4.2 换元积分法(1)

60 第二节 换元积分法（一）

一、填空题

1. 2e d d x x =21

()2x C +e 21d d x x =1()C x

-+

d x =

()C 1ln d d x x x =(ln ln )x C +

d x =

(2arctan )C 2e d d x x x -=21()2x C --+e 2. ()e x f x -=，则(ln )d f x x x '?＝1C x +. 提示：()e x f x -'=-，则ln 1(ln )e x f x x

-'=-=-代入求解即可． 3. 设()sin d f x x x C =+?，则2(1)d xf x x -?＝21sin(1)2

x C --+. 提示：由已知得()cos f x x =，则22(1)cos(1)f x x -=-代入求解即可．

4. 21e d e x x x +?＝arctan x C +e .

5. 2

2(arctan )1d x x x +?＝31(arctan )3

x C +. 6. 21ln (ln )d x x x x +?＝1ln C x x

-+. 7．1sin cos d x x x x -+?＝ln cos x x C ++. 二、求下列不定积分

1.x ?

解：33222221121(3)(3)(3)2233x x x C x C =-=?-+=-+?． 2. 123d x x

+?

解：11(23)23323d d x x x x +=++??1ln 233x C =++．

61 3.23sin cos d x x x ?

解：2322sin cos sin cos sin d d x x x x x x =??=()24sin sin sin d x x x -?

351

1

sin sin 35x x C =-+．

4. 3tan d x x ?

解：32tan tan (sec 1)d d x x x x x =-??=2tan sec tan d d x x x x x -??

sin tan tan cos d d x

x x x x =-??21

tan 2x =ln cos x C ++．

5．1

211e d x x x x +??- ???? 解：1

1

1

2111e d e d e x x x x x x x x C x x +++????-=+=+ ? ???????．

6

. ()2d x

解：(

)2

22)2)3d d x =

?2

2)3=-C +．

7

．arcsin x

x

解：arcsin arcsin 2(arcsin )d x

x x x =?arcsin 12ln 2x C =+．

8．21

cos (1tan )d x x x +? 解：221sec cos (1tan )1tan d d x x x x x x =++??(1tan )

1tan d x x +=+?ln 1tan x =+C +．

9

．x

62

解：(222x C ===+??． 10．211ln 11d x x x x

+--? 解：211ln 11d x x x x

+--?C x x x x x x +??? ??-+=-+-+=?211ln 4111ln d 11ln 21.