1-z-6-1-dz- Tinku Tara November 6, 2024 None 0 Comments FacebookTweetPin Question Number 213451 by issac last updated on 06/Nov/24 ∫1z6−1dz=?? Answered by Frix last updated on 06/Nov/24 ∫dzz6−1=∑4k=1IkI1=16∫dzz−1=ln∣z−1∣6I2=−16∫dzx+1=−ln∣z+1∣6I3=16∫z−2z2−z+1dz=ln(z2−z+1)12−36tan−12z−13I4=−16∫z+2z2+z+1dz=−ln(z2+z+1)12−36tan−12z+13∫dzz6−1==112ln(z−1)2(z2−z+1)(z+1)2(z2+z+1)+36tan−13zz2−1+V Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: Question-213504Next Next post: Question-213482 Leave a Reply Cancel replyYour email address will not be published. Required fields are marked *Comment * Name * Save my name, email, and website in this browser for the next time I comment.