⒈不定积分的方法与技巧

设F（x)是函数f(x)的一个原函数，我们把函数f(x)的所有原函数F(x) C（C为任意常数）叫做函数f(x)的不定积分，记作，即∫f(x)dx=F(x)+C。

其中∫叫做积分号，f(x)叫做被积函数，x叫做积分变量，f(x)dx叫做被积式，C叫做积分常数，求已知函数的不定积分的过程叫做对这个函数进行积分。

第一换元法（凑微分法）

　

 

例2.5.9求http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point4/images/p0205041001.gif .

解.因(1+x²)' =2x，与被积函数的分子只差常数倍数2，如果将分子补成2x,即可将原式变形：

原式=http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point4/images/p0205041002.gif            　(令u=1+x²)

=http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point4/images/p0205041003.gif. (代回u=1+x²)

注.此例解法的关键是凑了微分d(1+x²)．一般地

在F '(u)=f(u),u=j(x)可导,且j' (x)连续的条件下,我们有

 

第一换元公式(凑微分)：

                                                         u=j (x)      积分  代回 u=j (x)

∫f[j(x)]j' (x)dx =∫f[j(x)]dj(x)=∫f(u)du=F(u)+C=F[j(x)]+C

其中函数j(x)是可导的,且F(u)是f(u)的一个原函数．

从上述公式可看出凑微分法的步骤：

 

凑微分————→换元————→积分————→再换元

j' (x)dx=dj(x)   u=j(x)      得F(u)+C      得F[j(x)]+C

　

注.凑微分法的过程实质上是复合函数求导的逆过程．事实上，在

F '(u)=f(u)的前提下，上述公式右端经求导即得:

[F[j(x)]+C]' =F '[j(x)]j' (x)=f[j(x)]j' (x)

 

例2.5.10求∫(ax+b)^mdx.(m≠-1,a≠0)

解.原式=http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point4/images/p0205041004.gif      (凑微分d(ax+b))

=http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point4/images/p0205041005.gif                 (换元u=ax+b)

=http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point4/images/p0205041006.gif         (积分)

=http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point4/images/p0205041007.gif.    (代回u=ax+b)

例2.5.11 求http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point4/images/p0205041008.gif .

解.原式=http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point4/images/p0205041009.gif         (凑微分d(-x³)=-3x²dx)

=http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point4/images/p0205041011.gif                       

=http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point4/images/p0205041012.gif.            (换元u=-x³)

 

例2.5.12求∫tanxdx　.

解.原式=http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point4/images/p0205041013.gif                 

=-ln|cosx|+C .

同理可得

∫cotxdx=ln|sinx|+C .

例2.5.13求http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point4/images/p0205041014.gif.(a>0)

解.原式=http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point4/images/p0205041015.gif

=http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point4/images/p0205041016.gif  .

例2.5.14 求http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point4/images/p0205041017.gif(a>0).

解.原式=http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point4/images/p0205041018.gif

=http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point4/images/p0205041034.gif .

例2.5.15求http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point4/images/p0205041020.gif.

解.原式=http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point4/images/p0205041021.gif

=http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point4/images/p0205041022.gif

=http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point4/images/p0205041023.gif

=http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point4/images/p0205041024.gif.  

例2.5.16∫secxdx.

解.原式=http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point4/images/p0205041025.gif  (换元u=sinx)

=http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point4/images/p0205041026.gif

=http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point4/images/p0205041027.gif

=http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point4/images/p0205041028.gif (代回u=sinx)

=http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point4/images/p0205041029.gif

=http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point4/images/p0205041030.gif

=ln|secx+tanx|+C .

 公式:∫secxdx=ln|secx+tanx|+C .

例.2.5.17求∫cscxdx .

解.原式=http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point4/images/p0205041031.gif

=http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point4/images/p0205041032.gif

=ln|cscx-cotx|+C .

公式:∫cscxdx=ln|cscx-cotx|+C . 

        第二换元法

　

不定积分第一换元法的公式中核心部分是

∫f[j(x)]j'(x)dx=∫f(u)du

 

我们从公式的左边演算到右边，即换元：u=j(x)．与此相反，如果我们从公式的右边演算到左边，那么就是换元的另一种形式，称为第二换元法．即若f(u),u=j(x),j'(x)均连续,u=j(x)的反函数x=j^-1(u)存在且可导,F(x)是f[j(x)]j'(x)的一个原函数,则有

 

∫f(u)du =∫f[j(x)]j'(x)dx =F(x)+C =F[j^-1(u)]+C .

第二换元法常用于被积函数含有根式的情况．

例2.5.18求http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point5/images/p0205051001.gif

解.令http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point5/images/p0205051002.gif  (此处j(t)=t²)．于是

原式=http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point5/images/p0205051003.gif

=http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point5/images/p0205051004.gif

=http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point5/images/p0205051006.gif)

注.换元http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point5/images/p0205051007.gif=t的目的在于将被积函数中的无理式转换成有理式,然后积分．

第二换元法除处理形似上例这种根式http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point5/images/p0205051011.gif(a>0)的被积函数的积分．

 

被积函数含根式	换元方法	运用的三角公式
http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point5/images/p0205051012.gif	x=asect	sec2t-1=tan2t
http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point5/images/p0205051013.gif	x=atant	tan2t+1=sec2t
http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point5/images/p0205051014.gif	x=asint	1－sin2t=cos2t

 

例2.5.19求http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point5/images/p0205051015.gif. (a>0)

解.令x=asect，则dx=asect tant dt,于是

原式=http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point5/images/p0205051016.gif=∫sectdt

=ln|sect+tant|+C₁ .

到此需将t代回原积分变量x，用到反函数t=arcsechttp://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point5/images/p0205051018.gif满足：

sect=http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point5/images/p0205051019.gif

http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point5/images/p0205050001.gif 

由此，原式=ln|sect+tant|+C₁

= http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point5/images/p0205051020.gif

= http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point5/images/p0205051021.gif

  http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point5/images/p0205051023.gif.

注．C₁是任意常数,-lna是常数,由此C=C₁-lna仍是任意常数．

例2.5.20求http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point5/images/p0205051024.gif. (a>0)

解.令x=atant，则dx=asec²tdt，于是

原式=http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point5/images/p0205051025.gif=∫sectdt

=ln|sect+tant|+C₁ ．

http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point5/images/p0205050002.gif

图解换元得

原式=ln|sect+tant|+C₁

 

= http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point5/images/p0205051026.gif

http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point5/images/p0205051028.gif.

公式:http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point5/images/p0205051029.gif.  

例2.5.21求http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point5/images/p0205051030.gif. (a>0)

解.令x=asint，则dx=acostdt，于是

原式=http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point5/images/p0205051037.gif

=http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point5/images/p0205051032.gif

=http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point5/images/p0205051033.gif+C .

http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point5/images/p0205050003.gif

图解换元得：

原式=http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point5/images/p0205051034.gif+C

=http://www.math.nankai.edu.cn/~gdsxjxb/wlkj/windows/artsmath/chapter2/section5/point5/images/p0205051035.gif+C .

除了换元法积分外，还有一个重要的积分公式，即分部积分公式