- anonymous

How do I Integrate
(210000 E^t) / (4.1 + 1.2E^t)
latex coming...

- jamiebookeater

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- anonymous

\[ F(x) = \int\limits_{0}^{t} (210000 E^x) / (4.1 + 1.2E^x) dx \]
F[0] = 0

- dan815

|dw:1438639191274:dw|

- anonymous

I can see how I might convert it to this.. is there a product rule for integration?
\[ F(x) = \int\limits_{0}^{t} (210000 E^x) (4.1 + 1.2E^x)^-1 \]

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## More answers

- dan815

use that method i showed u up there

- anonymous

lol, dan, that makes no sense to me at all.

- dan815

u will have to do integration by parts for the equation in the form 1/(1+e^x)

- Astrophysics

Yeah, what exactly did you do danni boi

- dan815

well u can transform your equation into that form if u pull out ur constants

- Astrophysics

Show

- dan815

|dw:1438639475428:dw|

- Astrophysics

I mean I know you can pull the constants out

- Astrophysics

but the other part

- dan815

|dw:1438639533510:dw|

- freckles

he added in a 0

- dan815

do parts on the 2nd term

- Astrophysics

Hey, why can't we just say u = 1+e^x, du = e^x dx,

- dan815

yea thats fine too

- Astrophysics

Then we would just have \[\int\limits \frac{ 1 }{ u } du\]

- Astrophysics

I think that's much more straight forward

- anonymous

so we get ln[u] ?

- Astrophysics

\[\int\limits \frac{ e^x }{ 1+e^x } dx\]
\[u = 1+e^x \implies du = e^x dx \]
\[\int\limits \frac{ 1 }{ u } du\]

- dan815

hehe this was silly

- Astrophysics

:P

- dan815

the derivative of the bottom was on the top of the fraction you shuld notice that

- Astrophysics

When in doubt try u sub @hughfuve hehe

- dan815

|dw:1438639816201:dw|

- dan815

|dw:1438640062643:dw|

- freckles

I'm curious @hughfuve was the question really to integrate
I notice you wrote an definite integral and called it F(t)
\[F(t)=\int\limits_0^t \frac{210000e^x}{4.1+1.2e^x}dx \\ \text{ I mean we could evaluate the integral on the } \\ \text{ right hand side }\]
but depending on the question it is asking we may not have to

- anonymous

Hi freckles, well the actual problem I was trying to work out is
given that integral, how long would it take F[t] to reach the value of 625000 if F[0]=0.
I figured if I could get to an actual definition of F[t] that I could solve for F[t]=625000 and find t

- Astrophysics

Is it F(x) or F(t)

- freckles

should be F(t) if the upper limit is t

- Astrophysics

Yeah that's what I thought at first as well, but it says F(x) at the top just making sure

- anonymous

I think it would be F[t] given that integral, how much time t before we reach 625k

- anonymous

Unfortunately I have not been introduced to integration by parts.. (not sure if that is needed) I have only just been introduced to the Fundamental Theorem, and that's about all I know about integration.
\[\int\limits_{a}^{b} f(x) dx = F(b)-F(a)\]
That's the net extent of my knowledge so far.. so I'm looking at all this and wondering how it all relates. Up until now it has been all differentiation. And about all I know is that this equation results in an antiderivative. My class has no textbook and no lectures and no teacher.. so it's really bloody hard. I have to google my way to answer problems, but I do have mathematica installed, and I am allowed to use mathematica as a solution. But for some reason , it just doesn;t like this equation at all, so I have to prove it by hand.

- anonymous

If you can throw me a google link that makes this any easier.. I'd appreciate it.

- freckles

so you haven't actually talk about how to integrate
like substitution (this was the method used above)

- anonymous

have not seen anything about substitution in the course.. but I did use substitution to differentiate with the chain rule.. (which was also not in the course, lol) .. but I do have chain rule down pretty good.

- anonymous

so far the recommendation to integrate.. is just work out what you would do to differentiate your way here. Which is fine for most things that I can just look at a table of differentiation rules.. but this one threw me off..

- anonymous

They use Dsolve and integrate a lot in the course to show examples of what you are trying to achieve.. they dont actually use the long hand methods.

- Astrophysics

Youtube has many videos as the links I've sent you, in 10 minutes you can learn u - substitution, you really take advantage of these videos: https://www.youtube.com/watch?v=qclrs-1rpKI

- Astrophysics

should*

- anonymous

I have googled substitution rule.. doing some reading now.. I;ll close this out and come back to it later.. thanks everyone, this will get me there.

- anonymous

thanks astro

- Astrophysics

Go with the examples on youtube, as you can pause it try it yourself and then watch the person work it out, and once you get good you can also make it 2x speed and watch it in half the time. Youtube is great, take advantage!!

- anonymous

lol, I can't believe this bloody course.. lmao.. its so bad,

- Astrophysics

Your welcome :)

- Astrophysics

Haha, yeah it can be tedious but it's beautiful.

- freckles

\[\int\limits_{}^{} \frac{ a e^x}{b+c e^{x}} dx \\ \text{ \let } u=b+c e^{x} \\ \frac{du}{dx}=0+ce^x \\ \frac{du}{dx}=ce^{x} \\ du=c e^{x} dx \\ \text{ divide } c \text{ on both sides } \\ \frac{1}{c} du=e^x dx \\ \int\limits \frac{ae^x}{b+ce^{x}} dx=a \int\limits \frac{ 1}{b+ce^{x}} e^{x} dx =a \int\limits \frac{1}{u} \frac{1}{c} du \\ \frac{a}{c} \int\limits \frac{1}{u} du=\frac{a}{c} \ln|u|+K=\frac{a}{c}\ln|b+ce^{x}|+K\]
by the way a,b,c are constants that are you were given
and K is the constant of integration which won't matter for your case since you have a definite integral

- anonymous

nice one freckles.... that makes sense.

- freckles

as you see differentiating that answer should give us what we started with
the derivative of K is 0
so we are looking at:
\[(\frac{a}{c} \ln|b+ce^{x}|)' \\ \text{ by constant multiple rule } \\ \frac{a}{c}(\ln|b+ce^{x}|)' \\ \text{ by chain rule } \\ \frac{a}{c} \frac{0+ce^{x}}{b+ce^{x}} \\ \frac{a}{c} \frac{ce^{x}}{b+ce^{x}} \\ \text{ we know } \frac{c}{c}=1 \\ a \frac{e^{x}}{b+ce^x} \\ \frac{ae^{x}}{b+ce^{x}} \\ \text{ just if you wanted to check your integration }\]

- anonymous

sweet thanks.. okay I have to head out for a bit.. but this is very helpful.

- anonymous

I wish I could give you all medals

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