How to integrate (202.31)/(1+e^(3.938-0.314x)).

[ThomasJR]

Homework Statement

Calculate the antiderivative of the following:
(202.31)/(1+e^(3.938-0.314x)) using either U-substitution or Parts but NOT partial functions.

Homework Equations

please show to me your work

The Attempt at a Solution

I'm not sure how to approach this, I'm stuck

 

[MysticDude]

I don't know about Integration by Parts but can you not start by factoring out the constant making it $$202.31\int\frac{1}{1+e^{3.938-0.314x}}$$? I'm just giving you ideas for now :D

 

[ThomasJR]

Oh yes, i factored out the constant, then i set u= 3.938-0.314x, and du=-0.314
which gets me 1/(1+e^(u)... :( I'm not sure how to integrate this now! Please help me

 

[MysticDude]

Now just divide the outer constant by -.314 to get $$-644.299\int\frac{1}{e^{u}+1}du$$

I'm not too much of an expert but I used Wolfram Alpha to make sure I was going in the right direction. So you can use this to help you out. There is a button that says show steps and that will show you how to get the answer. But if you get confused, I'll do my best :D

 

[ThomasJR]

Yes i went to wolfram mathematica too, and got the same setp as you did here, however, i am not suppose to use partial function at this point, i am bounded by Parts and U-substitution :(
The next thing i did was multply the top and bottom by 1+e^u-e^u, and cancel out like terms which leads me to (1+e^u-e^u)/(1+e^u) and now i am stuck again :(

 

[MysticDude]

Oh gosh, I haven't studied integration by parts so I can't help you there if you have to use it (sorry). I'll try and think of a way though.

 

[ThomasJR]

It can be done without using parts!

 

[Dick]

It can be done without using parts!

The usual way to integrate 1/(1+e^u) is to substitute v=(1+e^u) and use partial fractions. Since you aren't supposed to do that, continue with what you were saying in post 5. 1/(1+e^u)=(1+e^u)/(1+e^2)-e^u/(1+e^u)=1-e^u/(1+e^u). The second term is easy to integrate using the substitution v=(1+e^u).

 

[ThomasJR]

ok so now I have 1-e^u/(1+e^u), if i make any more substitutions i will b totally lost. Can somebody please do this problem step by step

 

[quantumdude]

It's against the Forum rules for us to post complete solutions. We're here to help you, not do it for you. Follow Dick's suggestion, it's easy to implement. But allow me to clear up one potential point of confusion.

Dick said...

continue with what you were saying in post 5. 1/(1+e^u)=(1+e^u)/(1+e^2)-e^u/(1+e^u)=1-e^u/(1+e^u). The second term is easy to integrate using the substitution v=(1+e^u).

Dick's method is the right one, but it's not what you were saying in post 5. In that post you said multiply the top and the bottom by $1+e^u-e^u$. That's not what Dick is doing. He is simply using $1=1+e^u-e^u$ in the numerator alone. That's the way to do it.

 

[ThomasJR]

1=1+e^u-e^u,... wait where did this come from?

 

[quantumdude]

Dick introduced that substitution to take your one difficult integral and transform it into two easy ones. To do integration you have to have a certain measure of creativity.

 

[ThomasJR]

Let me get this right. So i have this original function (1/((e^x)+1), i multiply the top and bottom by 1+e^x-e^x, and then simplifying i have, (1+e^x-e^x)/((1+e^x)... So now i set u= 1+e^u-e^u ? is this what you are saying?

 

[Mark44]

No, you don't multiply by anything. All you are doing is rewriting the numerator (1) as 1 + e^x - e^x. If you add e^x and then subtract it back off again, you get exactly what you started with, which is 1 in this case.

 

[quantumdude]

No. Here's what I said:

Dick's method is the right one, but it's not what you were saying in post 5. In that post you said multiply the top and the bottom by $1+e^u-e^u$. That's not what Dick is doing. He is simply using $1=1+e^u-e^u$ in the numerator alone. That's the way to do it.

Don't multiply the top and bottom by anything. Just rewrite the numerator of the integrand as $1+e^u-e^u$ and split the integral into two integrals, like so:

$$\int\frac{1}{1+e^u}du=\int\frac{1+e^u-e^u}{1+e^u}du$$

$$\int\frac{1}{1+e^u}du=\int\frac{1+e^u}{1+e^u}du-\int\frac{e^u}{1+e^u}du$$

Dick told you what to do from there.

 

[ThomasJR]

OK so now i have (1+e^x-e^x)/(e^x+1), now i set u=e^x+1 du=e^x, now i set up my integral u-(u-1)/u and it gives me 1-1+Ln|u|... Please tell me if this is right... I feel suicidal right now because this thing is due Thursday :(

 

[ThomasJR]

OMG ok i got it... so now my final integral u-ln|u|?

 

[quantumdude]

OMG ok i got it... so now my final integral u-ln|u|?

Actually, I'm not sure because I've just noticed that you're using $x$ to represent 2 different variables. That's somewhat confusing. Your original integral was in terms of $x$. If you follow through the substitutions you have $u=3.938-0.314x$. If you look at my last post, you need to let $v=1+e^u$ in the second integral on the right side. You need to clean that up before I can say if you got it right.

 

[ThomasJR]

Sir, this is what started off with, now please tell me if i did everything correct. I will be thankful to you guys truly.
so here's what i started off with.
202.31/(1+e^(3.938-0.314t)... Now i set x=3.938-0.314t, du=-0.314, revised du= -644.299.
Next i set that integral up. -644.299 (1)/(1+e^(x))... Since -6.44 is a constant i didn't thought it was important to carry that out... The i integrated using you guys's method and ended off with u-ln|u|... Now i went back and plugged everything in.. But in wolfram mathematics, when i tell it to integrate the exact function, it tells me that the result is
http://www3.wolframalpha.com/input/?i=Integrate+%28202.31%29%2F%281%2Be^%283.938-0.314x%29%29
They used partial functions and i am not suppose to use that, i am not suppose to use u-substitution... Please help me know where did i go wrong... is it a Logarithmic property i am missing out...?

 

[ThomasJR]

I can fix that confusion but can you help me clarify my result.. Thanks a lot!

 

[quantumdude]

Sir, this is what started off with, now please tell me if i did everything correct. I will be thankful to you guys truly.
so here's what i started off with.
202.31/(1+e^(3.938-0.314t)... Now i set x=3.938-0.314t, du=-0.314, revised du= -644.299.

OK

Next i set that integral up. -644.299 (1)/(1+e^(x))...

NO. This is the problem that I just pointed out, you're using $x$ to represent 2 different variables. Your integrand should be in terms of $u$ at this point.

Since -6.44 is a constant i didn't thought it was important to carry that out...

You're right, it's not terribly important. But keeping your variables straight is important. Please make the correction that I advised and follow through with it. It's not possible to say if you're right if we're not sure of what you mean by "$u$".