## Integral help

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### Integral help

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

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
silvermace

Posts: 73
Joined: Mon Apr 20, 2009 7:56 pm UTC

### Re: Integral help

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).
++$_ Mo' Money Posts: 2370 Joined: Thu Nov 01, 2007 4:06 am UTC ### Re: Integral help ++$_ 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
silvermace

Posts: 73
Joined: Mon Apr 20, 2009 7:56 pm UTC

### Re: Integral help

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.
++$_ Mo' Money Posts: 2370 Joined: Thu Nov 01, 2007 4:06 am UTC ### Re: Integral help ++$_ 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
silvermace

Posts: 73
Joined: Mon Apr 20, 2009 7:56 pm UTC

### Re: Integral help

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.)
++$_ Mo' Money Posts: 2370 Joined: Thu Nov 01, 2007 4:06 am UTC ### Re: Integral help ++$_ 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
silvermace

Posts: 73
Joined: Mon Apr 20, 2009 7:56 pm UTC

### Re: Integral help

Spoilered for that math code stuff.
Spoiler:
\int\frac{xe^x}{\sqrt{1+e^x}}dx

Let:
u=1+e^x

\frac{du}{dx}=e^x

du=e^xdx

x=ln(u-1)

\int\frac{xe^x}{\sqrt{1+e^x}}dx

=\int\frac{ln(u-1)}{\sqrt{u}}e^xdx

=\int\frac{ln(u-1)}{\sqrt{u}}du

=\int\ln(u-1)u^\frac{-1}{2}du

...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.
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Annihilist

Posts: 70
Joined: Mon Mar 05, 2012 12:29 pm UTC
Location: Byron Bay, Australia

### Re: Integral help

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.
++$_ Mo' Money Posts: 2370 Joined: Thu Nov 01, 2007 4:06 am UTC ### Re: Integral help ++$_ 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.
I write reviews of movies, music, literature, and anything else weekly at http://bedroomcritic.blogspot.com/. Check it out if it's your kinda thing.

Annihilist

Posts: 70
Joined: Mon Mar 05, 2012 12:29 pm UTC
Location: Byron Bay, Australia

### Re: Integral help

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.

PM 2Ring

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Joined: Mon Jan 26, 2009 3:19 pm UTC
Location: Mid north coast, NSW, Australia

### Re: Integral help

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?
Ben-oni

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