Question

Integral of 1 / ( 1 + e ^ x) actual process please... :D

Answer 1

Dang you got 103 fans now!

Answer 2

Hi arman,

Try looking at it like this\[\int\limits_{}{}\frac{1}{1+e^x}dx=\int\limits_{}{}\frac{1+e^x-e^x}{1+e^x}dx=\int\limits_{}{}1-\frac{e^x}{1+e^x}dx\]

Answer 3

Haha, I know!

Answer 4

\[=x-\log (1+e^x) + c\]

Answer 5

log == ln always unless a base is made explicit

Answer 6

It's an arctrig function and x-ln(1+e^x) is e^x/(1+e^x) (chain rule)

Answer 7

II know that it's arctan of something ><

Answer 8

The second integral makes use of this fact:\[\int\limits_{}{}\frac{f'(x)}{f(x)}dx=\log f(x)+c\]

Answer 9

If f(x)=1+e^x, then f'(x)=e^x, which is the form you have after rearranging.

Answer 10

I'm on my fone so I didn't see half of those equations till after I posted my last post.... lemmy try that... yummy and sec

Answer 11

Gimmy a sec**

Answer 12

u substitution then partial fraction decomposition

Answer 13

I don't get why I can't think of these on my own -.-

Answer 14

i got an answer of ln(e^x/(e^x+1))+c

Answer 15

There's one way to check your answer - take the derivative and see if you get back to the integrand.

Answer 16

yea it does

Answer 17

just checked it

Answer 18

Nope, pfd is in the chapter after this (I slipped it previousy and went ahead) so Lucas answer makes perfect sense...

Answer 19

Lol

Answer 20

Good work

Answer 21

Oh boy, integral of e^sqrtx from 0 to 4...

Answer 22

one sec

Answer 23

I don't even know what to use on this one...

Answer 24

2(e^(sqrt(x))(sqrt(x)-1)|(0,4)

Answer 25

Why?

Answer 26

have u learned integration by parts yet?

Answer 27

Yep

Answer 28

and i take it you know u substitution?

Answer 29

start with your integral and set u equal to sqrt(x)

Answer 30

Ok...

Answer 31

u should get int(2u(e^u)du)

Answer 32

do you follow me so far?

Answer 33

You can put the dx in terms of u aswell? (Like I can write 2u instead of 2sqrtx? )

Answer 34

correct

Answer 35

from this point you can use integration by parts

Answer 36

Ok then that makes sense and i can use the uv vdu then.... -.- I'm retarded

Answer 37

lol its hard to see that without substituting first tho

Answer 38

No... I'm just retarded. And the answer is 2e^2

Answer 39

did u evaluate at zero also?

Answer 40

im getting 2(e^2)+2

Answer 41

Good point, further testing my retardation.

Answer 42

lol common mistake

Answer 43

When you uv vdu sub, can you choose which term is your u and v? I'm pretty sure you can

Answer 44

usually it depends on what you have

Answer 45

to remember what to take for u i use "lipet"

Answer 46

which gives you an order in which to take your u (Lograthims Inverse.trg Polynomials Exponetials Trigonometrics)

Answer 47

Well I got a badass equation here: integral of (lnx) /x^2 from 1 to 2

Answer 48

I know I gotta use uv vdu on it

Answer 49

judging by lipet take log to be your u

Answer 50

and 1/x^2 to be your dv

Answer 51

im getting (-ln(x)/x)-(1/x)+c

Answer 52

your evaluated integral should come out to be (-ln(2)/2)+1/2

Answer 53

were you able to solve it?