呼和浩特职业学院：《经济数学》课程教学资源（习题详解）第四章 不定积分

习题4.1详解 计算下列不定积分 1.∫x2d 解∫F=+c 2.「a+e')d 解∫I+e)k=x+e'+C 3.∫0-3x2) 解「1-3x2)=x-x2+C 4.∫0+cot2x)k 解∫0+cot2x)t=∫csc2xdk=-cotx+C 5.∫3+x) 解小e+h=+言+c 6∫x 解∫小乐=号+C 7.x+1)2-e] 解jx+2-e=写x2+r2+x-e+C & 解j-e-h=e-4c 1e4 解∫0本-=ja@s-m=n+cax4C

1 习题 4.1 详解 计算下列不定积分 1.  x dx 3 2 解 x dx = x + C  3 5 3 2 5 3 2.  + e dx x (1 ) 解 e dx x e C x x + = + +  (1 ) 3.  (1− 3x )dx 2 解 − x dx = x − x + C  2 3 (1 3 ) 4.  (1+ cot x)dx 2 解 + x dx = xdx = − x + C   (1 cot ) csc cot 2 2 5.  + x dx x (3 ) 5 解 x dx x C x x + = + +  5 6 6 1 3 ln 3 1 (3 ) 6.  x x dx 解 x x dx = x + C  4 7 7 4 7.  x + − e dx x [( 1) ] 2 解 x e dx x x x e C x x + − = + + − +  2 3 2 3 1 [( 1) ] 8.  + − dx e e x x 1 1 2 解 dx e dx e x C e e x x x x = − = − + + −   ( 1) 1 1 2 9.  + dx x x x cos sin cos 2 解 dx x x dx x x C x x x x dx x x x = − = + + + − = +    (cos sin ) sin cos cos sin cos sin cos sin cos2 2 2

1 10∫x0+r

2 10.  + dx x (1 x ) 1 2 2 解 x C x dx x x dx x x = − − + + = − +   arctan 1 ) 1 1 1 ( (1 ) 1 2 2 2 2

习题4.2详解 1.在下列格式的横线上填入适当的系数，使等式成立。解答： 0=5d5x+2) (2)x迹=d(r2+3) ()r=6d-2x) 回 间左=24+到 (6)k=d血x-) (7)sin3xdr=-d(cos3*) (图)es=de） (9)xcosdd(sinx) (0）4-4em2刘 2.填空： ()x迹=dr) (2)xdx=d(x) (3)Id=danx) )子=d- (③）店=d2同 (6)e=de) (7)cos6xdx=d(sin6x) (③)sin5xd=d-5cos写x) 9中子k=rn) (0)d() 3.计算下列不定积分： (∫(2x+3)d 解：j2x+3)=打2x+3)d(2x+3)=2（2x+3+C (2)∫sin3xk 解∫sin3k=,Jsin3d(3x)=-os3x+C

3 习题 4.2 详解 1. 在下列格式的横线上填入适当的系数，使等式成立。解答： ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 3 2 2 2 2 2 2 1 1 1 5 2 2 3 5 2 1 1 1 3 - 1 2 4 1 6 1 1 5 2 3 6 ln 1 1 1 7 sin 3 - cos 3 8 3 2 1 1 1 9 cos sin 10 arctan 2 2 1 4 2 x x dx d x xdx d x x dx d x dx d x x dx d x dx d x x x xdx d x e dx d e x x dx d x dx d x x = + = +   = − = −     = + = − = = = = + 2. 填空： ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 3 2 4 4 1 1 1 ( ) 2 ( ) 2 3 1 1 1 3 (ln ) 4 ( ) 1 1 5 (2 ) 6 ( ) 4 1 1 1 7 cos 6 ( sin 6 ) 8 sin ( 5cos ) 6 5 5 x x xdx d x x dx d x dx d x dx d x x x dx d x e dx d e x xdx d x xdx d x = = = = − = = = = − ( ) ( ) 2 2 1 1 9 (arctan ) 10 (arcsin ) 1 1 dx d x dx d x x x = = + − 3. 计算下列不定积分： ( ) ( ) 2 1 2 3 x dx +  ( ) ( ) ( ) ( ) 2 2 1 1 3 2 3 2 3 2 3 2 3 C 2 6 x dx x d x x + = + + = + + 解：   (2 sin 3 ) xdx  ( ) 1 1 sin 3 sin 3 3 cos 3 3 3 xdx xd x x C = = − + 解  

