Question

evaluate the definite integral....

from 1 to e^4

dx/(x(1+ln(x)))

Answer 1

Note d/dx (ln(x)) = 1/(x) and do it from inspection.

Answer 2

Or if you're a baby, let u = ln(x)

Answer 3

\[\int_1^{e^4} \frac{1}{x(1+1ln(x))} dx\]
\(\text{Let u} = ln(x) \implies du = \frac{1}{x}dx \implies dx = x\ du\)

Answer 4

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.

Answer 5

whats the final answer? i'm confused

Answer 6

Did you try it with the u substitution I suggested?

Answer 7

tryin now

Answer 8

r u left with 1/u?

Answer 9

\(\frac{1}{1+u}\)

Answer 10

Actually a better u sub might be let u = 1+lnx

Answer 11

Then you'll get 1/u

Answer 12

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. ..

Answer 13

ln f(x) = the result, my bad

Answer 14

so then how would i integrate 1/u? is it ln(u)?

Answer 15

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.

Answer 16

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.

Answer 17

Cambridge Part III Tripos would eat you alive, kiddo (polpak)

Answer 18

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=

Answer 19

sorry that didnt copy right

Answer 20

Polpak, I told him both ways to do it faster than you LaTeX'd one, and you're bitter. grow up.

Answer 21

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.

Answer 22

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.

Answer 23

poketjjax: did you get it solved?

Answer 24

yes i did. i have another one as well.