- #36
gabbagabbahey
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lubuntu said:Serious how did you even thing to start down that line of thinking?
It looked similar in form to your [tex]\int_0^{\pi/2} \frac{1}{1+\tan\theta^p}dx[/tex] integral...so, once I realized that [tex]\left[\frac{1}{1+\left(\frac{\ln(3+x)}{\ln(9-x)}\right)^b}+ \frac{1}{1+\left(\frac{\ln(9-x)}{\ln(3+x)}\right)^b}\right]=1[/tex] , it was just a matter of justifying that the two individual integrals were equal. So, I graphed both integrands and noticed that one was a reflection of the other.