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Question Number 74484 by mathmax by abdo last updated on 24/Nov/19
calculate∫0∞arctan(2x)x2+3dx
Commented by mathmax by abdo last updated on 26/Nov/19
letf(t)=∫0∞arctan(tx)x2+3dxwitht⩾0f′(t)=∫0∞x(1+t2x2)(x2+3)dx=tx=u∫0∞ut(1+u2)(u2t2+3)dut=∫0∞udu(u2+1)(u2+3t2)decompsitionofF(u)=1(u2+1)(u2+3t2)F(u)=au+bu2+1+cu+du2+3t2F(−u)=F(u)⇒−au+bu2+1+−cu+du2+3t2=F(u)⇒a=c=0⇒F(u)=bu2+1+du2+3t2limu→+∞u2F(u)=0=b+d⇒d=−b⇒F(u)=bu2+1−bu2+3t2F(0)=13t2=b−b3t2=(3t2−13t2)b⇒b=13t2−1⇒F(u)=13t2−1{1u2+1−1u2+3t2}⇒f′(t)=13t2−1∫0∞du1+u2−13t2−1∫0∞duu2+3t2∫0∞du1+u2=π2∫0∞duu2+3t2=u=3tz∫0∞3tdz3t2(1+z2)=13t×π2⇒f′(t)=π2(3t2−1)−π23t(3t2−1)⇒f(t)=π2∫dt3t2−1−π23∫dtt(3t2−1)+c....becontinued....
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