有理函数积分12例

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1、有理函数积分12例x−2例1：求dxx2−x+1解：由于x2−x+1的导数是2x−1用分子x−2去除2x−1就有x−21−3商余2x−12213也就是x−2=2x−1−22132x−1−22原式=dxx2−x+112x−131=dx−dx2x2−x+12x2−x+1123111x=lnx−x+1−dxdx=arctan22(x−1)2+(3)2x2+a2aa2212321=lnx−x+1−arctan(x−)22321221=lnx−x+1−3arctan(x−)2322x2+3x−1例2：求dxx3−x2x2+3x−12x2+3x−

2、1ABC解：==++x3−xxx−1(x+1)x−1xx+1两边同乘x-1得2x2+3x−1x−1x−1=A+B+Cx(x+1)xx+1带入x=1解得A=2，同理得到B=1、C=-12x2+3x+1211dx=+−dxx3−xx−1xx+1x(x−1)2=In

3、

4、x+12x2+x+1例3：求dxx(x2+1)2x2+x+1ABx+C解：=+x(x2+1)xx2+1两边乘以x得2x2+x+1x(Bx+C)=A+x2+1x2+1带x=0得A=12两边乘以x+1得2x2+x+1A(x2+1)=+(Bx+C)xx带入x=i得B=1、C=12x2+x+

5、11x+1dx=+dxx(x2+1)xx2+11x1=++dxxx2+1x2+1=Inxx2+1+arctanxx+9例4：求dxx3+2x−3x+9x+9ABx+C解：==+x3+2x−3x−1(x2+x+3)x−1x2+x+3容易解得A=222两边同时乘x+x+3，并令x+x+3=0有x+9=Bx+C(如果让左边没有分子就好办了)x−12为了利用x+x+3，做如下处理x+9(x+2)x2+x+3=Bx+C(x+2不是随便带入的，而是的商)x−1(x+2)x−1x2+11x+1810x+15得==Bx+Cx2+x−2−5解得B=-2、C=

6、-3x+922x+3x3+2x−3dx=x−1−x2+x+3dx22x+12=−−dx分母导数的倍数加常数x−1x2+x+3x2+x+32242x+1=In(x−1)−Inx+x+3−arctan1111(x−1)242x+1=In−arctanx2+x+311118x例5：求dx(x+1)2(x−1)8xABC解：=++(x+1)2(x−1)x−1(x+1)2(x+1)容易求的A=2、B=4两边同乘x得8x22x4xCx=++(x+1)2(x−1)x−1(x+1)2(x+1)令x→∞两边求极限得，0=2+C，也就是C=-28x242dx

7、=+−dx(x+1)2(x−1)x−1(x+1)2x+1242=In(x−1)−−In(x+1)x+1x−124=In()−x+1x+1x3−4x2−4x+23例6：求dx(x−3)4解：令y=x-3则原式化简为(分子是高次线性)x3−4x2−4x+23y3+5y2−y+21512==+−+(x−3)4y4yy2y3y4x3−4x2−4x+23512dx=Iny−−+(x−3)4y2y23y3512=Inx−3−−+x−32(x−3)23(x−3)31例7：求dx(x2+4)31x2+4−x2解：=dx降次4(x2+4)3111x2=d

8、x−dx4(x2+4)24(x2+4)3111x21=dx−d4(x2+4)24−4x(x2+4)2111x1=dx+[−dx]4(x2+4)216(x2+4)2(x2+4)231x=dx+16(x2+4)216(x2+4)23x2+4−x2x=dx+降次64(x2+4)216(x2+4)2313x2x=dx−dx+64x2+464(x2+4)216(x2+4)2313x21x=dx−(d)+64x2+464−2xx2+416(x2+4)2313x31x=dx+−dx+64x2+4128x2+4128x2+416(x2+4

9、)2313xx=dx++128x2+4128x2+416(x2+4)23x3xx=arctan++2562128(x2+4)16(x2+4)24x3+4x2+1例8：求dxx5+2x3+x4x3+4x2+1ABx+CDx+E解：=++x(x2+1)2x(x2+1)2x2+1容易解得A=1、B=3、C=-4两边同乘x4x3+4x2+1x3x2−4xDx2+Ex=++(x2+1)2x(x2+1)2x2+1然后求极限x→∞得D=-1，然后带入x=1得E=44x3+4x2+113x−4x−4x5+2x3+xdx=x+(x2+1)2−x2+1dx31

10、2×2x−4x−4=dx+dx−dxx(x2+1)2x2+1311x−4=Inx+−4dx−dx2x2+1(x2