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f-x-x-2-2x-5-find-f-x-f-1-x-dx-and-f-1-x-f-x-dx-




Question Number 147466 by mathmax by abdo last updated on 21/Jul/21
f(x)=x^2 −2x+5  find ∫ ((f(x))/(f^(−1) (x)))dx   and ∫  ((f^(−1) (x))/(f(x)))dx
f(x)=x22x+5findf(x)f1(x)dxandf1(x)f(x)dx
Answered by EDWIN88 last updated on 21/Jul/21
f(x)=(x−1)^2 +4 ⇒x=±(√(x−4))+1  f^(−1) (x)=1±(√(x−4))  (1) ∫ (((x−1)^2 +4)/(1+(√(x−4)))) dx =  let x−4=u^2  →dx=2u du  ∫ (((u^2 +3)^2 +4)/(1+u)) (2u du)=  2∫ ((u^4 +6u^3 +13u)/(u+1))=  2∫(u^3 +5u^2 −5u+18−((18)/(u+1)))du=  2((1/4)u^4 +(5/3)u^3 −(5/2)u^2 +18u−18ln (u+1))+c=  (1/2)(x−4)^2 +((10)/3)(x−4)^(3/2) −5(x−4)+36(√(x−4))−36 ln (1+(√(x−4)))+c
f(x)=(x1)2+4x=±x4+1f1(x)=1±x4(1)(x1)2+41+x4dx=letx4=u2dx=2udu(u2+3)2+41+u(2udu)=2u4+6u3+13uu+1=2(u3+5u25u+1818u+1)du=2(14u4+53u352u2+18u18ln(u+1))+c=12(x4)2+103(x4)3/25(x4)+36x436ln(1+x4)+c
Answered by mathmax by abdo last updated on 21/Jul/21
f(x)=y ⇒x=f^(−1) (y) ⇒x^2 −2x+5=y ⇒x^2 −2x+5−y=0  Δ^′  =1−(5−y)=y−4  for y≥4  we get  x_1 =1+(√(y−4))  and x_2 =1−(√(y−4))  ⇒f^(−1) (x)=1+(√(x−4)) or f^(−1) (x)=1−(√(x−4))  case1 ∫ ((f(x))/(f^(−1) (x)))dx=∫ ((x^2 −2x+5)/(1+(√(x−4))))dx =_(x−4=y^2 )   ∫  (((4+y^2 )^2 −2(4+y^2 )+5)/(1+y))2ydy  =2∫  ((16+8y^2  +y^4 −8−2y^2 +5)/(y+1))ydy  =2∫  ((y^5 +6y^3 +13y)/(y+1))dy =_(y+1=t)   2∫  (((t−1)^5  +6(t−1)^3  +13(t−1))/t)dt  =2∫   ((Σ_(k=0) ^5  C_5 ^k t^k (−1)^(5−k)  +6(t^3 −3t^2  +3t−1)+13t−13)/t)dt  =−2Σ_(k=0) ^5  C_5 ^(k ) (−1)^k ∫ t^(k−1) dt+12∫  (t^2 −3t+3−(1/t))dt+2∫(13−((13)/t)dt  =−2log∣t∣−2Σ_(k=1) ^5  C_5 ^k (−1)^k  (t^k /k) +12{(t^3 /3)−(3/2)t^2 +3t−log∣t∣)  +26t−26log∣t∣ +C  t=y+1=(√(x−4))+1 ⇒  ∫  ((f(x))/(f^(−1) (x)))dx=−2log∣(√(x−4))+1∣−2Σ_(k=1) ^5  C_5 ^k  (((−1)^k )/k)((√(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))∣ +C  rest to treat case 2 .....be continued...  =
f(x)=yx=f1(y)x22x+5=yx22x+5y=0Δ=1(5y)=y4fory4wegetx1=1+y4andx2=1y4f1(x)=1+x4orf1(x)=1x4case1f(x)f1(x)dx=x22x+51+x4dx=x4=y2(4+y2)22(4+y2)+51+y2ydy=216+8y2+y482y2+5y+1ydy=2y5+6y3+13yy+1dy=y+1=t2(t1)5+6(t1)3+13(t1)tdt=2k=05C5ktk(1)5k+6(t33t2+3t1)+13t13tdt=2k=05C5k(1)ktk1dt+12(t23t+31t)dt+2(1313tdt=2logt2k=15C5k(1)ktkk+12{t3332t2+3tlogt)+26t26logt+Ct=y+1=x4+1f(x)f1(x)dx=2logx4+12k=15C5k(1)kk(x4+1)k+4(1+x4)318(1+x4)2+36(1+x4)+3(1+x4)log1+x4+26(1+x4)26log1+x4+Cresttotreatcase2..becontinued=

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