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

1. anonymous

?

2. anonymous

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

3. anonymous

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

4. anonymous

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 :)

5. ganeshie8

$I=\int_2^4 \frac{\sqrt{\ln(9-x)} dx}{\sqrt{\ln(9-x)}+\sqrt{\ln(3+x)}} \\~\\ =\int_2^4 \frac{\sqrt{\ln(9-(2+4-x))} dx}{\sqrt{\ln(9-(2+4-x))}+\sqrt{\ln(3+(2+4-x))}}\\~\\ =\int_2^4 \frac{\sqrt{\ln(3+x)} dx}{\sqrt{\ln(3+x)}+\sqrt{\ln(9-x)}}$ Add first and last parts and get $2I = \int_2^41\,dx = 2 \implies I=1$

6. anonymous

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

7. anonymous

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: $I= \int\limits^{-1}_1\frac{\sqrt{\ln(9-(3-u)}(-du)}{\sqrt{\ln(9-(3-u)}+\sqrt{\ln(3+(3-u)}} \\ \ \ =-\int\limits^{-1}_1\frac{\sqrt{\ln(6-u)}du}{\sqrt{\ln(6-u)}+\sqrt{\ln(6-u)}} \\ \ \ = \int\limits^{1}_{-1} \frac{du}{2}=\frac{2}{2}=1$

8. anonymous

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

9. anonymous

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

10. ganeshie8

that's a clever substitution! :)