluated.}\\ & && \qquad \qquad \quad \ =x(\ln x)^2 - 2[x\ln x-x]+C\\ & && \ \int (\ln x)^2 dx = x [(\ln x)^2 -2 \ln x-2]+C\\\end{align*}$
::使用 :\\ udv=uvvduvdu.... Pass 2Choose: u=lnx dv=1dxForm: du=1xdxv=xvaluate:\\( xx)2- 2[xlnx=1xxxxxxxx] =x(xxx)2-2[xlnxxxx=xxxxxxx]-2[xxxxxxxxxxxxx] 。新的整体部分更简单, 并且可以评估。 =x(xxx)2- 2xx-xx2]+C。 =x(xx)2- 2nx-x2]+C

The evaluation of $\begin{align*}\int (\ln x)^2 dx\end{align*}$ took 2 passes before obtaining a successful evaluation.
:lnx)2dx的评价在获得成功评价之前,取了2张通行证。

Now, evaluate $\begin{align*}\int x^2 e^x dx\end{align*}$ .
::现在,评价 x2exdx。

$\begin{align*}& \text{Use:} && \int udv = uv -\int vdu. \qquad \qquad \ \ \ldots \text{Pass 1}\\ & \text{Choose:} && u=x^2 \qquad dv=e^x dx\\ & \text{Form:} && du = 2 xdx \quad v=e^x\\ & \text{Evaluate:} && \int x^2 e^x dx = x^2 e^x - \int 2e^x xdx \quad \ldots \text{The new integral is simpler, but}\\ & && \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \text{requires more work! Try another iteration.}\end{align*}$
::使用 : @ udv=uvvdu.... Pass 1Choose: u=x2dv=exdxForm:du=2xdxv=exEvaluate:\\x2exdx=x2ex=x2ex*2exdxx... 新的集成更简单, 但需要更多的工作! 尝试另一种循环。

As you can see, the integral $\begin{align*}\int 2e^x xdx\end{align*}$ is less complicated than the original $\begin{align*}\int x^2 e^x dx\end{align*}$ .
::如你所见,积分 2exxdx 不如 x2exdx 原版复杂。

This tells us that we have made the right choice. However, to evaluate $\begin{align*}\int 2e^x xdx\end{align*}$ we again perform integration by parts.
::这告诉我们,我们作出了正确的选择,然而,为了评估2exxdx,我们再次按部分进行整合。

$\begin{align*}& \text{Use:} && \int udv = uv - \int vdu. \qquad \qquad \ \ldots \text{Pass 2}\\ & \text{Choose:} && u=2x \quad \ \ dv = e^x dx \\ & \text{Form:} && du = 2dx \quad v=e^x\\ & \text{Evaluate:} && \int x^2 e^x dx = x^2 e^x - \int 2e^x xdx \quad \ldots \text{The new integral is simpler, and can be evaluated.}\\ & && \qquad \qquad \ = x^2 e^x - [2x e^x - \int 2e^x dx]\\ & && \qquad \qquad \ = x^2 e^x -[2xe^x - 2e^x]+C \\ & && \int x^2 e^x dx = [x^2 - 2x +2]e^x +C\end{align*}$
::使用 : @ udv=uvvdu.... Pass 2Choose: u=2x dv=exdxForm: du=2dxv=exEvaluate:\\ x2exdx=x2ex=x2ex=2ex=2exxxxxxxxxxxxxxxxxxxxxxxxxxx... 新的内装件更简单,可以评估。 =x2ex- [2xex=2ex2exx2xxxxxxxxxxxxxxxxxxxxxxxx2xxxxxxxxxxxxxxxxxxxxxxxxx2xxxxxxxxxxxx+C

The integration by parts method that we have just used to evaluate $\begin{align*}\int x^2 e^x dx\end{align*}$ works for any integral that has the form $\begin{align*}\int x^n e^x dx\end{align*}$ , where $\begin{align*}n\end{align*}$ is a positive integer $\begin{align*}\ge 2\end{align*}$ . The larger the value of $\begin{align*}n\end{align*}$ , the larger the number of passes required. In situations like these it is best to organize calculations to reduce the amount of tedious work and to avoid making unpredictable mistakes. A technique called tabular integration by parts can help do this.
::我们刚刚用来评估 {x2exdx } 的集成法是用来计算任何以 {xnexdx 为表单的集成,n是正整数 } =2。n 值越大,需要的通行证数就越多。在这样的情况下,最好组织计算,以减少烦琐的工作量,避免不可预测的错误。一种称为表格集成的方法可以帮助做到这一点。

