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Question Number 186911 by Spillover last updated on 11/Feb/23
∫0∞11+ax+ax2dx=1lna[ln3−π33]
Answered by witcher3 last updated on 13/Feb/23
ax2=t,a>1x=2ln(t)ln(a)dx=2ln(a)tdt⇔2ln(a)∫1∞dtt(1+t+t2)2ln(a)∫01ydy1+y+y2=2ln(a)∫01y+121+y+y2−1ln(a)∫01dy(y+12)2+34=2ln(a).12[ln(1+y+y2)]01−1ln(a)∫01.43dy1+(2y3+13)1ln(a)[ln(3)−23tan−1(3)+23tan−1(13))=1ln(a)[ln(3)−π33]
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