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Question Number 84809 by M±th+et£s last updated on 16/Mar/20

∫(x/((x^2 +1)^(3/2) arctan(x))) dx

x(x2+1)32arctan(x)dx

Commented by abdomathmax last updated on 18/Mar/20

A =∫    (x/((x^(2 ) +1)^(3/2)  arctanx)) changement arctanx=t  give A =∫  ((tant)/((1+tan^2 t)^(3/2) t))×(1+tan^2 t)dt  =∫  ((tant)/(t(√(1+tan^2 t))))dt =∫ ((sint)/(tcost))×cost dt =∫ ((sint)/t)dt

A=x(x2+1)32arctanxchangementarctanx=tgiveA=tant(1+tan2t)32t×(1+tan2t)dt=tantt1+tan2tdt=sinttcost×costdt=sinttdt

Answered by TANMAY PANACEA last updated on 16/Mar/20

x=tana  ∫((tana.sec^2 a)/(sec^3 a.a))da  ∫((sina)/(cosa.seca.a))da  ∫((sina)/a)da  ∫((a−(a^3 /(3!))+(a^5 /(5!))−(a^7 /(7!))+...)/a) da  ∫1−(a^2 /(3!))+(a^4 /(5!))−(a^6 /(7!))+...  a−(a^3 /(3×3!))+(a^5 /(5×5!))−(a^7 /(7×7!))+...∞  =Σ_(n=1) ^∞ (a^(2n−1) /((2n−1).(2n−1)!))  =Σ_(n=1) ^∞ (((tan^(−1) x)^(2n−1) )/((2n−1)(2n−1)!))

x=tanatana.sec2asec3a.adasinacosa.seca.adasinaadaaa33!+a55!a77!+...ada1a23!+a45!a67!+...aa33×3!+a55×5!a77×7!+...=n=1a2n1(2n1).(2n1)!=n=1(tan1x)2n1(2n1)(2n1)!

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