Tabular Integration by Parts
::按部件分列的表格集成

The technique of tabular integration enables you to perform successive integrations by parts on integrals of the form
::表格集成技术使您能够按表格整体部分进行连续集成

$\begin{align*}\int P(x) Q(x) dx \end{align*}$ , where $\begin{align*}u=p(x)\end{align*}$ and $\begin{align*}dv=Q(x) dx\end{align*}$
::P(x)Q(x)dx, 其中 u=p(x) 和 dv(x) dx

without having to deal with a lot of algebraic detail that can slow down the solution process.
::无需处理许多代数细节,这些细节可以减缓解决进程的速度。

Here is an example of how it can be used:
::以下是如何使用这一工具的例子:

Evaluate $\begin{align*}\int x^3 e^x dx\end{align*}$ using tabular integration.
::使用表格集成评价 x3exdx。

Begin by letting $\begin{align*}u=x^3\end{align*}$ and $\begin{align*}dv=e^x dx\end{align*}$ .
::以 u=x3 和 dv=exdx 开始。

Next, create a table that consists of the three columns, as shown below:
::下一张表格由三栏组成,如下所示:

**Tabular Integration by Parts**
Alternate signs ::替代符号	$\begin{align}u=P(x)\end{align}$ and its ::u=P(x)及其	*$\begin{align}dv[Q(x)]\end{align}$* and its antiderivatives ::dv[Q(x)]及其抗降解剂
+	$\begin{align}x^3 \searrow\end{align}$ ::x3	$\begin{align}e^x\end{align}$ ::以 un, un, un, un, un, un, un, un, un, un, un, un, un, un
-	$\begin{align}3x^2 \searrow\end{align}$ ::3x2	$\begin{align}e^x\end{align}$
+	$\begin{align}6x\searrow\end{align}$	$\begin{align}e^x\end{align}$
-	$\begin{align}6 \searrow\end{align}$	$\begin{align}e^x\end{align}$
+	0	$\begin{align}e^x\end{align}$ ::以 un, un, un, un, un, un, un, un, un, un, un, un, un, un

The solution for the integral will be a summation of terms generated by performing the following steps:
::整体部分的解决方案将是对执行下列步骤所产生的术语进行总结:

1st term: Pick the sign from the 1st row $\begin{align*}(+1)\end{align*}$ ; multiply it by $\begin{align*}u\end{align*}$ of the 1st row $\begin{align*}(x^3)\end{align*}$ and then multiply by the $\begin{align*}dv\end{align*}$ of the 2 ^nd row $\begin{align*}(e^x)\end{align*}$ (watch the direction of the arrows.)
::第一个术语:从第一行(+1)中选择符号;将符号乘以第一行(x3)的u;然后乘以第二行(ex)的 dv(注意箭头的方向)。

1 ^st term: $\begin{align*}x^3e^x\end{align*}$
::第一个学期: x3ex

2 ^nd term: Pick the sign from the 2nd row (-1): multiply it by the $\begin{align*}u\end{align*}$ of the 2nd row $\begin{align*}(3x^2)\end{align*}$ and then follow the arrow to multiply the product by the $\begin{align*}dv\end{align*}$ in the 3 ^rd row $\begin{align*}(e^x)\end{align*}$ .
::第二行:从第二行(-1)中选择符号:乘以第二行(3x2)的u,然后跟随箭头将产品乘以第三行(ex)的 dv。

2 ^nd term: $\begin{align*}-3x^2 e^x\end{align*}$
::第二学期: - 3x2ex

3 ^rd term: Pick the sign from the 3 ^rd row (+1): multiply it by the $\begin{align*}u\end{align*}$ of the 3 ^rd row $\begin{align*}(6x)\end{align*}$ and then follow the arrow to multiply the product by the $\begin{align*}dv\end{align*}$ in the 4 ^rd row $\begin{align*}(e^x)\end{align*}$ .
::第三行:从第三行(+1)中选择符号:乘以第三行(6x)的u,然后跟随箭头将产品乘以第四行(ex)的 dv。

3 ^rd term: $\begin{align*}6xe^x\end{align*}$
::第三学期: 6xex

4 ^th term: Pick the sign from the 4 ^th row (-1): multiply it by the $\begin{align*}u\end{align*}$ of the 4 ^th row $\begin{align*}(6)\end{align*}$ and then follow the arrow to multiply the product by the $\begin{align*}dv\end{align*}$ in the 5 ^th row $\begin{align*}(e^x)\end{align*}$ .
::第4行:从第4行(-1)中选择符号:乘以第4行(6)的u,然后跟随箭头将产品乘以第5行(ex)的 dv。