(B)∫cos2 解jcos本=2eos28=2sim+C 解-字0+)+)C e -e-c A 解=024C 仞 解∫本=nda)=hx+C (周4 解本-品m=邮如+C

4 (3 cos ) 2 x dx  cos 2 cos 2sin 2 2 2 2 x x x x dx d C   = = +     解   ( ) 2 2 4 1 x dx + x  ( ) ( ) 2 2 2 2 2 1 1 ln 1 1 1 x dx d x x C x x = + = + + + + 解   ( ) ( ) 2 2 5 3 x dx x −  ( ) ( ) ( ) ( ) 2 2 2 2 2 2 1 1 1 3 3 3 2 2 3 x dx d x C x x x = − = − + − − − 解   ( ) 1 6 1 dx + x  ( ) 1 1 1 2 1 C 1 1 dx d x x x x = + = + + + + 解   ( ) ln 7 x dx x  ( ) 2 ln 1 ln ln ln C 2 x dx xd x x x = = + 解   ( ) cos 8 sin x dx x  ( ) cos 1 sin ln sin sin sin x dx d x x C x x = = + 解   ( ) ( ) 2 1 9 x dx x + 

g0可网-c 0) 四-流 ￥流网2，-日产 dx -i(lmlx-I-Ink+2)+C c 2)* 解∫-F点-+anmx+c -In(+)-arctan x+C 4.计算下列不定积分： ()∫xe 解∫e=打e=e+c (2)∫x2sinx'dk 解∫sin=打sind=-cosx+C

5 ( ) ( ) ( ) ( ) 2 1 2 3 2 2 1 1 1 3 x dx x d x x C x + = + + = + + 解   ( ) 2 1 10 4 9 dx + x  2 2 1 1 1 3 1 3 arctan 4 9 6 2 6 2 3 1 2 dx d x x C x x   = = +   +     +     解   ( ) ( )( ) 11 1 2 dx x x − +  ( )( ) ( ) ( ) ( )( ) ( ) 1 1 1 1 2 1 1 2 3 1 2 3 1 2 1 ln 1 ln 2 3 1 1 ln 3 2 dx x x dx dx x x x x x x x x C x C x +−−   = = −   − + − + − +   = − − + + − = + + 解    ( ) 2 1 12 1 x dx x − +  ( ) ( ) 2 2 2 2 2 2 1 1 1 1 1 arctan 1 1 1 2 1 1 ln 1 arctan 2 x x dx dx dx d x x C x x x x x x C − = − = + − + + + + + = + − + 解     4. 计算下列不定积分： ( ) 2 1 x xe dx  2 2 2 1 1 2 C 2 2 x x x xe dx e dx e = = + 解   ( ) 2 3 2 sin x x dx  2 3 3 3 3 1 1 sin sin cos 3 3 x x dx x dx x C = = − + 解  

(3)∫cos(1+3x) 解∫cos(+3xk=cos+3x0+3x)=sin(1+3)+C + 解+=+++c 间∫ 解∫=ax=nh时4c 同j4 解j5女=-+子(e-r(osC (7)「sin2xdh 解小sm=2k=杰-o2 x-2x+C 图j 解jfe=i+e0+e)=n+e)+c (9)∫xF+idk 6

