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

- anonymous

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

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- anonymous

Dang you got 103 fans now!

- anonymous

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

- anonymous

Haha, I know!

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## More answers

- anonymous

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

- anonymous

log == ln always unless a base is made explicit

- anonymous

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

- anonymous

II know that it's arctan of something ><

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

Gimmy a sec**

- anonymous

u substitution then partial fraction decomposition

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

yea it does

- anonymous

just checked it

- anonymous

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

- anonymous

Lol

- anonymous

Good work

- anonymous

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

- anonymous

one sec

- anonymous

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

- anonymous

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

- anonymous

Why?

- anonymous

have u learned integration by parts yet?

- anonymous

Yep

- anonymous

and i take it you know u substitution?

- anonymous

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

- anonymous

Ok...

- anonymous

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

- anonymous

do you follow me so far?

- anonymous

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

- anonymous

correct

- anonymous

from this point you can use integration by parts

- anonymous

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

- anonymous

lol its hard to see that without substituting first tho

- anonymous

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

- anonymous

did u evaluate at zero also?

- anonymous

im getting 2(e^2)+2

- anonymous

Good point, further testing my retardation.

- anonymous

lol common mistake

- anonymous

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

- anonymous

usually it depends on what you have

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

I know I gotta use uv vdu on it

- anonymous

judging by lipet take log to be your u

- anonymous

and 1/x^2 to be your dv

- anonymous

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

- anonymous

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

- anonymous

were you able to solve it?

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