4 ^th term: $\begin{align*}-6e^x\end{align*}$
::第4个学期: - 6ex

Stop*, since $\begin{align*}u=0\end{align*}$ .
::停止*,从U=0开始。

The solution is the sum of the terms:
::解决办法是条件的总和:

$\begin{align*}\int x^3 e^x dx = x^3 e^x -3x^2 e^x + 6xe^x - 6e^x +C=(x^3-3x^2+6x-6)e^x + C\end{align*}$
::x3exdx=x3ex- 3x2ex+6xex-6ex+C=(x3- 3x2+6x-6)ex+C

* The technique enables integrations by parts on integrals of the form
::* 技术能够使表单整体件各部分的集成结合。

$\begin{align*}\int P(x)Q(x) dx\end{align*}$ , where $\begin{align*}u=P(x)\end{align*}$ and $\begin{align*}dv=Q(x) dx\end{align*}$ .
::P(x)Q(x)dx,其中u=P(x)和dv(x)dx。

If $\begin{align*}P(x)\end{align*}$ is a polynomial, like the example above, then there will only be a finite number of terms to sum because a derivative of $\begin{align*}u=P(x)\end{align*}$ will eventually go to 0.
::如果 P(x) 是一多边性, 如上例, 那么将只有一定数量的条件可以加起来, 因为 u=P(x) 的衍生物最终会达到 0 。

Examples
::实例

Example 1
::例1

Earlier, you were asked to perform the required integration by parts steps on the integral $\begin{align*}\int x^2 \sin 5xdx\end{align*}$ . Two passes are required to evaluate the integral, and the answer is :
::早些时候,您被要求在 x2sin5xxx 上按部件步骤进行所需的整合。需要两张通行证才能评价组件, 答案是 :

$\begin{align*}\int x^2 \sin 5xdx = -\frac{1}{5} x^2 \cos 5x+\frac{2}{25} x\sin 5x + \frac{2}{125} \cos 5x +C\end{align*}$
::x2sin5xx15x2cos5x+225xsin5x+2125cs5x+C

Example 2
::例2

Evaluate $\begin{align*}\int x^2 \sin 3xdx\end{align*}$ using tabular integration.
::使用表格集成评价 x2sin3xdx。

Begin by letting $\begin{align*}u=x^2\end{align*}$ and $\begin{align*}dv=\sin 3xdx\end{align*}$ .
::开始启用 u=x2 和 dv=sin3xdx 。

Next, create a table that consists of the three columns, as shown below:
::下一张表格由三栏组成,如下所示:

**Tabular Integration by Parts**
Alternate signs ::替代符号	$\begin{align}u\end{align}$ and its derivatives ::u 及其衍生物	*$\begin{align}dv\end{align}$* and its antiderivatives ::dv及其抗降解剂
+	$\begin{align}x^2\searrow\end{align}$ ::x2	$\begin{align}\sin 3x\end{align}$ ::罪恶3x
-	$\begin{align}2x\searrow\end{align}$ ::2x___________________________________________________________________________________________________	$\begin{align}-\frac{\cos 3x}{3}\end{align}$
+	$\begin{align}2\searrow\end{align}$	$\begin{align}-\frac{\sin 3x}{9}\end{align}$
-	0	$\begin{align}\frac{\cos 3x}{27}\end{align}$

The solution for the integral will be a summation of terms generated by performing the following steps:
::整体部分的解决方案将是对执行下列步骤所产生的术语进行总结:

1st term: Pick the sign from the 1st row (+1): multiply it by $\begin{align*}u\end{align*}$ of the 1st row $\begin{align*}(x^2)\end{align*}$ and then multiply by the $\begin{align*}dv\end{align*}$ of the 2 ^nd row $\begin{align*}\left(-\frac{\cos 3x}{3}\right)\end{align*}$ (watch the direction of the arrows.)
::第一个术语:从第一行(+1)中选择符号:乘以第一行(x2)的 u,再乘以第二行(- cos3x3) dv(注意箭头的方向)。

1 ^st term: $\begin{align*}x^2 \left(-\frac{\cos 3x}{3}\right)\end{align*}$
::第一个学期: x2(- cos3x3)