6 (3 cos 1 3 ) ( + x dx )  ( ) ( ) ( ) ( ) 1 1 cos 1 3 cos 1 3 1 3 sin 1 3 3 3 + = + + = + + x dx x d x x C 解   ( ) 3 4 1 2 x dx     +    3 3 4 1 1 2 1 1 1 2 2 2 2 2 x x x x dx d C                 + = + + = + +         解   ( ) 1 5 ln dx x x  1 1 ln ln ln ln ln dx d x x C x x x = = + 解   ( ) 2 sin 6 1 cos x dx + x  ( ) ( ) 2 2 sin 1 cos arctan cos 1 cos 1 cos x dx d x x C x x = − = − + + + 解   ( ) 2 7 sin xdx  2 1 cos 2 1 1 sin cos 2 2 2 2 1 1 sin 2 2 4 x xdx dx dx xdx x x C − = = − = − + 解     (8) 1 x x e dx + e  ( ) ( ) 1 1 ln 1 1 1 x x x x x e dx d e e C e e = + = + + + + 解   ( ) 2 9 1 x x dx + 

解jF+i-F+a(e+-F+可+c +)i+c o-4 eee可g是4 1 训六产-p-小20c 0j-454 gjgdag-24n-g-a 1 1 =arctan(x-2)+C 回 222小220 -+2x24+2+2-+0 =In(x2+2x+2)-2arctan(x+1)+C 5。计算下列不定积分： 7

7 ( ) ( ) ( ) 3 2 2 2 2 2 2 1 1 2 1 1 1 1 2 2 3 1 1 1 3 x x dx x d x x C x x C + = + + =  + + = + + + 解   ( ) 2 1 10 4 dx − x  ( )( ) ( ) ( ) ( )( ) ( ) 2 1 1 1 2 2 4 2 2 4 2 2 1 1 1 1 ln 2 ln 2 4 2 2 4 1 2 ln 4 2 x x dx dx dx x x x x x dx x x C x x x C x + + − = = − + − + −   = + = − − + + +     − + + = + −     解 ( ) 2 1 11 4 5 dx x x − +  ( ) ( ) ( ) 2 2 2 1 1 1 2 4 5 4 4 1 2 1 arctan 2 dx dx d x x x x x x x C = = − − + − + + − + = − + 解    ( ) 2 2 12 2 2 x dx x x + +  ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 2 2 2 1 2 2 1 1 ln 2 2 2arctan 1 x x x dx dx dx x x x x x x x x x dx dx x x x x d x x d x x x x x x x C + − +   = = −   + + + + + + + +   + = − + + + + = + + − + + + + + = + + − + +      解 5. 计算下列不定积分： ( ) 1 1 2 dx − x 

解j242u-2创=-2h-2c 回代-2W2-x+C ( 解ja点匹咖可- =u-ln(1+w）+C 回代W2x-ln(1+√2x)+C (3)∫x+xd 解∫ri+x直+x=u2r-lrd=2u-i)=子u-子r+C +闲-+可+C =后(3x-20+小+x+C ∫+ 解jF*世血=4=h =6可r-u+i-可n+=2w-3w+6u-6lu++c 回代2WF-3+6F-6In(F+1)+C (⑤)∫-r 8

8 ( ) 1 1 2 2 2 2 2 2 2 dx x u udu du u C x u x C − = − = − = − + − − − + 解    回代 ( ) 1 2 1 2 dx + x  ( ) ( ) 1 1 1 1 1 2 1 1 2 1 1 1 ln 1 2 ln 1 2 u dx x u udu du du x u u u u u C x x C + −   = = = −   + + + +   = − + + − + + 解     回代 (3 1 ) x xdx +  ( ) ( ) ( ) ( ) ( )( ) 2 2 4 2 5 3 5 3 2 2 1 1 2 1 2 5 3 2 2 1 1 5 3 2 3 2 1 1 15 x xdx x u u u du u u du u u C x x C x x x C + + = − = − = − + = + − + + = − + + + 解    ( ) 3 1 4 dx x x +  ( ) ( ) 5 3 3 6 3 3 2 2 3 2 3 6 6 1 6 1 1 6 6 1 1 1 6 1 6 2 3 6 6ln 1 1 2 3 6 6ln 1 u u u dx x u du du du x x u u u u u u du du u u u u C u x x x x C + − = = = + + + + = − + − = − + − + + + − + − + +       解 回代 ( ) 2 5 1− x dx 

