Re: Putnam 2005 -- some answers [SPOILER ALERT]

A5 I found:
The trig substituion x = tan(a), transforms the the integral to:

I = Int^(0,pi/4) log(tan(a) + 1) da.

Then a-->pi/4 - a gives

I = Int^(0,pi/4) log(tan(pi/4 - a) + 1)da = int^(0,pi/4) log(2) da - I
(by noting that tan(pi/4 - a) = (1 - tan a)/(1 + tan a)

And so 2I = pi/4*log(2)

so I = pi/8*log(2)

Of course..B6, the coolest problem ive seen so far had me baffled.

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