2 ^nd term: Pick the sign from the 2nd row (-1): multiply it by the $\begin{align*}u\end{align*}$ of the 2nd row $\begin{align*}(2x)\end{align*}$ and then follow the arrow to multiply the product by the $\begin{align*}dv\end{align*}$ in the 3 ^rd row $\begin{align*}\left(-\frac{\sin 3x}{9}\right)\end{align*}$ .
::第二行:从第二行(-1)中选择符号:乘以第二行(2x)的u,然后跟随箭头将产品乘以第三行(- sin3x9)的 dv。

2 ^nd term: $\begin{align*}-2x\left(-\frac{\sin 3x}{9}\right)\end{align*}$
::第二学期:-2x(- sin3x9)

Add this term to the previous term.
::在上个任期中添加此术语。

3 ^rd term: Pick the sign from the 3 ^rd row (+1): multiply it by the $\begin{align*}u\end{align*}$ of the 3 ^rd row $\begin{align*}(2)\end{align*}$ and then follow the arrow to multiply the product by the $\begin{align*}dv\end{align*}$ in the 4 ^rd row $\begin{align*}\left(\frac{\cos 3x}{27}\right)\end{align*}$ .
::第三行:从第三行(+1)中选择符号:乘以第三行的u(2),然后跟随箭头将产品乘以第四行(cos3x27)的 dv。

3 ^rd term: $\begin{align*}2 \left(\frac{\cos 3x}{27}\right)\end{align*}$
::第三学期: 2( 2( cos3x27)

Add this term to the previous term.
::在上个任期中添加此术语。

Stop when $\begin{align*}u=0\end{align*}$ .
::当U=0停止时停止。

The solution is the sum of the terms:
::解决办法是条件的总和:

$\begin{align*}\int x^2 \sin 3xdx=\frac{-1}{3}x^2 \cos 3x +\frac{2}{9}x \sin 3x +\frac{2}{27}\cos 3x + C\end{align*}$
::x2sin3xx13x2cos3x+29xsin3x+227cs3x+C

Example 3
::例3

Evaluate $\begin{align*}\int (\ln x)^3 dx\end{align*}$ using Integration by parts
::使用整合方式按部分进行 {( lnx) 3dx <button class="play-button btn btn-success" style="float: right;" value="@s"> 播放段落 </button> Here, it appears we only have one term in the integrand, ( ln x ) 3 <script id="MathJax-Element-111" type="math/tex"> \begin{align*}(\ln x)^3\end{align*} , but we can always assume that this term is multiplied by 1, i.e., we have $\begin{align*}\int (\ln x)^3 \cdot 1dx\end{align*}$ :
::这里,我们似乎在整数中只有一个词, (xx) 3, 但我们总是可以假设这个词乘以 1, 也就是说, 我们有 (xx) 3x1dx :

$\begin{align*}& \text{Use:} && \int udv =uv - \int vdu. \qquad \qquad \ldots \text{Pass 1}\\ & \text{Choose:} && u= (\ln x)^3 \qquad \quad \ dv=1 dx\\ & \text{Form:} && du = 3 \frac{1}{x}(\ln x)^2 dx \quad v=x \\ & \text{Evaluate:} && \int (\ln x)^3 dx= x(\ln x)^3 - \int 3x \frac{1}{x}(\ln x)^2 dx\\ & && \qquad \qquad \quad =x(\ln x)^3 -3 \int(\ln x)^2 dx \qquad \ldots \text{The new integral is simpler, but requires}\\ & && \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \text{more work! Try another iteration.}\end{align*}$
::使用: @ udv=uvvdu.... Pass 1Choose: u=( lnx3dv=1dxForm: du= 31x( lnx) 2dxv=xEvaluate: @ (lnx) 3_%3x1x( lnx) 2dx=x( lnx) 3- 3_( lnx)2dxx... 新的组件更简单, 但需要更多工作 ! 请尝试另一个迭代。

From earlier, we know that the evaluation of $\begin{align*}\int (\ln x)^2 dx\end{align*}$ takes 2 passes before obtaining the solution
::从早些时候开始,我们知道对(lnx)2dx的评估在获得解决方案之前需要2个通行证。

$\begin{align*}\int (\ln x)^2 dx = x[(\ln x)^2-2\ln x-2]+C.\end{align*}$
:

lnx)2dx=x[(lnx)2--2lnx-2]+C。

Using this results yields
::利用这一成果产生产量

$\begin{align*}\int (\ln x)^3 dx &= x(\ln x)^3 -3 \left(x[(\ln x)^2-2\ln x-2]+C\right)\\ &=x[(\ln x)^3-3(\ln x)^2 + 6\ln x+6]+C\end{align*}$
:

xx)3dx=x(xx)3-3-3(x[(xx)2-2-2)+C)=x[(xx)3-3-3(xx)2+6lnx+6]+C

Review
::回顾

For #1-8, evaluate the integrals using integration by parts. ( Remark: In some, you may need to use $\begin{align*}u\end{align*}$ -substitution along with integration by parts.)
::对于 # 1-8, 使用各部件的集成来评估集成体。 (备注: 在某些部分中, 您可能需要使用 u 替代和部件的集成。 )

