高等数学第六版上册(同济大学) 第四章答案(2)

1 =∫lntanxdlntanx=(lntanx)2+C. 2

(34)∫x2

a x22dx(a>0);

解 ∫令x=asinta2sin2t2221 cos2ttdt=atdt=adxcossin∫acost∫∫2dt, 22a xx2

1

2a2a2xx22 =at sin2t+C=arcsin a x+C. 242a2

(35)∫dx

xx 1

dx

xx2 12; 解 ∫令

x=sect11 ttdt=dt=t+C=+C. sectanarccos∫sect tant∫x

或 ∫dx

xx2 1=∫1x2 1

x2dx= ∫1 11=arccos+C. x1xx2

(36)∫dx

(x+1)

dx

(x+1)2323; 解 ∫令x=tant1(tant+1)23tant=∫costdt=sint+C=xx+12

+C.

(37)∫x2 9dx; x

x2 9令x=3sect9sec2t 9(3sect)=3∫tan2tdt dx∫3sectx 解 ∫

=3∫(321)3tan393arccos dt=t t+C=x +C. 2xcost1

(38

)∫

解 dx1+2xdx

2x; 令2x=t11= tdt(1∫1+t∫1+tdt=t ln(1+t)+C=2x ln(1+2x)+C. ∫1+

(39)∫dx1+ x

dx2; 解 ∫令x=sint

1+ x21112tcostdt=(1 )dt=(1 secdt ∫1+cost∫1+cost∫22

txsint+C=arcsinx +C. =t tan+C=t 221+cost1+ x

(40)∫dx

x+ x

dx

x+ x22. 解 ∫令x=sint∫sint+cost costdt=2∫11cost+sint+cost sintdt sint+cost

=11111dt+∫d(sint+cost)=t+ln|sint+cost|+C ∫22sint+cost22

11 =arcsinx+ln| x2+x|+C. 22

习题4 3

求下列不定积分:

1. ∫xsinxdx;

解 ∫xsinxdx= ∫xdcosx= xcosx ∫cosxdx= xcosx+sinx+C. 2. ∫lnxdx;

解 ∫lnxdx=xlnx ∫xdlnx=xlnx ∫dx=xlnx x+C.

3. ∫arcsinxdx;

解 ∫arcsinxdx=xarcsinx ∫xdarcsinx

=xarcsinx ∫x

x2dx

=xarcsinx+ x2+C.

4. ∫xe xdx;

解 ∫xe xdx= ∫xde x= xe x+∫e xdx

= xe x e x+C= e x(x+ 1)+C.

5. ∫x2lnxdx;

解 ∫x2lnxdx=111lnxdx3=x3lnx ∫x3dlnx ∫333

1111 =x3lnx ∫x2dx=x3lnx x3+C. 3339

6. ∫e xcosxdx;

解 因为

∫e xcosxdx=∫e xdsinx=e xsinx ∫sinxde x=e xsinx+∫e xsinxdx =e xsinx ∫e xdcosx=e xsinx e xcosx+∫cosxde x

=e xsinx e xcosx ∫e xcosxdx,

11所以 ∫e xcosxdx=(e xsinx e xcosx)+C=e x(sinx cosx)+C. 22

x 7. ∫e 2xsindx; 2

解 因为

xxxx ∫e 2xsindx=2∫e 2xdcos=2e 2xcos 2∫cosde 2x 2222

xxxx =2e 2xcos+4∫e 2xcosdx=2e 2xcos+8∫e 2xdsin 2222

xxx =2e 2xcos+8e 2xsin 8∫sinde 2x 222

xxx =2e 2xcos+8e 2xsin+16∫e 2xsindx, 222

x2xx所以 ∫e 2xsindx= e 2x(cos+4sin+C. 21722

x 8. ∫xcosdx; 2

xxxxxx 解 ∫xcosdx=2∫xdsin=2xsin 2∫sindx=2xsin+4cos+C. 222222

9. ∫x2arctanxdx;

解 ∫x2arctanxdx=1131313xdx=xx x dx arctanarctan3∫33∫1+x2

131x211213 =xarctanx ∫dxxarctanx(1dx2 = ∫22361+x361+x

111 =x3arctanx x2+ln(1+x2)+C. 663

10. ∫xtan2xdx

1 解 ∫xtan2xdx=∫x(sec2x 1)dx=∫xsec2xdx ∫xdx= x2+∫xdtanx 2

11 = x2+xtanx ∫tanxdx= x2+xtanx+ln|cosx|+C. 22

11. ∫x2cosxdx;

解 ∫x2cosxdx=∫x2dsinx=x2sinx ∫sinx 2xdx=x2sinx+2∫xdcosx

=x2sinx+2xcosx 2∫cosxdx=x2sinx+2xcosx 2sinx+C.

12. ∫te 2tdt;

解 ∫te 2tdt= 11 2t1 2t 2ttde= te+∫edt 2∫22

1111 = te 2t e 2t+C= e 2t(t++C. 2422

13. ∫ln2xdx;

1 解 ∫ln2xdx=xln2x ∫x 2lnx dx=xln2x 2∫lnxdx x

1 =xln2x 2xlnx+2∫x dx=xln2x 2xlnx+2x+C. x

14. ∫xsinxcosxdx;

解 ∫xsinxcosxdx=1111xsin2xdx= xdcos2x= xcos2x+cos2xdx 2∫4∫44∫

11 = xcos2x+sin2x+C. 48

15. ∫x2cos2

解 ∫x2cos2xdx; 2x11111dx=∫x2(1+cosx)dx=x3+∫x2dsinx=x3+x2sinx ∫xsinxdx 226262

1111 =x3+x2sinx+∫xdcosx=x3+x2sinx+xcosx ∫cosxdx 6262

11 =x3+x2sinx+xcosx sinx+C. 62

16. ∫xln(x 1)dx;

解 ∫xln(x 1)dx=1121212ln(x 1)dx=xln(x 1) x dx 2∫22∫x 1

111 =x2ln(x 1) ∫(x+1+ dx 22x 1

1111 =x2ln(x 1) x2 x ln(x 1)+C. 2422

17. ∫(x2 1)sin2xdx;

解 ∫(x2 1)sin2xdx= 111(x2 1)dcos2x= (x2 1)cos2x+∫cos2x 2xdx ∫222

11 = (x2 1)cos2x+∫xdsin2x 22

111 = (x2 1)cos2x+xsin2x ∫sin2xdx 222

111 = (x2 1)cos2x+xsin2x+cos2x+C. 224

18. ∫

解 ln3x2x2

11131 = ln3x 3∫ln2xd= ln3x ln2x+3∫dln2x xxxxx∫xln3xdx; 11111dx= ∫ln3xd= ln3x+∫dln3x= ln3x+3∫2ln2xdx xxxxx

131131 = ln3x ln2x+6∫2lnxdx= ln3x ln2x 6∫lnxd xxxxxx

1361 = ln3x ln2x lnx+6∫2dx xxxx

1366 = ln3x ln2x lnx +C. xxxx

19. ∫e

解 ∫e3xdx; 令x=txdx∫t2etdt=3∫t2det

=3t2et 6∫tetdt=3t2et 6∫tdet

=3t2et 6tet+6∫etdt

=3t2et 6tet+6et+C

=3e3x(x2 2x+2)+C.

20. ∫coslnxdx;

解 因为

1 ∫coslnxdx=xcoslnx+∫x sinlnx dx x

1 =xcoslnx+∫sinlnxdx=xcoslnx+xsinlnx ∫x coslnx dx x