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let-f-x-dt-x-t-t-2-x-2-1-determine-a-explicit-form-of-f-x-2-determine-dt-x-2-t-2-4-and-dt-x-1-t-2-1-




Question Number 59576 by maxmathsup by imad last updated on 12/May/19
let f(x) =∫       (dt/((x+t)(√(t^2 −x^2 ))))  1) determine a explicit form of f(x)  2) determine ∫     (dt/((x+2)(√(t^2 −4))))  and  ∫      (dt/((x+1)(√(t^2 −1))))
letf(x)=dt(x+t)t2x21)determineaexplicitformoff(x)2)determinedt(x+2)t24anddt(x+1)t21
Commented by tanmay last updated on 12/May/19
∫(dt/((t+2)(√(t^2 −4)))) and ∫(dt/((t+1)(√(t^2 −1))))  ∫(dt/((t+a)(√(t^2 −a^2 ))))  k=(1/(t+a))  t+a=(1/k)→dt=((−dk)/k^2 )  ∫(k/( (√(((1/k))((1/k)−2a)))))×((−dk)/k^2 )  ∫((−dk)/(k(√((1−2ak)/k^2 ))))   p^2 =1−2ak   →2pdp=−2adk  ∫((pdp)/(a×p))  (1/a)×p+c  =((√(1−2ak))/a)+c  =((√(1−((2a)/(t+a)) ))/a)+c  now put a=2    ((√(1−(4/(t+2))))/2)+c....(ans for Qno 1)  now put a=1  ((√(1−(2/(t+1))))/1)+c....(ans for Qno 2)
dt(t+2)t24anddt(t+1)t21dt(t+a)t2a2k=1t+at+a=1kdt=dkk2k(1k)(1k2a)×dkk2dkk12akk2p2=12ak2pdp=2adkpdpa×p1a×p+c=12aka+c=12at+aa+cnowputa=214t+22+c.(ansforQno1)nowputa=112t+11+c.(ansforQno2)
Commented by maxmathsup by imad last updated on 12/May/19
2)the Q.is determine ∫  (dt/((t+2)(√(t^2 −4))))  and ∫  (dt/((t+1)(√(t^2 −1))))
2)theQ.isdeterminedt(t+2)t24anddt(t+1)t21
Commented by maxmathsup by imad last updated on 14/May/19
changement t =x ch(u) give f(x)=∫  ((xsh(u))/((x+xch(u))∣x∣sh(u))) du  =ξ(x) ∫   (du/(x(1+ch(u)))) =((ξ(x))/x) ∫   (du/(1+((e^u  +e^(−u) )/2))) =((2ξ(x))/x) ∫   (du/(2 +e^u  +e^(−u) ))  =_(e^u =α) ((2ξ(x))/x) ∫   (1/(2 +α +α^(−1) )) (dα/α) =((2ξ(x))/x) ∫    (dα/(2α +α^2  +1))  =((2ξ(x))/x) ∫  (dα/((α+1)^2 )) =−((2ξ(x))/x) (1/(α+1)) +c =−(2/x)ξ(x) (1/(e^u  +1))  but  u =argch((t/x)) =ln((t/x)+(√((t^2 /x^2 )−1))) ⇒e^u  =(t/x)+(√((t^2 /x^2 )−1)) ⇒  f(x) =−(2/x)ξ(x) (1/((t/x)+((√(t^2 −x^2 ))/(∣x∣)))) +c =−((2ξ(x))/(t +ξ(x)(√(t^2 −x^2 )))) +c  with  ξ(x) =1 if x>0  and ξ(x)=−1 if x<0 so   f(x) =((−2)/(t+(√(t^2 −x^2 )))) +c  if x>0  f(x)=(2/(t−(√(t^2 −x^2 ))))+c if x<0
changementt=xch(u)givef(x)=xsh(u)(x+xch(u))xsh(u)du=ξ(x)dux(1+ch(u))=ξ(x)xdu1+eu+eu2=2ξ(x)xdu2+eu+eu=eu=α2ξ(x)x12+α+α1dαα=2ξ(x)xdα2α+α2+1=2ξ(x)xdα(α+1)2=2ξ(x)x1α+1+c=2xξ(x)1eu+1butu=argch(tx)=ln(tx+t2x21)eu=tx+t2x21f(x)=2xξ(x)1tx+t2x2x+c=2ξ(x)t+ξ(x)t2x2+cwithξ(x)=1ifx>0andξ(x)=1ifx<0sof(x)=2t+t2x2+cifx>0f(x)=2tt2x2+cifx<0
Commented by maxmathsup by imad last updated on 14/May/19
2) ∫   (dt/((t+2)(√(t^2 −4)))) =f(2) =((−2)/(t +(√(t^2 −4)))) +c  ∫   (dt/((t+1)(√(t^2 −1)))) =f(1) =((−2)/(t +(√(t^2 −1)))) +c
2)dt(t+2)t24=f(2)=2t+t24+cdt(t+1)t21=f(1)=2t+t21+c

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