anonymous
  • anonymous
Integral of 1 / ( 1 + e ^ x) actual process please... :D
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Dang you got 103 fans now!
anonymous
  • 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
  • anonymous
Haha, I know!

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anonymous
  • anonymous
\[=x-\log (1+e^x) + c\]
anonymous
  • anonymous
log == ln always unless a base is made explicit
anonymous
  • anonymous
It's an arctrig function and x-ln(1+e^x) is e^x/(1+e^x) (chain rule)
anonymous
  • anonymous
II know that it's arctan of something ><
anonymous
  • anonymous
The second integral makes use of this fact:\[\int\limits_{}{}\frac{f'(x)}{f(x)}dx=\log f(x)+c\]
anonymous
  • anonymous
If f(x)=1+e^x, then f'(x)=e^x, which is the form you have after rearranging.
anonymous
  • 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
  • anonymous
Gimmy a sec**
anonymous
  • anonymous
u substitution then partial fraction decomposition
anonymous
  • anonymous
I don't get why I can't think of these on my own -.-
anonymous
  • anonymous
i got an answer of ln(e^x/(e^x+1))+c
anonymous
  • anonymous
There's one way to check your answer - take the derivative and see if you get back to the integrand.
anonymous
  • anonymous
yea it does
anonymous
  • anonymous
just checked it
anonymous
  • anonymous
Nope, pfd is in the chapter after this (I slipped it previousy and went ahead) so Lucas answer makes perfect sense...
anonymous
  • anonymous
Lol
anonymous
  • anonymous
Good work
anonymous
  • anonymous
Oh boy, integral of e^sqrtx from 0 to 4...
anonymous
  • anonymous
one sec
anonymous
  • anonymous
I don't even know what to use on this one...
anonymous
  • anonymous
2(e^(sqrt(x))(sqrt(x)-1)|(0,4)
anonymous
  • anonymous
Why?
anonymous
  • anonymous
have u learned integration by parts yet?
anonymous
  • anonymous
Yep
anonymous
  • anonymous
and i take it you know u substitution?
anonymous
  • anonymous
start with your integral and set u equal to sqrt(x)
anonymous
  • anonymous
Ok...
anonymous
  • anonymous
u should get int(2u(e^u)du)
anonymous
  • anonymous
do you follow me so far?
anonymous
  • anonymous
You can put the dx in terms of u aswell? (Like I can write 2u instead of 2sqrtx? )
anonymous
  • anonymous
correct
anonymous
  • anonymous
from this point you can use integration by parts
anonymous
  • anonymous
Ok then that makes sense and i can use the uv vdu then.... -.- I'm retarded
anonymous
  • anonymous
lol its hard to see that without substituting first tho
anonymous
  • anonymous
No... I'm just retarded. And the answer is 2e^2
anonymous
  • anonymous
did u evaluate at zero also?
anonymous
  • anonymous
im getting 2(e^2)+2
anonymous
  • anonymous
Good point, further testing my retardation.
anonymous
  • anonymous
lol common mistake
anonymous
  • anonymous
When you uv vdu sub, can you choose which term is your u and v? I'm pretty sure you can
anonymous
  • anonymous
usually it depends on what you have
anonymous
  • anonymous
to remember what to take for u i use "lipet"
anonymous
  • anonymous
which gives you an order in which to take your u (Lograthims Inverse.trg Polynomials Exponetials Trigonometrics)
anonymous
  • anonymous
Well I got a badass equation here: integral of (lnx) /x^2 from 1 to 2
anonymous
  • anonymous
I know I gotta use uv vdu on it
anonymous
  • anonymous
judging by lipet take log to be your u
anonymous
  • anonymous
and 1/x^2 to be your dv
anonymous
  • anonymous
im getting (-ln(x)/x)-(1/x)+c
anonymous
  • anonymous
your evaluated integral should come out to be (-ln(2)/2)+1/2
anonymous
  • anonymous
were you able to solve it?

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