f-x-x-2-2x-5-find-f-x-f-1-x-dx-and-f-1-x-f-x-dx- Tinku Tara June 4, 2023 Relation and Functions 0 Comments FacebookTweetPin Question Number 147466 by mathmax by abdo last updated on 21/Jul/21 f(x)=x2−2x+5find∫f(x)f−1(x)dxand∫f−1(x)f(x)dx Answered by EDWIN88 last updated on 21/Jul/21 f(x)=(x−1)2+4⇒x=±x−4+1f−1(x)=1±x−4(1)∫(x−1)2+41+x−4dx=letx−4=u2→dx=2udu∫(u2+3)2+41+u(2udu)=2∫u4+6u3+13uu+1=2∫(u3+5u2−5u+18−18u+1)du=2(14u4+53u3−52u2+18u−18ln(u+1))+c=12(x−4)2+103(x−4)3/2−5(x−4)+36x−4−36ln(1+x−4)+c Answered by mathmax by abdo last updated on 21/Jul/21 f(x)=y⇒x=f−1(y)⇒x2−2x+5=y⇒x2−2x+5−y=0Δ′=1−(5−y)=y−4fory⩾4wegetx1=1+y−4andx2=1−y−4⇒f−1(x)=1+x−4orf−1(x)=1−x−4case1∫f(x)f−1(x)dx=∫x2−2x+51+x−4dx=x−4=y2∫(4+y2)2−2(4+y2)+51+y2ydy=2∫16+8y2+y4−8−2y2+5y+1ydy=2∫y5+6y3+13yy+1dy=y+1=t2∫(t−1)5+6(t−1)3+13(t−1)tdt=2∫∑k=05C5ktk(−1)5−k+6(t3−3t2+3t−1)+13t−13tdt=−2∑k=05C5k(−1)k∫tk−1dt+12∫(t2−3t+3−1t)dt+2∫(13−13tdt=−2log∣t∣−2∑k=15C5k(−1)ktkk+12{t33−32t2+3t−log∣t∣)+26t−26log∣t∣+Ct=y+1=x−4+1⇒∫f(x)f−1(x)dx=−2log∣x−4+1∣−2∑k=15C5k(−1)kk(x−4+1)k+4(1+x−4)3−18(1+x−4)2+36(1+x−4)+3(1+x−4)−log∣1+x−4∣+26(1+x−4)−26log∣1+x−4∣+Cresttotreatcase2…..becontinued…= Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: Question-16392Next Next post: if-n-gt-2-n-N-then-prove-that-2n-1-n-2n-n-lt-2n-1-n- Leave a Reply Cancel replyYour email address will not be published.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.