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Question Number 104028 by mathmax by abdo last updated on 19/Jul/20

calculate  ∫_(20) ^(+∞)   (dx/((x−18)^(19) (x−19)^(18) ))

calculate20+dx(x18)19(x19)18

Answered by Dwaipayan Shikari last updated on 19/Jul/20

∫_(20) ^(+∞) (((x−19)dx)/((x^2 −37x+342)^(19) ))  (1/2)∫_(20) ^(+∞) ((2x−37)/((x^2 −37x+342)^(19) ))dx−(1/((x^2 −37+342)^(19) ))  −(1/(36))[(1/((x^2 −37x+342)^(18) ))]_(20) ^(+∞) −(1/2)∫(1/((x−18)^(19) (x−19)^(19) ))  (1/(36)).(1/2^(18) )−{(1/2)∫_(20) ^(+∞) (1/((x−18)^(19) (x−19)^(19) ))}→I_a   (1/(36.2^(18) ))−I_a     continue......

20+(x19)dx(x237x+342)191220+2x37(x237x+342)19dx1(x237+342)19136[1(x237x+342)18]20+121(x18)19(x19)19136.1218{1220+1(x18)19(x19)19}Ia136.218Iacontinue......

Answered by mathmax by abdo last updated on 19/Jul/20

A =∫_(20) ^(+∞)  (dx/((x−18)^(19) (x−19)^(18) ))  changement x−19 =t give  A =∫_1 ^(+∞)  (dt/((t+1)^(19) t^(18) )) =∫_1 ^∞  (dt/(((t/(t+1)))^(18)  (t+1)^(37) ))  we do the changement (t/(t+1)) =u ⇒  t =ut +u ⇒(1−u)t =u ⇒t =(u/(1−u)) ⇒(dt/du) =((1−u+u)/((1−u)^2 )) =(1/((1−u)^2 ))  t+1 =(u/(1−u)) +1 =((u+1−u)/(1−u)) =(1/(1−u)) ⇒  A =∫_(1/2) ^1    (du/((1−u)^2  u^(18) ((1/(1−u)))^(37) ))  =∫_(1/2) ^1  (((1−u)^(37) )/((1−u)^2  u^(18) ))du   =∫_(1/2) ^1  (((1−u)^(35) )/u^(18) ) du =−∫_(1/2) ^1  (((u−1)^(35) )/u^(18) )du =−∫_(1/2) ^1  ((Σ_(k=0) ^(35 )  C_(35) ^k  u^k (−1)^(35−k) )/u^(18) )du  =Σ_(k=0) ^(35 )  (−1)^k  C_(35) ^k   ∫_(1/2) ^1  u^(k−18)  du  =Σ_(k=0 and k≠17) ^(35) (−1)^k  C_(35) ^k  [(1/(k−17)) u^(k−17) ]_(1/2) ^1   −C_(35) ^(17)  [ln∣u∣]_(1/2) ^1   A=Σ_(k=0and k≠17) ^(35)   (((−1)^k  C_(35) ^k )/(k−17)){1−(1/2^(k−17) )} −ln(2)C_(35) ^(17)

A=20+dx(x18)19(x19)18changementx19=tgiveA=1+dt(t+1)19t18=1dt(tt+1)18(t+1)37wedothechangementtt+1=ut=ut+u(1u)t=ut=u1udtdu=1u+u(1u)2=1(1u)2t+1=u1u+1=u+1u1u=11uA=121du(1u)2u18(11u)37=121(1u)37(1u)2u18du=121(1u)35u18du=121(u1)35u18du=121k=035C35kuk(1)35ku18du=k=035(1)kC35k121uk18du=k=0andk1735(1)kC35k[1k17uk17]121C3517[lnu]121A=k=0andk1735(1)kC35kk17{112k17}ln(2)C3517

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