Can you solve this for me? (simple math)

[VetLegion]

[VetLegion]

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.

[VetLegion]

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

[VetLegion]

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.

[DeathByTheSword]

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

[Urban Ranger]

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).

[Petek]

My TI-89 gives the answer as

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

[Odin]

SIMPLE math?

[Whoha]

Re: Can you solve this for me? (simple math)

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.

[Az]

SIMPLE math?

well, it isn't difficult.

[Lul Thyme]

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)...

[Lul Thyme]

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.

[Az]

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