08-04-2005 05:33									#1
	Can you solve this for me? (simple math) Support Apolyton buy from Amazon

I wouldn't bug you folks but it's 2:30 AM here and I can't go wake up someone just for this.

Solve this:

integral( (squareroot( x ) / (x+2))dx )


I hope you understand it, its quite simple equation and I can't draw it so I wrote it that way.

I need the procedure of solving, I have the solution.

Anyone?

	08-04-2005 05:46									#2
	Can you solve this for me? Support Apolyton, buy Civilization 2

no.


but i can bump it.

	08-04-2005 05:49									#3
	Avatar Enlargement: We've got the solution

no free lunch tickets dude. Wikipedia will let you on track about integrals in case you have been sleeping in class.



PS. it's not too difficult

And now it's time for me to sleep as well. Ta-ta

	08-04-2005 05:51									#4
	Support Apolyton

It ought to be quite simple. I'm frustrated for not being able to find the solution. Substitution and partial integration failed me, and I've been trying for an hour.

	08-04-2005 05:52									#5
	Tired of ads?

Pekka you bastard you can't do it either!

	08-04-2005 05:55									#6
	Lose 30 kilos (of popups)

if I couldn't, I'd be a pretty sucky math student, now wouldn't I? The reason I won't give you the answer is because this is not complicated, if you know integrals, you know this. Thus I think you have some holes in your study so you need to figure it out yourself, even wikipedia info could be sufficient.

Only an hour? Dude, that's nothing. Keep on going.

I go to sleep now. I'll check this thread out tomorrow in case you have tried hard but not succeeded. But you got to put some thinking here so I knwo you tried hard, and if you just can't get it, I show you the way.

Now to sleep.

	08-04-2005 06:05									#7
	Support Apolyton, buy GURPS/ Alpha Centauri

Hahahahahahahahahha, right sport

You know and I know that you know the solution

	08-04-2005 06:08									#8
	Avatar Enlargement: We've got the solution

I got something now, but it's different than what http://integrals.wolfram.com/ tells me. Hmmmm....

	08-04-2005 06:20									#9
	Browse Apolyton AD-FREE

Define y = sqrt(x/2). Then, dy = dx/(4sqrt(x)), or dx = 4ydy

Then you need the integral of 4ydy*y/(2y^2 + 2), or 2y^2dy/(y^2 + 1) (let's call this *)

Now define u = y and dv = 2ydy/(y^2 + 1).

Then * is the integral of udv = uv|evaluated at boundaries - integral of vdu

The integral of dv = ln(y^2 + 1). While du = dy.

So * = yln(y^2 + 1)|@boundary - integral of ln(y^2 + 1)dy.

And so on. Find a clever substitution, compute, rinse and repeat. That's how it's done.

	08-04-2005 06:20									#10
	Support Apolyton, buy Alpha Centauri

You try trigometric subsitution?

	08-04-2005 06:31									#11
	Get a bigger avatar today!

quote: So * = yln(y^2 + 1)|@boundary - integral of ln(y^2 + 1)dy. And so on. Find a clever substitution, compute, rinse and repeat. That's how it's done.

Thanks for the effort Ramo

But... it's still not solved. What is the integral of ln(y^2 + 1)dy?

All I know how to do is substitute t=y^2+1, but that gets me nowhere. It gets me to integral of t/sqrt(t+1)dt, which I can't solve.

	08-04-2005 06:34									#12
	Help yourself to an AD-FREE life

quote: Originally posted by Bill3000 You try trigometric subsitution?

Nope. I'm not good at it. Care to try?

	08-04-2005 07:49									#13
	Increase Your PM Length

I can't quite see how wolfram gets there either, and i totally don't follow the substition Ramo suggests - i see the derivitave of sqrt(x/2) as 1/2*(1/sqrt(x))*(1/sqrt 2). (Do you mean sqrt(x)/2 ??), and i totally fail to see how y=sqrt(x/2) is helpful in the least, as it's sqrt(x)/(x+2) not sqrt (x/2), but maybe i'm misreading it?

Anywho, I have to say from first glance it looked as if logs are the way to go, and wolfram seems to agree - but it's been 5 years since i even thought about calculus, so i'm not going to remember any of that. ln(sqrt(x)/(x+2)) = ln(sqrt(x)) - ln(x+2) = 1/2(ln x) - ln(x+2), which leads to a fairly simple ln integral. I just don't remember when and how you're allowed to use ln like that ... and i could be totally smoking something But i recall how much i luvved my ln when i was back in college

	08-04-2005 10:15									#14
	Lose 30 kilos (of popups)

I also totally don't follow what Ramo did.