$\begin{align*}\int x(\ln x)^2 dx\end{align*}$
::x( lnx) 2dx

$\begin{align*}\int x^2 e^{-x}dx\end{align*}$
::x2e - xdx

$\begin{align*}\int x^2 \sin 2 xdx\end{align*}$
::x2sin2xdx

$\begin{align*}\int \theta^5 \cos (\theta^2)d \theta\end{align*}$ (Hint: Make a substitution before applying integration by parts).
::5cos(2)d(提示:在按部分进行整合之前先进行替换)。

$\begin{align*}\int \sec^3 xdx\end{align*}$
::sec3xdx

$\begin{align*}\int\limits_{0}^{1} (x^2 + 3)e^{-2x} dx\end{align*}$
::10( x2+3) e- 2xxx

$\begin{align*}\int \frac{(\ln x)^2}{x} dx\end{align*}$
: lnx) 2xdx

$\begin{align*}\int\limits_{0}^{1}x^2 e^x dx\end{align*}$
::10x2exdx

Use integration by parts to show that $\begin{align*}\int (\ln x)^n dx= x (\ln x)^n-n \int(\ln x)^{n-1}dx\end{align*}$
::使用部件集成来显示 ( lnx) ndx= x( lnx) n- n ( lnx) n- 1dx

For #10-15, use the method of tabular integration by parts to evaluate the integrals.
::对于第10-15号,使用按部分分列的表格集成方法来评价集成物。

$\begin{align*}\int x^2 e^{5x}dx\end{align*}$
::x2e5xxx

$\begin{align*}\int \frac{x^2+3}{\sqrt[3]{3x+7}}dx\end{align*}$
::x2+333x+7dx

$\begin{align*}\int\limits_{0}^{1} x^2 e^{-x}dx\end{align*}$
::10x2e - xdx

$\begin{align*}\int (y^2 -3y-7)e^y dy\end{align*}$
::*(y2-3y-7)eydy(y2-3y-7)

$\begin{align*}\int x^4 e^{-3x}dx\end{align*}$
::x4e-3xdx

$\begin{align*}\int (x^3 +4x^2 +3) \sin 2 xdx\end{align*}$
:x3+4x2+3)sin2xdx

Review (Answers)
::回顾(答复)

Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
::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。

Alternate signs ::替代符号	$\begin{align}u=P(x)\end{align}$ and its ::u=P(x)及其	*$\begin{align}dv[Q(x)]\end{align}$* and its antiderivatives ::dv[Q(x)]及其抗降解剂
+	$\begin{align}x^3 \searrow\end{align}$ ::x3	$\begin{align}e^x\end{align}$ ::以 un, un, un, un, un, un, un, un, un, un, un, un, un, un
-	$\begin{align}3x^2 \searrow\end{align}$ ::3x2	$\begin{align}e^x\end{align}$
+	$\begin{align}6x\searrow\end{align}$	$\begin{align}e^x\end{align}$
-	$\begin{align}6 \searrow\end{align}$	$\begin{align}e^x\end{align}$
+	0	$\begin{align}e^x\end{align}$ ::以 un, un, un, un, un, un, un, un, un, un, un, un, un, un

Alternate signs ::替代符号	$\begin{align}u\end{align}$ and its derivatives ::u 及其衍生物	*$\begin{align}dv\end{align}$* and its antiderivatives ::dv及其抗降解剂
+	$\begin{align}x^2\searrow\end{align}$ ::x2	$\begin{align}\sin 3x\end{align}$ ::罪恶3x
-	$\begin{align}2x\searrow\end{align}$ ::2x___________________________________________________________________________________________________	$\begin{align}-\frac{\cos 3x}{3}\end{align}$
+	$\begin{align}2\searrow\end{align}$	$\begin{align}-\frac{\sin 3x}{9}\end{align}$
-	0	$\begin{align}\frac{\cos 3x}{27}\end{align}$

章节大纲

常规

按部分集成: 多个通道

Integration by Parts: Multiple Passes ::按部分集成: 多个通道

Tabular Integration by Parts ::按部件分列的表格集成

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Review ::回顾

Review (Answers) ::回顾(答复)