Can you solve this for me? (simple math)

[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?

[Zkribbler]

Can you solve this for me?

no.


but i can bump it.

[Pekka]

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

[VetLegion]

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.

[VetLegion]

Pekka you bastard you can't do it either!

[Pekka]

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.

[VetLegion]

Hahahahahahahahahha, right sport

You know and I know that you know the solution

[VetLegion]

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

[Ramo]

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.

[Bill3000]

You try trigometric subsitution?

[VetLegion]

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.

[VetLegion]

Originally posted by Bill3000
You try trigometric subsitution?

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

[snoopy369]

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

[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.

[Ramo]

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.