find-ln-x-x-1-dx- Tinku Tara June 4, 2023 Integration 0 Comments FacebookTweetPin Question Number 38719 by maxmathsup by imad last updated on 28/Jun/18 find∫ln(x+x+1)dx Answered by behi83417@gmail.com last updated on 29/Jun/18 I=xln(x+x+1)−∫x12x+12x+1x+x+1dx==do−∫x2xx+1dx=do−∫x2x+1dxx=tg2t⇒dx=2tgt(1+tg2t)dt⇒∫xx+1dx=∫tgt.2tgt(1+tg2t)dtsect==∫2tg2t.sectdt=2∫(sec2t−1)sectdt==2∫(sec3t−sect)dt==2(sect.tgt2+12∫sectdt−∫sectdt)==sect.tgt−ln(sect+tgt)+const⇒I=x.ln(x+x+1)−12x.x+1−−12ln(x+x+1)+const.I=12[(2x−1).ln(x+x+1)−x2+x]+const. Commented by maxmathsup by imad last updated on 30/Jun/18 thankyousirBehi. Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: 1-find-f-x-0-pi-ln-2-x-cos-d-2-calculate-0-pi-ln-2-cos-d-Next Next post: let-f-x-1-2x-2-x-2-3-1-calculate-lim-x-f-x-and-lim-x-f-x-2-calculate-lim-x-f-x-x-and-lim-x-f-x-x-3-give-the-assymtote-to-graph-C-f-4-give-the-assy Leave a Reply Cancel replyYour email address will not be published.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.
![I=xln((√x)+(√(x+1)))−∫x(((1/(2(√x)))+(1/(2(√(x+1)))))/( (√x)+(√(x+1))))dx= =do−∫(x/(2(√x)(√(x+1))))dx=do−∫((√x)/(2(√(x+1))))dx x=tg^2 t⇒dx=2tgt(1+tg^2 t)dt ⇒∫((√x)/( (√(x+1))))dx=∫((tgt.2tgt(1+tg^2 t)dt)/(sect))= =∫2tg^2 t.sectdt=2∫(sec^2 t−1)sectdt= =2∫(sec^3 t−sect)dt= =2(((sect.tgt)/2)+(1/2)∫sectdt−∫sectdt)= =sect.tgt−ln(sect+tgt)+const ⇒I=x.ln((√x)+(√(x+1)))−(1/2)(√x).(√(x+1))− −(1/2)ln((√x)+(√(x+1)))+const. ■ I=(1/2)[(2x−1).ln((√x)+(√(x+1)))−(√(x^2 +x))]+const.](https://www.tinkutara.com/question/Q38791.png)
