有理函數(shù)積分12例

1、有理函數(shù)積分12例例1：求x-2x2-x+1dx解：由于x2-x+1的導(dǎo)數(shù)是2x-1用分子x-2去除2x-1就有x-22x-1商12余-32也就是x-2=122x-1-32原式=122x-1-32x2-x+1dx=122x-1x2-x+1dx-321x2-x+1dx=12lnx2-x+1-321(x-12)2+(32)2dx1x2+a2dx=1aarctanxa=12lnx2-x+1-3223arctan(x-12)=12lnx2-x+1-3arctan23(x-12)例2：求2x2+3x-1x3-xdx解：2x2+3x-1x3-x=2x2+3x-1xx-1(x+1)=Ax-1+Bx+Cx+1

2、兩邊同乘x-1得2x2+3x-1x(x+1)=A+Bx-1x+Cx-1x+1帶入x=1解得A=2，同理得到B=1、C=-12x2+3x+1x3-xdx= 2x-1+1x-1x+1dx=In|x(x-1)2x+1|例3：求2x2+x+1x(x2+1)dx1 / 10解：2x2+x+1x(x2+1)=Ax+Bx+Cx2+1兩邊乘以x得2x2+x+1x2+1=A+x(Bx+C)x2+1帶x=0得A=1兩邊乘以x2+1得2x2+x+1x=A(x2+1)x+(Bx+C)帶入x=i得B=1、C=12x2+x+1x(x2+1)dx=1x+x+1x2+1dx=1x+xx2+1+1x2+1dx=Inxx2+1+

3、arctanx例4：求x+9x3+2x-3dx解：x+9x3+2x-3=x+9x-1(x2+x+3)=Ax-1+Bx+Cx2+x+3容易解得A=2兩邊同時(shí)乘x2+x+3，并令x2+x+3=0有x+9x-1=Bx+C (如果讓左邊沒(méi)有分子就好辦了)為了利用x2+x+3，做如下處理x+9(x+2)x-1(x+2)=Bx+C (x+2不是隨便帶入的，而是x2+x+3x-1的商)得x2+11x+18x2+x-2=10x+15-5=Bx+C解得B=-2、C=-3x+9x3+2x-3dx=2x-1-2x+3x2+x+3dx=2x-1-2x+1x2+x+3-2x2+x+3dx分母導(dǎo)數(shù)的倍數(shù)加常數(shù)=In(x-

4、1)2-Inx2+x+3-411arctan2x+111=In(x-1)2x2+x+3-411arctan2x+111例5：求8x(x+1)2(x-1)dx解：8x(x+1)2(x-1)=Ax-1+B(x+1)2+C(x+1)容易求的A=2、B=4兩邊同乘x得8x2(x+1)2(x-1)=2xx-1+4x(x+1)2+Cx(x+1)令x兩邊求極限得，0=2+C，也就是C=-28x(x+1)2(x-1)dx=2x-1+4(x+1)2-2x+1dx=In(x-1)2-4x+1-In(x+1)2=In(x-1x+1)2-4x+1例6：求x3-4x2-4x+23(x-3)4dx解：令y=x-3則原式化

5、簡(jiǎn)為(分子是高次線性)x3-4x2-4x+23(x-3)4=y3+5y2-y+2y4=1y+5y2-1y3+2y4x3-4x2-4x+23(x-3)4dx=Iny-5y-12y2+23y3=Inx-3-5x-3-12(x-3)2+23(x-3)3例7：求1(x2+4)3dx解：=14x2+4-x2(x2+4)3dx降次=141(x2+4)2dx-14x2(x2+4)3dx=141(x2+4)2dx-14x2-4xd1(x2+4)2=141(x2+4)2dx+116x(x2+4)2-1(x2+4)2dx=3161(x2+4)2dx+x16(x2+4)2=364x2+4-x2(x2+4)2dx+x

6、16(x2+4)2降次=3641x2+4dx-364x2(x2+4)2dx+x16(x2+4)2=3641x2+4dx-(364x2-2xd1x2+4)+x16(x2+4)2=3641x2+4dx+3x128x2+4-31281x2+4dx+x16(x2+4)2=31281x2+4dx+3x128x2+4+x16(x2+4)2=3256arctanx2+3x128(x2+4)+x16(x2+4)2例8：求4x3+4x2+1x5+2x3+xdx解：4x3+4x2+1x(x2+1)2=Ax+Bx+C(x2+1)2+Dx+Ex2+1容易解得A=1、B=3、C=-4兩邊同乘x4x3+4x2+1(x2+

