Shoutbox

How do you know it is solvable?

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Originally posted by John Anderton
x^2 = (sin)^2

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lol thats the same than x=sen

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Originally posted by John Anderton

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Originally posted by andrewdodd13
Well... the square root of x^2 1/x^2 is just x + 1/x isn't it?

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wtf?
If that were true most of the maths problems in the world would be solved in a few steps

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Yeah, wasn't really thinking straight when I wrote that...

I used an on-line integrator (its the first hit for "Online integrator" on google ) and it gave (3x^4 - 1) ^-1 . I don't think it's right - doesn't differentiate to the correct result as far as I can tell:

- (12x^3) / ((3x^4)^2)

I'll have another bash it via substitution, but I think integration by parts is the way to go... but I can't do that yet. ^_^.

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Originally posted by Chrono

quote:

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Originally posted by John Anderton
x^2 = (sin)^2

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lol thats the same than x=sen

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I think he was thinking of doing something with De'Mouive's but I don't think it will work in this case.  What I did to get as far as I did was I had to first expand the equations and then use a reversal of the product rule and you should then try to simplify what you get.  It is the only way I can think of doing this problem.

EDIT:  Just expanding out the equation I got it to    x^3+1/x+x/(x^4+1)    you should be able to integrate it in parts now.

for some reason maple is giving me 1/4*x^4+ln(x)+1/2*arctan(x^2)  but I didn't think the arctan was right, I thought it was meant to be a ln   and for some reason it is different to my other expression   I guess it might be that I messed up somewhere last time, I could get rid of the squareroot which i couldn't before.

I don't know, lets hope I'm not as dodgy in my maths tutorial this afternoon, I'll ask my teacher to help if I remember

ah maple, im glad i dont need to use that thing anymore . i really disliked that program for some reason, even though it was pretty useful for Diferential equations

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Originally posted by J-Thread
How do you know it is solvable? 

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its from a set of question from a pretty respectable book.

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Originally posted by markee
EDIT:  Just expanding out the equation I got it to    x^3+1/x+x/(x^4+1)    you should be able to integrate it in parts now.

for some reason maple is giving me 1/4*x^4+ln(x)+1/2*arctan(x^2)

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x^3+1/x+x/(x^4+1) does not equal (x+1/x)*sqrt(x^2+1/x^2) which is why maple gave you the wrong answer.

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Originally posted by andrewdodd13
I used an on-line integrator (its the first hit for "Online integrator" on google ) and it gave (3x^4 - 1) ^-1 . I don't think it's right - doesn't differentiate to the correct result as far as I can tell:

- (12x^3) / ((3x^4)^2)

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lol I did that too, got this:

don't know if it's correct, looks far too complicated for me

edit: image is gone, will try to get it back
edit2: got it

never did this type of problem b4, and substitution technique doesnt seem to work for it so maybe that's how it goes:
we know that :
(x^2 + 1/x^2)^1/2 = (x + 1/x)
so we are left with:
S (x + 1/x) (x + 1/x)
multiply them we get:
S (x^2 + 2 + x^-2)
so the answer is:

1/3 x^3 + 2x - 1/x


Again Im not sure and didnt look for other ways to solve it, but who knows maybe I'm right

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Originally posted by s7a5
never did this type of problem b4, and substitution technique doesnt seem to work for it so maybe that's how it goes:
we know that :
(x^2 + 1/x^2)^1/2 = (x + 1/x)

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Really? So... (5^2 + 1/6^2)^1/2 = (5 + 1/6)? No. If that were true, then (x + 1/x)^2 = (x^2 + 1/x^2), but if you work that out, it's (x^2 + 2 + 1/x^2)