Integral help

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silvermace
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Integral help

Postby silvermace » Wed Apr 04, 2012 2:06 am UTC

I'm completely stumped on how to start this integra, any help would be much appreciated!!

Image

edit: no hyperbolic trig since we haven't touched on that yet.

i tried usubstituting sqrt(1+e^x) and then doing integration by parts but i ended up being stuck with the integral (u^2 / (u^2 - 1)) which i couldn't solve

other than that I have no luck

++$_
Mo' Money
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Re: Integral help

Postby ++$_ » Wed Apr 04, 2012 2:19 am UTC

I started with u = sqrt(1+ex).

That got me to (1/2)log(u2-1) (unless I messed up). That should be doable. (Just factor u2-1).

silvermace
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Re: Integral help

Postby silvermace » Wed Apr 04, 2012 2:22 am UTC

++$_ wrote:I started with u = sqrt(1+ex).

That got me to (1/2)log(u2-1) (unless I messed up). That should be doable. (Just factor u2-1).


yes, then id do integration by parts from there. But that ends me up with a u^2 / (u^2-1) which i have no idea about how to do

++$_
Mo' Money
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Re: Integral help

Postby ++$_ » Wed Apr 04, 2012 2:25 am UTC

If you want to do u2/(u2-1), just add and subtract 1 from the numerator to get 1 + 1/(u2-1). That you should know how to do.

silvermace
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Re: Integral help

Postby silvermace » Wed Apr 04, 2012 2:34 am UTC

++$_ wrote:If you want to do u2/(u2-1), just add and subtract 1 from the numerator to get 1 + 1/(u2-1). That you should know how to do.


could you please expand on the adding and subtracting 1 to the numerator?
If you add and subtract you'd get; (u2+1-1)/(u2-1); i can split it up to integral 1 + integral 1/(u2-1). but the 2nd integral would be a hyperbolic trig which i can't do

++$_
Mo' Money
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Re: Integral help

Postby ++$_ » Wed Apr 04, 2012 2:35 am UTC

You can't integrate 1/(u^2-1)?

Try partial fractions. (Yes, you could also do it with hyperbolic trigonometric functions, but that will just get you the same answer in a more obscure form.)

silvermace
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Re: Integral help

Postby silvermace » Wed Apr 04, 2012 2:38 am UTC

++$_ wrote:You can't integrate 1/(u^2-1)?

Try partial fractions. (Yes, you could also do it with hyperbolic trigonometric functions, but that will just get you the same answer in a more obscure form.)


ah. yes. I understand now, thank you very much

Annihilist
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Re: Integral help

Postby Annihilist » Wed Apr 04, 2012 10:02 am UTC

Spoilered for that math code stuff.
Spoiler:
[math]\int\frac{xe^x}{\sqrt{1+e^x}}dx[/math]
Let:
[math]u=1+e^x[/math]
[math]\frac{du}{dx}=e^x[/math]
[math]du=e^xdx[/math]
[math]x=ln(u-1)[/math]

[math]\int\frac{xe^x}{\sqrt{1+e^x}}dx[/math]
[math]=\int\frac{ln(u-1)}{\sqrt{u}}e^xdx[/math]
[math]=\int\frac{ln(u-1)}{\sqrt{u}}du[/math]
[math]=\int\ln(u-1)u^\frac{-1}{2}du[/math]

...Now I'm not sure what to do from here - I don't know how to integrate log functions, even after a quick google search.

This is xkcd. I'm sure someone will know.
Last edited by Annihilist on Sun Apr 08, 2012 12:21 pm UTC, edited 1 time in total.

++$_
Mo' Money
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Re: Integral help

Postby ++$_ » Wed Apr 04, 2012 12:09 pm UTC

Your last line has an error (it should be u-1/2). It's strongly advised that you now put v = sqrt(u).

One way to integrate log(x) is by parts. One part is log(x). The other part is dx.

Annihilist
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Joined: Mon Mar 05, 2012 12:29 pm UTC

Re: Integral help

Postby Annihilist » Sun Apr 08, 2012 12:22 pm UTC

++$_ wrote:Your last line has an error (it should be u-1/2). It's strongly advised that you now put v = sqrt(u).

One way to integrate log(x) is by parts. One part is log(x). The other part is dx.
Fixed, thanks.

I haven't learned how to do integration by parts. I'm leaving it to others in this forum, if anyone can be bothered.

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PM 2Ring
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Location: Sydney, Australia

Re: Integral help

Postby PM 2Ring » Sat Apr 14, 2012 8:48 am UTC

Annihilist wrote:I haven't learned how to do integration by parts. I'm leaving it to others in this forum, if anyone can be bothered.

Integration by parts is fun - it's just using the product rule for derivatives in reverse.
d(uv)/dx = vdu/dx + udv/dx
Therefore, vdu/dx = d(uv)/dx - udv/dx
Or in differential form,
vdu = d(uv) - udv
So the integral of vdu = uv - the integral of udv

To actually use this to do integrals can be a little tricky, since you have to find appropriate v and du/dx by guesswork. But with practice you soon get to recognise likely suspects.

Sometimes, the application of integration by parts doesn't lead to an integral that's immediately solvable but it does give you a relationship that can be used to solve (or at least simplify) the original integral.

Ben-oni
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Re: Integral help

Postby Ben-oni » Sat Apr 14, 2012 10:23 am UTC

PM 2Ring wrote:d(uv)/dx = vdu/dx + udv/dx

While correct for the trivial case of u,v : R -> R, I would highly discourage remembering this form. The product rule is, and shall always be, "d(uv)/dx = (du/dx) v + u (dv/dx)". I actually checked this in my analysis text, and found they had the "wrong" form there, too. While the difference may seem pedantic, it will eventually matter. For instance, what if u,v : R -> Mnxn(R)? Does it matter which order you write it in that case? If so, what does the quotient rule look like?


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