Question

find the indefinite integral

∫ 1/x^2 (e ^-1/x) dx

Answer 1

The ambiguity kills...
\[\huge \int\limits_{}^{}\frac{1}{x^2(e^{-\frac{1}{x}})}dx\]
Is this it?

Answer 2

nope the (e ^-1/x) is not in the denominator. but it's right :)

Answer 3

It's not in the denominator, but it's right? It's wrong!!!!
But seriously.. :D
\[\huge \int\limits_{}^{}\frac{e^{-\frac{1}{x}}}{x^2}dx\]

This one then?

Answer 4

i guess it could be placed that way.. but then it's kinda just multiplied to it :))

Answer 5

Okay, since you're so particular :P
\[\huge \int\limits_{}^{}\frac{1}{x^2}(e^{-\frac{1}{x}})dx \]
This one better? :D

Answer 6

perfect. haha now how do we answer that ? haha

Answer 7

Well, always look for a way to simplify things...
Too bad there is no clear-cut way to integrate, it's not like trying to find the derivative, where you always know what you're doing, this takes a bit of guesswork :D

So... is there any bit of that integrand (fancy name for the function being integrated) whose derivative is also in the integrand?

Answer 8

My rule of thumb here is... always suspect the most complicated looking function :D

Answer 9

Stumped? Let me know :)

Answer 10

Suspicious looking complicated function. That's actually a pretty good advice! (-:
\[\Large \checkmark \]

Answer 11

:D
I just thought it up now... :/

Answer 12

So it has only been a rule of thumb for a few minutes, thanks though :D
@alyannahere
Are you still here?

Answer 13

here ! sorry. i think the e ^-1/x looks pretty complicated

Answer 14

Cool, I thought so :P

Answer 15

i think i get it ? but then i'm not sure how it works.. like i know there s a property but i can't seem to apply it

Answer 16

Go, you can tell me anything :D
I don't guarantee I'll liste..... LOL What was your solution? :)

Answer 17

there's this property stating...
∫e^f(x) X f'(x) X dx = e^f(x) + C

X is multiplied

Answer 18

Let me try to rid myself of ambiguities again....
\[\huge \int\limits e^{f(x)}Xf'(x)Xdx = e^{f(x)}+C\]
I don't quite get it... what's with the big X's? :/

Answer 19

X is multiply :)) haha times :))

Answer 20

/head scratching/
\[\huge \int\limits e^{f(x)}f'(x)dx = e^{f(x)} + C\]
is it?

Answer 21

yup! that's one of the general indefinite integral formulas

Answer 22

You know how to apply it?

Answer 23

Maybe you can simply let u = -1/x
du = 1/(x^2) dx
Then, the intergral becomes pretty easy

Answer 24

@RolyPoly how did du become 1/(x^2) dx?

Answer 25

Simply take the derivative of -1/x

Answer 26

\[\huge \frac{d}{dx}x^k=kx^{k-1}\],

Answer 27

And you know,
\[\huge -\frac{1}{x}=-x^{-1}\]is no exception :P

Answer 28

for k \(\ne\) 0, I guess?

Answer 29

Always missing those things :/
Yeah, for a nonzero k.

Answer 30

i still don't get it

Answer 31

:(

Answer 32

@alyannahere
Do you know how to differentiate 1/x ?

Answer 33

Use this formula instead, to be more general:
\[ \huge \frac{d}{dx}ax^k=akx^{k-1}\]

Answer 34

oooh. i see i get it now... thank you guysss!!!!!!!!!!!

Answer 35

Try to solve the problem. If you have any questions, please let us know :)

Answer 36

Well, I'm bored, I'm going to play a bit with this...
\[\huge \int\limits\limits e^{f(x)}f'(x)dx\]Let
\[\large u=f(x)\]\[\large du = f'(x) dx\]
\[\huge \int\limits\limits e^{f(x)}f'(x)dx = \int\limits e^udu\]\[\huge \int\limits e^u du = e^u + C\]Resusbstituting...
\[\huge e^{f(x)} + C\]
Voila D

Answer 37

i did. the answer is e^-1/x +C ;)

Answer 38

right? haha

Answer 39

Yup! Congrats :)

Answer 40

Nice work :)

Answer 41

how about this problem @RolyPoly

find the antiderivative of

dp/dx = e^x + e^-x / (e^x - e^-x)^2 ?

Answer 42

well i made u = e^x + e^-x
du = e^x - e^-x

then i got stuck

Answer 43

Basically trying to integrate...
\[\huge \int\limits \frac{e^x + e^{-x}}{(e^x-e^{-x})^2}\]
OKAAAY
Second rule that I just made up...
If you got two functions that are both suspiciously complicated, go for the one in the denominator... that's probably it :P

Answer 44

@alyannahere That's correct. Then, what will the integral become?

Answer 45

@RolyPoly
Are you sure it's correct?
I'd let
u = e^x - e^-x

Answer 46

hahahaah i like your rules @PeterPan & yup! :)

Answer 47

well @RolyPoly it will become u (u^-2)

Answer 48

@PeterPan 100% sure. The asker let u = e^x + e^(-x), not u = e^x - e^(-x)