find-x-2-x-2-1-dx- Tinku Tara June 4, 2023 Integration 0 Comments FacebookTweetPin Question Number 36424 by abdo.msup.com last updated on 01/Jun/18 find∫x2x2−1dx Commented by prof Abdo imad last updated on 03/Jun/18 changementx=ch(t)giveI=∫ch2tsh(t)sh(t)dt=∫(ch(t)sh(t))2dt=14∫sh2(2t)dt=18∫(ch(4t)−1)dt=−t8+132sh(4t)=−18argch(x)+132sh{4argch(x)}=−18ln(x+x2−1)+132sh{4ln(x+x2−1)}butsh{4ln(x+x2−1)=sh{ln(x+x2−1)2}=(x+x2−1)2+(x+x2−1)−22I=164{(x+x2−1)2+(x+x2−1)−2}−18ln(x+x2−1). Commented by abdo mathsup last updated on 04/Jun/18 I=164{(x+x2−1)2−(x+x2−1)−2}−18ln(x+x2−1). Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: find-dt-t-t-2-t-1-Next Next post: Question-101961 Leave a Reply Cancel replyYour email address will not be published.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.