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anonymous
 one year ago
Tough Calculus Question: Let's see who can get it first!
anonymous
 one year ago
Tough Calculus Question: Let's see who can get it first!

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[ \int_2^4 \frac{\sqrt{ln(9x)} dx}{\sqrt{ln(9x)}+\sqrt{ln(3+x)}}=? \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0In case you are curious this is a Putnam Problem.... Hint.... It isn't nearly as "tough" as it looks

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If no one gets it in a day or two I'll post the answer. I just figured my fellow maths aficionados would find this one as fun as I did when I came across it :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\[I=\int_2^4 \frac{\sqrt{\ln(9x)} dx}{\sqrt{\ln(9x)}+\sqrt{\ln(3+x)}} \\~\\ =\int_2^4 \frac{\sqrt{\ln(9(2+4x))} dx}{\sqrt{\ln(9(2+4x))}+\sqrt{\ln(3+(2+4x))}}\\~\\ =\int_2^4 \frac{\sqrt{\ln(3+x)} dx}{\sqrt{\ln(3+x)}+\sqrt{\ln(9x)}} \] Add first and last parts and get \[2I = \int_2^41\,dx = 2 \implies I=1\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hmmmm interesting.... yes correct :D Im glad I found a taker though you method is different from mine:

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I noted the arguments of the log went from 5 to 7 and 7 to 5 depending on either the 9x or x+3 so this suggested the substitution x=u6... this gives: \[I= \int\limits^{1}_1\frac{\sqrt{\ln(9(3u)}(du)}{\sqrt{\ln(9(3u)}+\sqrt{\ln(3+(3u)}} \\ \ \ =\int\limits^{1}_1\frac{\sqrt{\ln(6u)}du}{\sqrt{\ln(6u)}+\sqrt{\ln(6u)}} \\ \ \ = \int\limits^{1}_{1} \frac{du}{2}=\frac{2}{2}=1\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Darnit I meant the subsitution 3u :/ but im sure thats obvious :D

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I guess I wound up getting lucky last night stopping prematurely because I see my error now

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2that's a clever substitution! :)
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