find-dx-x-x-1-x-2-x- Tinku Tara June 4, 2023 Integration 0 Comments FacebookTweetPin Question Number 122205 by mathmax by abdo last updated on 14/Nov/20 find∫dxx(x+1)x2+x Commented by liberty last updated on 15/Nov/20 ∫dx(x2+x)32=∫dxx3(1+x−1)32=∫x−2dxx(1+x−1)32;[u23=1+1x]⇒[23u−13du=−x−2dx]J=∫(u23−1)(−23u−13du)(1u)J=−23∫(u23−1)(u−43)duJ=−23∫(u−23−u−43)duJ=−23[3u13+3u−13]+cJ=−2x+1x−2xx+1+cJ=−2(x+1+xx2+x)+c=−4x+2x2+x+c Commented by benjo_mathlover last updated on 15/Nov/20 letJ(x)=−4x+2x2+xdJ(x)dx=−[4x2+x−(4x+2)(2x+12x2+x)x2+x]=−[4x2+x−(2x+1)2x2+xx2+x]=−[4x2+4x−4x2−4x−1(x2+x)x2+x]=1x(x+1)x2+xcorrectsirLiberty. Answered by TANMAY PANACEA last updated on 14/Nov/20 ∫(x+1)−xx(x+1)x2+xdx∫dxxx2+x−∫dx(x+1)x2+xa=1x→dx=−1a2dab=1x+1→dx=−1b2db∫−daa2×1a×1a2+1a−∫−dbb2×1b(1b−1)2+(1b−1)∫−da1+a2−∫dbb×1b2−2b+1+1b−1∫−da1+a2−∫db1−b∫−da1+a2+∫d(1−b)1−b−ln(a+1+a2)+(1−b)1212+C−ln(1x+1+1x2+2(1−1x+1)12+c Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: Question-122203Next Next post: how-do-i-expand-p-q-r-3-i-need-general-principle- Leave a Reply Cancel replyYour email address will not be published.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.