Question

Tough Calculus Question: Let's see who can get it first!

Answer 1

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Answer 2

\[ \int_2^4 \frac{\sqrt{ln(9-x)} dx}{\sqrt{ln(9-x)}+\sqrt{ln(3+x)}}=? \]

Answer 3

In case you are curious this is a Putnam Problem.... Hint.... It isn't nearly as "tough" as it looks

Answer 4

If 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 :)

Answer 5

\[I=\int_2^4 \frac{\sqrt{\ln(9-x)} dx}{\sqrt{\ln(9-x)}+\sqrt{\ln(3+x)}} \\~\\
=\int_2^4 \frac{\sqrt{\ln(9-(2+4-x))} dx}{\sqrt{\ln(9-(2+4-x))}+\sqrt{\ln(3+(2+4-x))}}\\~\\
=\int_2^4 \frac{\sqrt{\ln(3+x)} dx}{\sqrt{\ln(3+x)}+\sqrt{\ln(9-x)}}
\]

Add first and last parts and get
\[2I = \int_2^41\,dx = 2 \implies I=1\]

Answer 6

Hmmmm interesting.... yes correct :D Im glad I found a taker though you method is different from mine:

Answer 7

I noted the arguments of the log went from 5 to 7 and 7 to 5 depending on either the 9-x or x+3 so this suggested the substitution x=u-6... this gives:
\[I= \int\limits^{-1}_1\frac{\sqrt{\ln(9-(3-u)}(-du)}{\sqrt{\ln(9-(3-u)}+\sqrt{\ln(3+(3-u)}} \\ \ \ =-\int\limits^{-1}_1\frac{\sqrt{\ln(6-u)}du}{\sqrt{\ln(6-u)}+\sqrt{\ln(6-u)}} \\ \ \ = \int\limits^{1}_{-1} \frac{du}{2}=\frac{2}{2}=1\]

Answer 8

Darnit I meant the subsitution 3-u :/ but im sure thats obvious :D

Answer 9

I guess I wound up getting lucky last night stopping prematurely because I see my error now

Answer 10

that's a clever substitution! :)