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Question Number 98679 by  M±th+et+s last updated on 15/Jun/20

prove that   ∫_0 ^∞ ((3+2(√x))/(x^2 +2x+5))dx=4.13049

provethat03+2xx2+2x+5dx=4.13049

Commented by MJS last updated on 15/Jun/20

t=(√x) → dx=2(√x)dt  2∫((t(2t+3))/(t^4 +2t^2 +5))dt  now decompose (2 square factors) and solve.  as many times before, easy path but horrible  constants...  2∫((t(2t+3))/(t^4 +2t^2 +5))dt=  =((√(−2+2(√5)))/4)(∫((2t+3)/(t^2 −(√(−2+2(√5)))t+(√5)))dt−       −∫((2t+3)/(t^2 +(√(−2+2(√5)))t+(√5)))dt)=  =((√(−2+2(√5)))/4)×(       ∫((2t−(√(−2+2(√5))))/(t^2 −(√(−2+2(√5)))t+(√5)))dt+       +(3+(√(−2+2(√5))))∫(dt/(t^2 −(√(−2+2(√5)))t+(√5))))−       −∫((2t+(√(−2+2(√5))))/(t^2 +(√(−2+2(√5)))t+(√5)))dt−       −(3−(√(−2+2(√5))))∫(dt/(t^2 +(√(−2+2(√5)))t+(√5)))  )  ...and tbese are easy to solve  (I hope I made no error of sign in front of  the “∫”s...)

t=xdx=2xdt2t(2t+3)t4+2t2+5dtnowdecompose(2squarefactors)andsolve.asmanytimesbefore,easypathbuthorribleconstants...2t(2t+3)t4+2t2+5dt==2+254(2t+3t22+25t+5dt2t+3t2+2+25t+5dt)==2+254×(2t2+25t22+25t+5dt++(3+2+25)dtt22+25t+5)2t+2+25t2+2+25t+5dt(32+25)dtt2+2+25t+5)...andtbeseareeasytosolve(IhopeImadenoerrorofsigninfrontofthes...)

Commented by  M±th+et+s last updated on 15/Jun/20

thank you sir

thankyousir

Answered by mathmax by abdo last updated on 15/Jun/20

I =∫_0 ^∞  ((3+2(√x))/(x^2  +2x+5)) dx changement (√x)=t give   I =∫_0 ^∞  ((3+2t)/(t^4  +2t^(2 ) +5))(2t)dt =2 ∫_0 ^∞  ((3t+2t^2 )/(t^4  +2t^2  +5))dt  let decompose  F(t) =((2t^2  +3t)/(t^4  +2t^2  +5))  t^4  +2t^2  +5 =0 →u^2  +2u +5 =0 →Δ^′  =−4 ⇒u_1 =−1+2i and u_2 =−1−2i  ⇒u^2 +2u +5 =(u−u_1 )(u−u_2 ) =(t^2 −u_1 )(t^2 −u_2 ) ⇒  I =∫_0 ^∞  ((2t^2  +3t)/((t^2 −u_1 )(t^2 −u_2 )))dt  =(1/(4i))∫_0 ^∞ (2t^2  +3t)((1/(t^2 −u_1 ))−(1/(t^2 −u_2 )))dt  =(1/(4i))∫_0 ^∞  ((2t^2  +3t)/(t^2 −u_1 )) dt−(1/(4i))∫_0 ^∞  ((2t^2  +3t)/(t^2 −u_2 ))dt  =(1/(4i))∫_0 ^∞  ((2(t^2 −u_1 )+3t+2u_1 )/(t^2 −u_1 ))dt −(1/(4i))∫_0 ^∞  ((2(t^2 −u_2 )+3t+2u_2 )/(t^2 −u_2 ))dt  =(1/(4i)){ ∫_0 ^∞   ((3t+2u_1 )/(t^2 −u_1 )) dt −∫_0 ^∞  ((3t+2u_2 )/(t^2 −u_2 ))dt}  =(1/(4i)){ (3/2) [ln(((t^2 −u_1 )/(t^2 −u_2 )))]_0 ^(+∞)  +2 ∫_0 ^∞  (u_1 /(t^2 −u_1 )) dt−2 ∫_0 ^∞  (u_2 /(t^2 −u_2 ))dt}  =(3/(8i))ln((u_1 /u_2 )) +(1/(2i)){ u_1 ∫_0 ^∞  (dt/(t^2 −u_1 )) −u_2 ∫_0 ^∞  (dt/(t^2 −u_2 ))}....be continued...

I=03+2xx2+2x+5dxchangementx=tgiveI=03+2tt4+2t2+5(2t)dt=203t+2t2t4+2t2+5dtletdecomposeF(t)=2t2+3tt4+2t2+5t4+2t2+5=0u2+2u+5=0Δ=4u1=1+2iandu2=12iu2+2u+5=(uu1)(uu2)=(t2u1)(t2u2)I=02t2+3t(t2u1)(t2u2)dt=14i0(2t2+3t)(1t2u11t2u2)dt=14i02t2+3tt2u1dt14i02t2+3tt2u2dt=14i02(t2u1)+3t+2u1t2u1dt14i02(t2u2)+3t+2u2t2u2dt=14i{03t+2u1t2u1dt03t+2u2t2u2dt}=14i{32[ln(t2u1t2u2)]0++20u1t2u1dt20u2t2u2dt}=38iln(u1u2)+12i{u10dtt2u1u20dtt2u2}....becontinued...

Commented by  M±th+et+s last updated on 15/Jun/20

thanks sir

thankssir

Commented by mathmax by abdo last updated on 16/Jun/20

you are welcome.

youarewelcome.

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