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Question Number 74484 by mathmax by abdo last updated on 24/Nov/19

calculate ∫_0 ^∞   ((arctan(2x))/(x^2 +3))dx

calculate0arctan(2x)x2+3dx

Commented by mathmax by abdo last updated on 26/Nov/19

let f(t)=∫_0 ^∞   ((arctan(tx))/(x^2  +3))dx    with t≥0  f^′ (t) =∫_0 ^∞   (x/((1+t^2 x^2 )(x^2 +3)))dx  =_(tx=u)   ∫_0 ^∞    (u/(t(1+u^2 )((u^2 /t^2 )+3)))(du/t)  =∫_0 ^∞    ((udu)/((u^2 +1)(u^2 +3t^2 ))) decompsition of F(u)=(1/((u^2 +1)(u^2 +3t^2 )))  F(u)=((au+b)/(u^2 +1)) +((cu +d)/(u^2  +3t^2 ))  F(−u)=F(u) ⇒((−au+b)/(u^2  +1)) +((−cu+d)/(u^2  +3t^2 )) =F(u) ⇒a=c=0 ⇒  F(u) =(b/(u^2 +1)) +(d/(u^2  +3t^2 ))  lim_(u→+∞) u^2 F(u) =0 =b+d ⇒d=−b ⇒F(u)=(b/(u^2  +1))−(b/(u^2  +3t^2 ))  F(0)=(1/(3t^2 )) =b−(b/(3t^2 )) =(((3t^2 −1)/(3t^2 )))b ⇒b=(1/(3t^2 −1)) ⇒  F(u) =(1/(3t^2 −1)){(1/(u^2 +1))−(1/(u^2  +3t^2 ))} ⇒  f^′ (t)=(1/(3t^2 −1)) ∫_0 ^∞   (du/(1+u^2 ))−(1/(3t^2 −1))∫_0 ^∞   (du/(u^2  +3t^2 ))  ∫_0 ^∞   (du/(1+u^2 )) =(π/2)  ∫_0 ^∞   (du/(u^2  +3t^2 )) =_(u=(√3)t z)  ∫_0 ^∞   (((√3)t dz)/(3t^2 (1+z^2 ))) =(1/((√3)t))×(π/2) ⇒  f^′ (t) =(π/(2(3t^2 −1)))−(π/(2(√3)t(3t^2 −1))) ⇒  f(t)=(π/2) ∫  (dt/(3t^2 −1))−(π/(2(√3))) ∫  (dt/(t(3t^2 −1))) +c ....be continued ....

letf(t)=0arctan(tx)x2+3dxwitht0f(t)=0x(1+t2x2)(x2+3)dx=tx=u0ut(1+u2)(u2t2+3)dut=0udu(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=0F(u)=bu2+1+du2+3t2limu+u2F(u)=0=b+dd=bF(u)=bu2+1bu2+3t2F(0)=13t2=bb3t2=(3t213t2)bb=13t21F(u)=13t21{1u2+11u2+3t2}f(t)=13t210du1+u213t210duu2+3t20du1+u2=π20duu2+3t2=u=3tz03tdz3t2(1+z2)=13t×π2f(t)=π2(3t21)π23t(3t21)f(t)=π2dt3t21π23dtt(3t21)+c....becontinued....

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