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1、44高等数学同济大学第六本 习题4-4 求下列不定积分: x3dx; 1. x+3x3x3+27-27(x+3)(x2-3x+9)-27dx=dx=dx 解 x+3x+3x+3 =(x2-3x+9)dx-271dx x+313 =x3-x2+9x-27ln|x+3|+C. 32 2. 解 x2+3x-10dx; 2x+3122dx=x2+3x-10x2+3x-10d(x+3x-10)=ln|x+3x-10|+C. 2x+3x5+x4-8dx; 3. x3-xx5+x4-8x2+x-82dx=(x+x+1)dx+3dx 解 x3-xx-x11843 =x3+x2+x+dx-dx-dx 32xx+
2、1x-111 =x3+x2+x+8ln|x|-4ln|x+1|-3ln|x-1|+C. 323x3+1dx; 31-x+2112x-131 解 3dx=(+2)dx=(-2+2)dx x+1x-x+1x+12x-x+12x-x+1x+111311 =ln|x+1|-2d(x2-x+1)+d(x-) 2x-x+12213(x-)2+222 4. =ln|x+1|x2-x+1+3arctan2x-1+C. 3 5. 解 xdx(x+1)(x+2)(x+3); xdx1413=(-(x+1)(x+2)(x+3)2x+2x+1x+3)dx 1 =(ln|x+2|-3ln|x+3|-ln|x+1|)+C
3、. 2x2+1 6. dx; (x+1)2(x-1)x2+111111dx=+- 解 2x+12x-1(x+1)2dx (x+1)2(x-1)111 =ln|x-1|+ln|x-1|+C 22x+111 =ln|x2-1|+C. 2x+11dx; 7. x(x2+1) 解 8. 解 1x1dx=(-)dx=ln|x|-ln(1+x2)+C. x(x2+1)x1+x221(x2+1)(x2+x); (x2+1)(x2+x)=(x-2x2+1-2x+1)dx dx11x+111dx11x+1 =ln|x|-ln|x+1|-2dx 22x+1112x11 =ln|x|-ln|x+1|-2dx-2dx
4、 24x+12x+1111 =ln|x|-ln|x+1|-ln(x2+1)-arctanx+C. 242dx 9. 2; (x+1)(x2+x+1) 解 dxx+1x=(-(x2+1)(x2+x+1)x2+x+1x2+1)dx =12x+1111+dx-ln(x2+1) 222x+x+12x+x+121111 =ln|x2+x+1|-ln(x2+1)+2dx 222x+x+11132x+1arctan+C. =ln|x2+x+1|-ln(x2+1)+22331x4+1dx; 11dx 解 4dx=2x+1(x+2x+1)(x2-2x+1) 10. 2121x+-x+2dx+42dx =242x
5、+2x+1x-2x+11212(2x+2)+(2x-2)-222dx-222dx =2244x+2x+1x-2x+12d(x2+2x+1)d(x2-2x+1)1dxdx-2+(2+2) =284x+2x+1x-2x+1x+2x+1x-2x+12x2+2x+122ln|2|+arctan(2x+1)+arctan(2x-1)+C. =84x-2x+14-x2-2dx; 11. 2(x+x+1)2-x2-2x-11dx=dx- 解 2(x2+x+1)2x2+x+1dx (x+x+1)2 =12x+1311dx-dx-x2+x+1dx 2(x2+x+1)22(x2+x+1)211311-2dx-dx
6、, =-2222x+x+12(x+x+1)x+x+1因为 1212x+122x+1dx=d=arctan, x2+x+131+(2x+1)23333而 (x2+x+1)2dx=dx1113(x+)2+2222xdx 由递推公式 得 (x2+a2)n=2a2(n-1)(x2+a2)n-1+(2n-3)(x2+a2)n-1, 1(x2+x+1)2dx=113(x+)2+2222dx 1dx =2(132)21dx12x+1222x+1(22+2)=2+arctan, 33x+x+1x+x+1x+x+133x+1112x+122x+122x+1-x2-2=-arctan-arctan+C 所以 2d
7、x2x2+x+12x2+x+1(x+x+1)23333 =- 12. 解 x+1x+x+12-43arctan2x+13+C. 3+sin2x; 3+sin2xdx=14-cos2xdx=1dxdtanx 4tan2x+31112tanx =dtanx=arctan+C. 43233tan2x+22 13. 3+cosxdx; 1xd11dx2 解 dx=xxx3+cosx21+cos2cos2(1+sec2)222xxtan2=1arctan2+C. =x222+tan22dtan13+cosxx令u=tan2dx12du 2u1+u23+1+u2或 = 14. 1u2+(2)2du=12a
8、rctanu2+C=12tanarctanx2+C. 212+sinxdx; xd1dx2 解 dx=xxxxx2+sinx2+2sincossin2(csc2+cot)22222xx1d(cot)d(cot+)222=- =- xxx13cot2+cot+1(cot+)2+222222x2cot+122+C. =-arctan3312+sinxx令u=tan2dx12du 2u1+u22+1+u2或 =1u+u+12du=113(u+)2+222du x2tan+122u+122+C. =arctan+C=arctan3333 15. dx1+sinx+cosx; xd(tan)dx1dx2
9、=ln|tanx|+C. 解 =xxx1+sinx+cosx22cos2(1+tan)1+tan222dx令u=tanx211+2u1+u21+u2+1-u2或 1+sinx+cosx21+u2du = 16. 1xdu=ln|u+1|+C=ln|tan+1|+C. u+12dx2sinx-cosx+5; dx2sinx-cosx+5令u=tanx2 解 1-+51+u21+u24u1-u221+u2du=13u2+2u+2du =13115(u+)2+233x3tan+113u+112+C. du=arctan+C=arctan5555或 -+51+u21+u2111 =2du=du 33u
10、+2u+215(u+)2+233dx2sinx-cosx+5令u=tanx214u1-u221+u2du x3tan+113u+112+C. =arctan+C=arctan5555 17. 1+31x+1dx; 解 1+3x+11dx=令3x+1=u1123udu=3(u-1+)du 1+u1+u33 =u2-3u+3ln|1+u|+C=3(x+1)2-33x+1+3ln(1+3x+1)+C. 22 18. (x)3+1x+1dx; 3 解 (x)3+112dx=(x)2-x+1dx=x2-x2+x+C. 23x+1dx; 19. x+1-1x+1+1 解 令x+1=uu-12dx2udu=
11、2(u-2+)du u+1u+1x+1+1x+1-11 =2(u2-2u+2ln|u+1|)+C 2 =(x+1)-4x+1+4ln(x+1+1)+C. 20. dxx+4x; 1 解 dxx+x4令x=u4u2+u4u3du =4(u-1+1)du=2u2-4u+4ln|1+u|+C 1+u =2x-4x+4ln(1+4x)+C. 1-xdx; 1+xx 21. 1-u2-4u1-xdx=du, 解 令, =u, 则x=1+x1+u2(1+u2)21-xdx1+u2-4u11=udu=2(+)du 222221+xx1-u(1+u)u-11+u =ln|u-1|+2arctanu+C u+11-x-1+x1-x+1+x|+2arctan1-x+C. 1+x =ln| 22. 3dx(x+1)(x-1)24. 解 令3u3+1x+16u2=u, 则x=3, dx=-3, 代入得 x-1u-1(u-1)23dx(x+1)2(x-1)4=-3333x+1du=-u+C=-+C. 222x-1
