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Question Number 86302 by john santu last updated on 28/Mar/20

∫ (dx/((x^2 +1)(√(x^2 +4)))) =?

dx(x2+1)x2+4=?

Commented by abdomathmax last updated on 28/Mar/20

I =∫   (dx/((x^2  +1)(√(x^2  +4)))) changement x =2sh(t) give  I =∫  ((2ch(t) dt)/((1+4sh^2 t)2ch(t))) =∫   (dt/(1+4×((ch(2t)−1)/2)))  =∫    ((2dt)/(2+4ch(2t)−4)) =∫   (dt/(2ch(2t)−1))  =∫  (dt/(2×((e^(2t) +e^(−2t) )/2)−1)) =∫   (dt/(e^(2t)  +e^(−2t) −1))  =_(e^t =z)    ∫      (dz/(z(z^2  +z^(−2) −1))) =∫   (dz/(z^3 −z^(−1) −z))  =∫  ((zdz)/(z^4 −1−z^2 )) =∫   ((zdz)/(z^4 −z^2 −1)) let decompose  F(z)=(z/(z^4 −z^2 −1))  z^4 −z^2 −1 =0⇒t^2 −t−1=0  (t=z^2 )  Δ=1+4 =5 ⇒t_1 =1+(√5) and t_2 =1−(√5)  ⇒F(z) =(z/((z^2 −t_1 )(z^2 −t_2 ))) =(z/(2(√5)))((1/(z^2 −t_1 ))−(1/(z^2 −t_2 )))  ⇒∫ F(z)dz =(1/(2(√5)))ln(((z^2 −t_1 )/(z^2 −t_2 ))) +C  =(1/(2(√5)))ln(((e^(2t) −t_1 )/(e^(2t) −t_2 )))+C  t=argsh((x/2)) =ln((x/2)+(√(1+(x^2 /4)))) ⇒  e^(2t)  =((x/2)+(√(1+(x^2 /4))))^2  ⇒  I =(1/(2(√5)))ln(((((x/2)+(√(1+(x^2 /2))))^2 −1−(√5))/(((x/2)+(√(1+(x^2 /4))))^2 −1+(√5)))) +C

I=dx(x2+1)x2+4changementx=2sh(t)giveI=2ch(t)dt(1+4sh2t)2ch(t)=dt1+4×ch(2t)12=2dt2+4ch(2t)4=dt2ch(2t)1=dt2×e2t+e2t21=dte2t+e2t1=et=zdzz(z2+z21)=dzz3z1z=zdzz41z2=zdzz4z21letdecomposeF(z)=zz4z21z4z21=0t2t1=0(t=z2)Δ=1+4=5t1=1+5andt2=15F(z)=z(z2t1)(z2t2)=z25(1z2t11z2t2)F(z)dz=125ln(z2t1z2t2)+C=125ln(e2tt1e2tt2)+Ct=argsh(x2)=ln(x2+1+x24)e2t=(x2+1+x24)2I=125ln((x2+1+x22)215(x2+1+x24)21+5)+C

Answered by TANMAY PANACEA. last updated on 28/Mar/20

x=(1/t)→dx=((−dt)/t^2 )  ∫((−dt)/(t^2 ((1/t^2 )+1)(√((1/t^2 )+4))))  ∫((−tdt)/((1+t^2 )(√(1+4t^2 ))))  t^2 =k→dk=2tdt  ∫((−dk)/(2(1+k)(√(1+4k))))  ((−1)/2)∫(dk/((1+k)(√(1+4k))))  1+4k=p^2   4dk=2pdp  ((−1)/2)∫((pdp)/(2×(1+((p^2 −1)/4))p))  =((−1)/4)∫((4dp)/((p^2 +3)))=((−1)/(√3))tan^(−1) ((p/(√3)))+c  =((−1)/(√3))tan^(−1) (((√(1+4k))/(√3)))+c  =((−1)/(√3))tan^(−1) (((√(1+4t^2 ))/(√3)))+c  =((−1)/(√3))tan^(−1) (((√(1+(4/x^2 )))/(√3)))+c

x=1tdx=dtt2dtt2(1t2+1)1t2+4tdt(1+t2)1+4t2t2=kdk=2tdtdk2(1+k)1+4k12dk(1+k)1+4k1+4k=p24dk=2pdp12pdp2×(1+p214)p=144dp(p2+3)=13tan1(p3)+c=13tan1(1+4k3)+c=13tan1(1+4t23)+c=13tan1(1+4x23)+c

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