Math Challenge

[Tree-Mac]

How the heck do you integrate this?

x dx/ (x^3 - 1) ...... (x^3 means "x cubed")

I spent like 3 hours trying to solve this problem but couldn't do it. Thanks in advance. :confused:

[Royals Ego]

I don't know. But can I ask what you did in those 3 hours?

[Stevierebel]

Nevermind. I might work on it if someone else doesn't give you an answer..

[D-Lite]

(x^4) / 4 - x + C

lol i dont think its that simple

[D-Lite]

(x^4) / 4 - x + C

EDIT..

you did not see the x*dx DIVIDED BY (x^3-1)


in any case.. i used an online integrator and that answer that popped up was too complicated.. im not gonna touch this one

[Tree-Mac]

I don't know. But can I ask what you did in those 3 hours?

From first look, I thought it was partial fractions, but it's not. And obviously substitution doesn't work either. I'm thinking maybe trigonometric substitution, but how do you make it into a^2 + x^2 form? I don't know.

[D-Lite]

How the heck do you integrate this?

x dx/ (x^3 - 1) ...... (x^3 means "x cubed")

I spent like 3 hours trying to solve this problem but couldn't do it. Thanks in advance. :confused:

alright nvm ill try, how about trig sub:


x^2=sec(x)?

then use the identity tan(x)^2=sec(x)^2-1

dunno if that works

[D-Lite]

yea this should work, remember to find the new dx as well

[Tree-Mac]

yea this should work, remember to find the new dx as well

Can you elaborate? This is what I got if I substitute like you said:

Sqrt (sec x) dx/ ((Sqrt (sec x) * (tan^2(x) - 1)-1) ........

This still doesn't work because how do you take care of the denominator? It's even messier than the original problem.

[D-Lite]

ahh my bad, its way too early in the morning.

I messed up in the beginning. What i wrote does not work at all. Back to the drawing board because i have no clue what to do.
:mad:

[Mr. Clutch]

:confused:


Does anyone know what these guys are talking about?

[Andy3000]

What the hell is "x"?

[D-Lite]

its been a while since ive done this stuff..

im gonna let you try this one if you want, because i have to go

x^3-1 factors into (x-1)*(X^2+x+1)

there is a trick in integration where

x/a= (x-1+1)/a

so by applying that

(x-1+1)/[(x-1)*(X^2+x+1)]

split them up into two different fractions

(x-1)/[(x-1)*(X^2+x+1)] + 1/[(x-1)*(X^2+x+1)]

in the first term, the (x-1) cancel out

you can try this

peace.

[TheRealist137]

I thought you needed to use substitution.

u = x^3 -1
du = 3x^2 dx

Take the 3 out

so it becomes

x^2 dx / (x^3 -1)

or

du/u

which makes it

3 ln (x^3-1) + C

I think that's what the answer is, not entirely sure that it's right.

[TheRealist137]

I thought you needed to use substitution.

u = x^3 -1
du = 3x^2 dx

Take the 3 out

so it becomes

x^2 dx / (x^3 -1)

or

du/u

which makes it

3 ln (x^3-1) + C

I think that's what the answer is, not entirely sure that it's right.

Actually, I don't think it's right, I had x^2 dx on the top while the question was only x dx.

My bad.

[Xcellerator]

On my ti-89 I get:

[ln( (x - 1)^2 / abs(x^2 + x + 1) ) / 6] + [ sqrt(3) * invtan(sqrt(3) *(2x + 1)) / 3]

[Northside Storm]

I would guess factoring the x^3-1 into an x^2 form, then using u substitution with the factor being 1/2.

I'm too lazy to actually check if that works though...and it's probably a bit too simple considering the answers getting churned out lol.

[Tree-Mac]

Thanks D-Lite! What is that x/a trick? Anyone else can do this problem?

[MadMax]

I got her numbah.

How ya like them apples???

[JuanValdez]

Bad thread title. This isn't a challenge so much as help doing your homework.

[Tree-Mac]

Bad thread title. This isn't a challenge so much as help doing your homework.

I'm not doing my homework. I'll graduate from college this summer. I was reviewing calculus because I'm planning on going to graduate school.

It's been years since I took calculus. Been too busy to re-study.

[Tb-Cain]

wow, i took calculus 1 thru 3, differential equations, calculus based physics, etc. etc. to get my b.s. in c.s.

15 years later, i haven't got a clue. :confused:

[FlyerFanatic]

I got her numbah.

How ya like them apples???

hahahaha :D

[WhoMikeJames]

What the hell is "x"?

This made me laugh really hard.

[Steve_Francis_rules]

How the heck do you integrate this?

x dx/ (x^3 - 1) ...... (x^3 means "x cubed")

I spent like 3 hours trying to solve this problem but couldn't do it. Thanks in advance. :confused:

You integrate this by looking it up in an integral table. At least that's what I would do.

[Franchise3]

I am now realizing that I don't remember anything from calculus class. :(

[thelasik]

On my ti-89 I get:

[ln( (x - 1)^2 / abs(x^2 + x + 1) ) / 6] + [ sqrt(3) * invtan(sqrt(3) *(2x + 1)) / 3]

LOL I love my TI-89

[D-Lite]

Thanks D-Lite! What is that x/a trick? Anyone else can do this problem?

I posted it on the first page:


x/a= (x-1+1)/a

so by applying that

(x-1+1)/[(x-1)*(X^2+x+1)]

split them up into two different fractions

(x-1)/[(x-1)*(X^2+x+1)] + 1/[(x-1)*(X^2+x+1)]

in the first term, the (x-1) cancel out


I have no clue what to do next, but its a way of simplifying it

and looking at the final answer the term (X^2+x+1) is in it.. which is in my simplification