Make a substitution to express the integrand as a rational function and then evaluate the integral. Integrate[sqrt x/(x-36), {x, 4, 9}]

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Make a substitution to express the integrand as a rational function and then evaluate the integral. Integrate[sqrt x/(x-36), {x, 4, 9}]

Mathematics
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\[\int\limits_{4}^{9} \sqrt x/(x-36) dx\]
|dw:1328291169857:dw|
Substitute \(u=\sqrt{x} \implies x=u^2 \implies dx=2udu \) to get \[\int_4^9 \frac{u}{u^2-36}2udu=2\int_4^9 \frac{u^2}{u^2-36}du=\cdots\] Can you take it from here?

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|dw:1328291206918:dw|
go math! :)
I did mine like math's and each time I keep coming up with 2-6 log(3)+log(64)
And thats not an option,
sqrtx=t=>x=t^2, dx=2tdt x=4=>t=2 x=9=>t=3 Integrate[2t^2/(t^2-36), {t, 2, 3}] 6ln2-6ln3+2
Oh sorry, I made a mistake. Change the limits.
At x=9 u=3, and at x=4 u=2. So your limits should be from 2 to 3.
See, either way I do it, if I completely solve my solution it comes out negative, and the options I get to choose from are positive...
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Also, please note that between 4 and 9 the expression x-36 is not null (otherwise your integral would be at a different level)
\[2\int \frac{u^2-36+36}{u^2-36}du=2\int(1+\frac{3}{u-6}-\frac{3}{u+6})du=2u+6\ln{\frac{6-u}{6+u}}.\]
Now plug the limits 2 to 3, and you should find your answer.
There was supposed to be an integral sign on that last park, correct?
No, not really.

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