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0-1-1-a-x-a-x-2-dx-1-ln-a-ln-3-pi-3-3-




Question Number 186911 by Spillover last updated on 11/Feb/23
  ∫_0 ^∞ (1/(1+a^x +a^(x/2) ))dx = (1/(ln a))[ln 3−(π/(3(√3)))]
011+ax+ax2dx=1lna[ln3π33]
Answered by witcher3 last updated on 13/Feb/23
a^(x/2) =t,a>1  x=2((ln(t))/(ln(a)))  dx=(2/(ln(a)t))dt  ⇔(2/(ln(a)))∫_1 ^∞ (dt/(t(1+t+t^2 )))  (2/(ln(a)))∫_0 ^1 y(dy/(1+y+y^2 ))  =(2/(ln(a)))∫_0 ^1 ((y+(1/2))/(1+y+y^2 ))−(1/(ln(a)))∫_0 ^1 (dy/((y+(1/2))^2 +(3/4)))  =(2/(ln(a))).(1/2)[ln(1+y+y^2 )]_0 ^1 −(1/(ln(a)))∫_0 ^1 .(4/3)(dy/(1+(((2y)/( (√3)))+(1/( (√3))))))  (1/(ln(a)))[ln(3)−(2/( (√3)))tan^(−1) ((√3))+(2/( (√3)))tan^(−1) ((1/( (√3)))))  =(1/(ln(a)))[ln(3)−(π/( 3(√3)))]
ax2=t,a>1x=2ln(t)ln(a)dx=2ln(a)tdt2ln(a)1dtt(1+t+t2)2ln(a)01ydy1+y+y2=2ln(a)01y+121+y+y21ln(a)01dy(y+12)2+34=2ln(a).12[ln(1+y+y2)]011ln(a)01.43dy1+(2y3+13)1ln(a)[ln(3)23tan1(3)+23tan1(13))=1ln(a)[ln(3)π33]

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