7、1)2=xx+3x2-4x(x2+1)2+Dx2+Exx2+1然后求極限x得D=-1，然后帶入x=1得E=44x3+4x2+1x5+2x3+xdx=1x+3x-4(x2+1)2-x-4x2+1dx=1xdx+32×2x-4(x2+1)2dx-x-4x2+1dx=Inx+321x2+1-41(x2+1)2dx-x-4x2+1dx=Inx-321x2+1+2arctanx-2xx2+1-12In(x2+1)=12In(x2x2+1)-2x+1.5x2+1+2arctanx例9：求x3+1x2-2x+3dx解：利用多項(xiàng)式除法x3+1=x(x2-2x+3)+2(x2-2x+3)+x-5x3+

8、1x2-2x+3=x+2+x-5x2-2x+3x-5x2-2x+3dx=122x-2x2-2x+3dx-4(x-1)2+22dx分離出一次和一個(gè)常數(shù)=12lnx2-2x+3-42arctanx-12x3+1x2-2x+3dx=12x2+2x+12lnx2-2x+3-22arctanx-12例10：求5x+6x3+4x2+5x+2dx解：x=-1是x3+4x2+5x+2的一個(gè)根x3+4x2+5x+2x+1=x2+3x+2=x+1(x+2)x3+4x2+5x+2=(x+1)2(x+2)5x+6x3+4x2+5x+2=A(x+1)2+Bx+1+Cx+2很容易求得A=1、C=-4兩邊同時(shí)乘x有5x2+

9、6xx3+4x2+5x+2=x(x+1)2+Bxx+1+-4xx+2求x時(shí)的極限有0=0+B-4得B=4原式=-1x+1+4lnx+1-4ln|x+2|例11：求xx+1(x+2)2(x+3)3dx解：Ax+1+Bx+2+C(x+2)2+Dx+3+E(x+3)2+F(x+3)3容易求的是最高次極點(diǎn)的系數(shù)A=-18、C=2、F=32兩邊乘x然后求x有：0=-18+B+D(1)帶入x=00=B2+D3+E9+3172(2)帶入x=111152=B3+D4+E16+2111152(3)由(1)、(2)、(3)解得B=-5D=418E=134xx+1(x+2)2(x+3)3=-18(x+1)-5x+2

10、+2(x+2)2+418(x+3)+134(x+3)2+32(x+3)3xx+1(x+2)2(x+3)3dx=-18lnx+1-5lnx+2-2x+2+418lnx+3-134(x+3)-3(x+3)2例12：求dxx2+4x+4(x2+4x+5)2解：1x2+4x+4(x2+4x+5)2=1(x+2)2(x2+4x+5)21x2+4x+4(x2+4x+5)2=Ax+2+B(x+2)2+Cx+Dx2+4x+5+Ex+F(x2+4x+5)2容易求得B=1由1i2+4i+4=Ei+F解得E=-425、F=325兩邊乘x，并求x得：0=A+C(1)帶入x=0得1100=A2+14+D5+3625(2

11、)帶入X=-1得14=A+1+-C+D2+7100(3)由(1)、(2)、(3)解得A=-104125C=104125D=1071251x2+4x+4(x2+4x+5)2=-104125(x+2)+1(x+2)2+104x+107125(x2+4x+5)+-4x+325(x2+4x+5)2dxx2+4x+4(x2+4x+5)2=-104125lnx+2-1x+2+104250250x+500-101125(x2+4x+5)dx-425x-34(x2+4x+5)2dx=-104125lnx+2-1x+2+52125ln(125x2+500x+625)-101125arctanx+2-425122

12、x+4-114(x2+4x+5)2dx=-104125lnx+2-1x+2+52125ln(125x2+500x+625)-101125arctanx+2+450(x2+4x+5)-1125(x+2)2+1-(x+2)2(x+2)2+12dx=-104125lnx+2-1x+2+52125ln(125x2+500x+625)-101125arctanx+2+450(x2+4x+5)-1125ln(x2+4x+5)+1125(x+2)2(x+2)2+12dx=-104125lnx+2-1x+2+52125ln(125x2+500x+625)-101125arctanx+2+450(x2+4x+5)-1125lnx2+4x+5+1125(x+2)2(x+2)2+12/-2(x+2)(x+2)2+12d1(x+2)2+1=-104125lnx+2-1x+2+52125ln(125x2+500x+625)-101125arctanx+2+450(x2+4x+5)-1125lnx2+4x+5-1150(x+2)d1(x+2)2+1=-104125lnx+2-1x+2+52125ln(125x2+500x+625)-101125arctanx+2