解子uuo咖-0+ow2h-+n2a+c 回代arcsin.+x-子+C 安 解j不字=wah-小ew咖-小 =∫加-j-m -0-n先aim 0+n1-indsin-h:m吵h-n小+c 坚立c=啡c网c 1+ 0阿 解jk=u黑utm咖-恤-ecu-h =amu-+C回代F-i-arccos+C 周 解三器m-h=c 9

9 ( ) 2 2 2 1 1 1 1 sin cos 1 cos 2 sin 2 2 2 4 1 1 arcsin 1 2 2 x dx x u udu u du u u C x x x C − = = + = + + + − + 解    回代 ( ) 2 1 6 1 dx + x  ( )( ) ( ) ( ) ( )( ) ( ) ( ) 2 2 2 2 2 1 1 1 tan sec sec 1 sec cos cos 1 sin cos 1 sin 1 1 1 sin 1 sin sin sin 1 sin 1 sin 2 1 sin 1 sin 1 1 1 1 1 sin ln 1 sin ln 1 sin 2 1 sin 1 sin 2 2 1 1 1 ln 2 1 1 dx x u udu udu du x u u u du d u u u u u d u d u u u u u d u u u C u u x x x x = = = + = = − + + − = = + − + −   = + = + − − +     + − + + − +          解 回代 ( ) 2 2 + = + + + C x x C ln 1 ( ) 2 1 7 x dx x −  ( ) 2 2 2 2 1 tan sec sec tan tan sec 1 sec 1 tan 1 arccos x u dx x u u udu udu u du x u u u C x C x − = = = − = − + − − + 解     回代 ( ) 2 1 8 1 dx x x −  2 1 sec tan sec 1 sec tan 1 arccos u u dx x u du du u C x x u u C x = = = + − + 解    回代

习题4.3详解 1∫arctanxdx 解arctand=xarctan-∫本=xarctan x-l+r)+C 2「x2edk 解∫x2edk=x2e-∫2xedk=x2e-2(xe-∫e'dk)=x2e-2xe+2e+C 3∫arccosxdx 解∫=aos-子本oas-看-子+C 4∫x3nx 解∫rhx=子xhx-∫xd)=xhx-子)+C 5∫xsin xdx 解「xsin xdx=-xCOSx+[cosxdx=-xcOSx+sinx+C 6∫e'cos xd 解 [e'cosxdx=[e'd(sin x)=e*sin x-[e*sin xdx =e*sin x+e*cosx-[e*cosxdx 将上式整理，再添上任意常数，得 fe*cosxd=(sn x+cosx)e*+C

10 习题 4.3 详解 1  arctan xdx 解 dx x x x C x x xdx x x = − + + + = −   ln(1 ) 2 1 arctan 1 arctan arctan 2 2 2  x e dx 2 x 解 x e dx x e x e dx x e x e e dx x e x e e C x x x x x x x x x = − = − − = − + +    2 2( ) 2 2 2 2 2 2 3  arccos xdx 解 dx x x x C x x xdx x x = − − + − = − −   2 2 arccos 1 1 arccos arccos 4  x ln xdx 3 解 dx x x x C x x xdx = x x − x = − +   ) 4 1 ( ln 4 1 ) 1 ( ln 4 1 ln 3 4 4 4 4 5  xsin xdx 解 x xdx = −x x + xdx = −x x + x +C   sin cos cos cos sin 6  e xdx x cos 解  e xdx x cos  = e d(sin x) x  = e x − e xdx x x sin sin  = e x + e x − e xdx x x x sin cos cos 将上式整理，再添上任意常数，得 e xdx x x e C x x = + +  (sin cos ) 2 1 cos