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1-z-6-1-dz-




Question Number 213451 by issac last updated on 06/Nov/24
∫  (1/(z^6 −1)) dz=??
1z61dz=??
Answered by Frix last updated on 06/Nov/24
∫(dz/(z^6 −1))=Σ_(k=1) ^4 I_k   I_1 =(1/6)∫(dz/(z−1))=((ln ∣z−1∣)/6)  I_2 =−(1/6)∫(dz/(x+1))=−((ln ∣z+1∣)/6)  I_3 =(1/6)∫((z−2)/(z^2 −z+1))dz=((ln (z^2 −z+1))/(12))−((√3)/6)tan^(−1)  ((2z−1)/( (√3)))  I_4 =−(1/6)∫((z+2)/(z^2 +z+1))dz=−((ln (z^2 +z+1))/(12))−((√3)/6)tan^(−1)  ((2z+1)/( (√3)))    ∫(dz/(z^6 −1))=  =(1/(12))ln (((z−1)^2 (z^2 −z+1))/((z+1)^2 (z^2 +z+1))) +((√3)/6)tan^(−1)  (((√3)z)/(z^2 −1)) +V
dzz61=4k=1IkI1=16dzz1=lnz16I2=16dzx+1=lnz+16I3=16z2z2z+1dz=ln(z2z+1)1236tan12z13I4=16z+2z2+z+1dz=ln(z2+z+1)1236tan12z+13dzz61==112ln(z1)2(z2z+1)(z+1)2(z2+z+1)+36tan13zz21+V

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