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anonymous
 5 years ago
evaluate the definite integral....
from 1 to e^4
dx/(x(1+ln(x)))
anonymous
 5 years ago
evaluate the definite integral.... from 1 to e^4 dx/(x(1+ln(x)))

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Note d/dx (ln(x)) = 1/(x) and do it from inspection.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Or if you're a baby, let u = ln(x)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int_1^{e^4} \frac{1}{x(1+1ln(x))} dx\] \(\text{Let u} = ln(x) \implies du = \frac{1}{x}dx \implies dx = x\ du\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Or if you're a douche you can learn all this stuff before hand then come online and make fun of other people who don't know how to do it yet.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0whats the final answer? i'm confused

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Did you try it with the u substitution I suggested?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Actually a better u sub might be let u = 1+lnx

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Honestly, polpak, I know you hate me, but if you can't see this is clearly a case to use: \[\int\limits \frac{f'(x)}{f(x)} \mathbb{d}x = f(x) + c\] you are silly. Ah well, I guess less intelligent people who learn maths by rote always fall back on subs. ..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ln f(x) = the result, my bad

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so then how would i integrate 1/u? is it ln(u)?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It's not a matter of whether or not I can see it. It's a matter of someone learning it for the first time needs to understand a bit about the process and how to do things even when it's somthing not so nice. And to do that, they have to first practice on nice things.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Newton, nobody cares if you can do it. This is a place for helping others learn, not for trying to impress people by being awsome at math.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Cambridge Part III Tripos would eat you alive, kiddo (polpak)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Let −90f(x)dx=5 −9−6f(x)dx=7 −30f(x)dx=4 . Find −6−3f(x)dx= and −3−6(5f(x)−7)dx=

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry that didnt copy right

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Polpak, I told him both ways to do it faster than you LaTeX'd one, and you're bitter. grow up.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm not bitter. I wouldn't have had a problem if you'd given him(her?) helpful advice, and then went on to explain how to do it easier. But to give a trite one line explanation and then tell him(her?) to do it by inspection is just you showing off, not trying to help. Then you go on to give him(her?) a way he might understand, while simultaneously insulting him(her?). You're just being an retriceat, and you've been at it all day. How about you go find the Cambridge math club and have yourself a nice circlejerk.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I was taught to integrate things of that form by inspection. Maybe I shouldn't assume everyone was (though they should be)  and that is long before University, it is how it is taught to everyone who does Maths in England. Substitutions waste time.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0poketjjax: did you get it solved?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes i did. i have another one as well.
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