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Question Number 84809 by M±th+et£s last updated on 16/Mar/20
∫x(x2+1)32arctan(x)dx
Commented by abdomathmax last updated on 18/Mar/20
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.sec2asec3a.ada∫sinacosa.seca.ada∫sinaada∫a−a33!+a55!−a77!+...ada∫1−a23!+a45!−a67!+...a−a33×3!+a55×5!−a77×7!+...∞=∑∞n=1a2n−1(2n−1).(2n−1)!=∑∞n=1(tan−1x)2n−1(2n−1)(2n−1)!
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