find-0-1-dx-x-1-x- Tinku Tara June 4, 2023 Integration 0 Comments FacebookTweetPin Question Number 42435 by maxmathsup by imad last updated on 25/Aug/18 find∫01dxx+1−x Commented by maxmathsup by imad last updated on 25/Aug/18 letI=∫01dxx+1−xchangementx=sin2tgiveI=∫0π22sintcostdtsint+cost=∫0π2sin2t+cos2t+2sintcost−1sint+costdt=∫0π2(sint+cost)2−1sint+costdt=∫0π2(sint+cost)dt−∫0π2dtsint+cost=[sint−cost]0π2−∫0π2dtsint+cost=2−∫0π2dtsint+costchangementtan(t2)=ugive∫0π2dtsint+cost=∫0112u1+u2+1−u21+u22du1+u2=2∫01du2u+1−u2=−2∫01duu2−2u−1=−2∫01du(u−1)2−1=u−1=−α−2∫10−dαα2−1=−2∫01dαα2−1=−∫01(1α−1−1α+1)=−[ln∣α−1α+1∣01=+∞sothisintegraldiverges!.. Answered by tanmay.chaudhury50@gmail.com last updated on 25/Aug/18 I=x=sin2αdx=2sinαcosα∫0Π22sinαcosαsinα+cosαdα∫0Π2sin2α+cos2α+2sinαcosα−1sinα+cosαdα∫0Π2sinα+cosαdα−∫0Π2dαsinα+cosα∫0Π2sinα+cosαdα−∫0Π2sec2α22tanα2+1−tan2α2dα∫0Π2sinα+cosαdα+2∫0Π2d(tanα2)tan2α2−2tanα2+1−2∫0Π2sinα+cosαdα+2∫0Π2d(tanα2)(tanα2−1)2−1∫0Π2sinα+cosαdα−2∫0Π2d(tanα2)1−(tanα2−1)2∫0Π2sinα+cosαdα−2∫0Π2d(tanα2−1)(tanα2−1)2−1→∫dxx2−a2=∣−cosα+sinα∣0Π2−2×12ln∣tanα2−1−1tanα2−1+1∣0Π2={(−0+1)−(−1+0)}−{ln∣1−1−11−1+1∣−ln∣−20∣}=2−{ln∣−1∣−ln∣−∞∣} Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: solve-y-x-1-y-2x-2x-1-x-1-2-0-and-x-x-2-x-1-y-x-2-1-y-x-2-x-1-3-2-Next Next post: JS-1-lim-x-0-sin-tan-x-tan-sin-x-x-sin-x-2-lim-x-x-2-1-cos-2-x-1-cos-2-x-1-cos-2-x- Leave a Reply Cancel replyYour email address will not be published.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.