If you set y = sqrt(x) then original equation becomes y/(y² + 2). Something like this. Then it's just integration by parts, I think.

	08-04-2005 10:22									#15
	Help yourself to an AD-FREE life

Obviously, I left off a root-2 factor in my first derivitive (which doesn't change much). And yes, I did mean to do that substitution - to get rid of the 2. Because the integral of dy/(y^2 + 1) = arctan(y). I didn't finish 'cuz integration is tedious. You just have to hammer away at different substitutions to find the solution manually.

	08-04-2005 12:23									#17
	Support Apolyton, buy Galactic Civilizations

quote: Originally posted by Urban Ranger I also totally don't follow what Ramo did. If you set y = sqrt(x) then original equation becomes y/(y2 + 2). Something like this. Then it's just integration by parts, I think.

Tried that, but it gets me to integrate: 1/sqrt(2)*arctan(y/sqrt(2)) which is not in the tables and not something I know to do either.

	08-04-2005 12:26									#18
	Full PM-box? Change here!

quote: ln(sqrt(x)/(x+2)) = ln(sqrt(x)) - ln(x+2) = 1/2(ln x) - ln(x+2), which leads to a fairly simple ln integral. I just don't remember when and how you're allowed to use ln like that

I think you're not, still, thanks for the idea

	08-04-2005 12:29									#19
	Support Apolyton, buy Call to Power 2

quote: I didn't finish 'cuz integration is tedious. You just have to hammer away at different substitutions to find the solution manually.

Which I was unable to do. Anyway, off to school soon. I'll find someone to solve it for me there. Thanks everyone who tried to help (NOT Pekka obviously ). I'll post the result later.

	08-04-2005 12:37									#20
	Support Apolyton buy from Amazon

i am really stupid or is it 1/8x^2+1/4x+ln(x) and then just fill in the borders to get the answer...

you make the sqrt(x) and x and then just partial intergration

	08-04-2005 14:34									#21
	Browse Apolyton AD-FREE

quote: Originally posted by VetLegion Tried that, but it gets me to integrate: 1/sqrt(2)*arctan(y/sqrt(2)) which is not in the tables and not something I know to do either.

You could always try substituting u = 1/sqrt(2).

	08-04-2005 21:37									#22
	Increase Your PM Length

My TI-89 gives the answer as

2*sqrt(x) - 2*sqrt(2)*arctan(sqrt(2*x)/2)

	08-04-2005 22:52									#23
	Get a bigger avatar today!

SIMPLE math?

	08-04-2005 22:58									#24
	Re: Can you solve this for me? (simple math) Help yourself to an AD-FREE life

quote: Originally posted by VetLegion I wouldn't bug you folks but it's 2:30 AM here and I can't go wake up someone just for this. Solve this: integral( (squareroot( x ) / (x+2))dx ) I hope you understand it, its quite simple equation and I can't draw it so I wrote it that way. I need the procedure of solving, I have the solution. Anyone?

its probably some very non-simple partial integral crap that I don't want to mess with.

	08-04-2005 23:02									#25
	Suffering from ads?

quote: SIMPLE math?

well, it isn't difficult.

	08-04-2005 23:32									#26
	Support Apolyton, buy Civilization: The Boardgame

I will admit that I have a major in math, and tried this for 5-10 minutes and wasnt able to do it.
Though I havnt done any integration in a couple of years...
I would be very surprised if it could not be done by elementary methods, though (simple substition, partial fractions, trigo, the uv-int v du that I cant remember the name etc...)
(I am pretty sure partial fractions in the complex numbers would have worked, but then Im not sure the original poster would have understood)...

	08-04-2005 23:39									#27
	Support Apolyton, buy Civilization: The Boardgame

Ha ha
got it
do the first substition ramo gave u=(x^1/2)
you should get 2 (u^2)/((u^2)+2) du
do polynomial long division you get
(2 - (4/(u^2 +2)) du
divide into 2 integrals, the second one is an arctan

do it properly and ull get what petek said...

a couple lines, very easy, not sure where Ramo was going with his solution



I actually remembered the rule when you have a quotient of polynomials :
If the degree of numerator is higher or equal to denominator, ALWAYS do long division first.
Once I remembered that, it was clear.

Last edited by Lul Thyme on 08-04-2005 at 23:45

	08-04-2005 23:45									#28
	Lose 30 kilos (of popups)

funny, I am retaking the diff. eq. course, and we've had a similar integral in it.