can-solve-dx-x-17-1-via-elementary-calculus- Tinku Tara June 3, 2023 Integration 0 Comments FacebookTweetPin Question Number 77422 by john santu last updated on 06/Jan/20 cansolve∫dxx17−1viaelementarycalculus? Commented by aliesam last updated on 07/Jan/20 ∫1x17−1dx=−∫11−x17(xx17)(x17x)dx=−∫(1−x)17−1(x)1−17(x)17−1dx=−∫(1−x)17−1(x17)117−1(x)17−1dx=−x2F1(1,117;1817;x17)+c Answered by mind is power last updated on 06/Jan/20 X17−1=∏16k=0(x−e2ikπ17)∫dxx17−1=∫dx∏16k=0(x−e2ikπ17)=∫{∑16k=0akx−e2ikπ17}dxak=117e2ikπ17.16=e−2ikπ1717⇒∫dxx17−1=∫∑16k=0e−2ikπ1717(x−e2ikπ17).dxwecanseethat∑16k=0e−2ikπ17x−e2ikπ17=(1x−1+∑8k=1(e−2ikπ17x−e2ikπ17+e−2iπ(17−k)17x−e2i(17−k)π17))=1x−1+∑8k=1(e−2ikπ17x−e2ikπ17+e2ikπ17x−e−2ikπ17)=1x−1+∑8k=12(xcos(2kπ17)−cos(2kπ17))x2−2cos(2πk17)x+1∫2xcos(2kπ17)−cos(2kπ17)x2−2cos(2kπ17)x+1dx=∫{cos(2kπ17)2x−2cos(2kπ17)x2−2cos(2kπ17)x+1+2cos2(2kπ17)−2cos(2kπ17)x2−2cos(2kπ17)x+1}dxsince∣cos(2kπ17)∣<1,∀k∈{1,…..8}∫cos(2kπ17)2x−2cos(2kπ17)x2−2xcos(2πk17)+1=cos(2kπ17)ln(x2−2xcos(2kπ17)x+1)∫2cos2(2kπ17)−2cos(2kπ17)x2−2cos(2kπ17)x+1=(2cos2(2kπ17)−2cos(2kπ17))∫dx(x−cos(2kπ17)2+sin2(2kπ17)=(2cos2(2kπ17)−2cos(2kπ17))arctan(xsin(2kπ17)−cot(2kπ17))weGet∫dxx17−1=117∫{1x−1+∑8k=1cos(2kπ17)2x−2cos(2kπ17)x2−2cos(2kπ17)x+1+(2cos2(2kπ17)−2cos(2kπ17)).1x2−2xcos(2kπ17)+1}dx=117{ln∣x−1∣+∑8k=1[cos(2kπ17)ln(x2−2xcos(2πk17)+1)+(2cos2(2kπ17)−2cos(2kπ17))arctan(xsin(2kπ17)−cot(2kπ17))]}+cc∈Rconstant Commented by john santu last updated on 06/Jan/20 thanksyousir Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: dy-dt-3t-2-y-t-2-y-0-1-y-t-Next Next post: 0-e-e-x-1-t-A-dx-A-and-t-are-constant- Leave a Reply Cancel replyYour email address will not be published.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.