求不定积分的一种新方法-分解积分法

:19990402:(1963),(),,:19993(,253023):,,:;,,,X=f(x)dx,X=a1X1+a2X2+arXr(1),ai(i=1,2,,r),xi(i=1,2,,r),,X1,X2,Xr:a11X1+a12X2++a1rXr=b1(2)a21X1+a22X2++a2rXr=b2ar1X1+ar2X2++arrXr=br:aijR(i,j=1,2,r)biR(i=1,2,r),aijû0,(2)X1,X2,Xr,X(1),X=f(x)dx,X,X1,X2,Xrr(2),,,1:sinx+2cosx2sinx+3cosxdx:01©1994-2011ChinaAcademicJournalElectronicPublishingHouse.Allrightsreserved.=sinx+2cosx2sinx+3cosxdx,X1=sinx2sinx+3cosxdx,X2=cosx2sinx+3cosxdx,2X1+3X2=2sinx+3cosx2sinx+3cosxdx=x+C1(3)-3X1+2X2=2cosx-3sinx2sinx+3cosxdx=lnû2sinx+3cosxû+C2(4)3(3)+2(4),(9+4)X2=3x+2lnû2sinx+3cosxû+3C1+2C2X2=313x+213lnû2sinx+3cosxû+3C1+2C2132(3)-3(4),(4+9)X1=2x-3lnû2sinx+3cosxû+2C1-3C2X1=213x-313lnû2sinx+3cosxû+2C1-3C2X1=213x-313lnû2sinx+3cosxû+2C1-3C113X=X1+2X2=813x+113lnû2sinx+3cosxû+C(C=8C1+C213)2:3+tgx2+5tgxdxX=3+tgx2+5tgxdx,X1=dx2+5tgx,X2=tgx2+5tgxdx,,2X1+5X2=dx=x+C1(5)5X1-2X2=5-2tgx2+5tgxdx=52-tgx1+52tgxdx,tg(arctg52-x)dx=lnûcos(arctg52-x)û+C2(6)2(5)+5(6)(4+25)X1=2x+5lnûcos(arctg25-x)û+2C1+5C2X1=229x+529lnûcos(arctg25-x)û+2C1+5C229X2=529-229lnûcos(arctg52-x)û+5C1-2C229X=3X1+X2=1129x-1329lnûcos(arctg52-x)û+C(C=11C1-13C229)3:a1sin2x+2b1sinxcosx+e1cos2xasinx+bcosxdxX=a1sin2x+2b1sinxcosx+e1cos2xasinx+bcosxdx,X1=sin2xasinx+bcosxdx,X2=2sinxcosxasinx+bcosxdx,X3=cos2xasinx+bcosxdx,11©1994-2011ChinaAcademicJournalElectronicPublishingHouse.Allrightsreserved.=dxasinx+bcosx=1a2+b2lntgx+arctgba2+C1(7)a2X1-b2X3=(asinx-bcosx)dx=-asinx-bsinx+C2(8)a2X1+abX2+b2X3=(asinx+bcosx)dx=-acosx+bsinx+C3(9)a2(7)-(8),X3=a2(a2+b2)3ö2lntgx+arctgba2+aa2+b2cosx+ba2+b2sinx+a2C1-C1a2+b2b2(7)+(8),X1=a2(a2+b2)3ö2lntgx+arctgba2-aa2+b2cosx-ba2+b2sinx+b2C1+C2a2+b2(9)-(8),X2=2a(sinx-bX3)+C3-C2ab=-2ab(a2+b2)3ö2lntgx+arctgba2+2aa2+b2sinx-2ba2+b2cosx+C3-C2ab:X=a1X1+b1X2+e1X3=a1b2+a2e1-ab1ab(a2+b2)3ö2lntgx+arctgba2+e1b-a1b+2b1aa2+b2sinx+e1a-a1a-2b1ba2+b2cosx+C:C=a1b2c1+a1c2a2+b2+b1C3-b1C2ab+e1a2C1-e1C2a2+b2,,X,X1,X2,XrX,:,,198521©1994-2011ChinaAcademicJournalElectronicPublishingHouse.Allrightsreserved.