clculate-0-1-x-x-2-2x-2-dx- Tinku Tara June 4, 2023 Integration 0 Comments FacebookTweetPin Question Number 31069 by abdo imad last updated on 02/Mar/18 clculate∫01xx2−2x+2dx Commented by abdo imad last updated on 03/Mar/18 letputI=∫01xx2−2x+2dxI=∫01x(x−1)2+1dxthech.x−1=shtI=∫argsh(−1)o(1+sht)chtchtdt=−∫0ln(−1+2)ch2tdt−∫0ln(−1+2)shtch2tdtbutwehave∫0ln(−1+2)ch2tdt=∫0ln(−1+2)1+ch(2t)2dt=12ln(−1+2)+14[sh(2t)]0ln(−1+2)=12ln(−1+2)+14sh(2ln(−1+2))and∫0ln(−1+2)shtch2tdt=13[ch3t]0ln(−1+2)=13(ch3(ln(−1+2)−1)byusingshu=eu−e−u2andchu=eu+e−u2tbevalueofIisdetermined. Answered by Joel578 last updated on 02/Mar/18 I=∫01(x−1+1)x2−2x+2dx=∫01(x−1)x2−2x+2dx+∫01(x−1)2+1dxI1=∫(x−1)x2−2x+2dxu=x2−2x+2→du=2(x−1)dxI1=12∫(x−1)u.dux−1=13uu=13(x2−2x+2)32+CI2=∫(x−1)2+1dxx−1=tanθ→dx=sec2θdθI2=∫tan2θ+1.sec2θdθ=∫sec3θdθ=tanθsecθ2−12ln∣secθ+tanθ∣+C=(x−1)x2−2x+22−12ln∣x2−2x+2+x−1∣+CI=[I1+I2]01=[13(x2−2x+2)32+(x−1)x2−2x+22−12ln∣x2−2x+2+x−1∣]01=(13+0−0)−(223−22−12ln(2−1))=1−223+12(2+ln(2−1)) Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: lim-n-1-n-k-1-n-k-Next Next post: Please-how-will-you-evaluate-dx- Leave a Reply Cancel replyYour email address will not be published.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.