anonymous
  • anonymous
evaluate the definite integral.... from 1 to e^4 dx/(x(1+ln(x)))
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Note d/dx (ln(x)) = 1/(x) and do it from inspection.
anonymous
  • anonymous
Or if you're a baby, let u = ln(x)
anonymous
  • anonymous
\[\int_1^{e^4} \frac{1}{x(1+1ln(x))} dx\] \(\text{Let u} = ln(x) \implies du = \frac{1}{x}dx \implies dx = x\ du\)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Or 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
  • anonymous
whats the final answer? i'm confused
anonymous
  • anonymous
Did you try it with the u substitution I suggested?
anonymous
  • anonymous
tryin now
anonymous
  • anonymous
r u left with 1/u?
anonymous
  • anonymous
\(\frac{1}{1+u}\)
anonymous
  • anonymous
Actually a better u sub might be let u = 1+lnx
anonymous
  • anonymous
Then you'll get 1/u
anonymous
  • anonymous
Honestly, 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
  • anonymous
ln f(x) = the result, my bad
anonymous
  • anonymous
so then how would i integrate 1/u? is it ln(u)?
anonymous
  • anonymous
It'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
  • anonymous
Newton, 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
  • anonymous
Cambridge Part III Tripos would eat you alive, kiddo (polpak)
anonymous
  • anonymous
Let −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
  • anonymous
sorry that didnt copy right
anonymous
  • anonymous
Polpak, I told him both ways to do it faster than you LaTeX'd one, and you're bitter. grow up.
anonymous
  • anonymous
I'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
  • anonymous
I 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
  • anonymous
poketjjax: did you get it solved?
anonymous
  • anonymous
yes i did. i have another one as well.

Looking for something else?

Not the answer you are looking for? Search for more explanations.