can someone please solve this integration problem for me? i have been trying it since a week, gone mad trying to solve it. please help.
(ps: its solvable. answer exists.)


edit: "x.x" means x square.

could some one do it?
plz reply.

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

So...

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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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Ok well I'm quite sure you have to use integration by parts. I will update this post when I get the answer for you.

EDIT: I had a go and it was terribly off, probably some little mistake somewhere.  Anyway instead of trying again I used Maple and it only gave me:

2*(x^3+1/x+x/(x^4+1))^(3/2)/(3*x^2-1/x^2+1/(x^4+1)-4*x^4/(x^4+1)^2)

That was after a bit of work of course.  Seemed it missed being able to do a reverse chain rule .  I tried simplifying it but I couldn't find anything, maybe someone else can help.  I hope this helps you. 

quote:

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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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i dont think this can be done by parts. i tried using it, but the expression we get inside the second integration sign is even more complicated than the question itself.
i tried substituting different forms of the expression, that hasnt helped me yet. i tried adding and substrating different terms like (x^3+1/x^3) to get some sort of differentiable term, but with not much success.
i even tried substituting trigonometric properties, but that doesnt help either.

have you tried changing variables like sen( u) = x ? Or maybe the good old Tan(x/2) ? i dunno it just looks like it could be possible with one of those, im too lazy to try now though maybe tomorrow

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Originally posted by Chrono
have you tried changing variables like sen( u) = x

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quote:

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Originally posted by Chrono
Or maybe the good old Tan(x/2) ?

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nups, i already tried various trigo properties, none work.

How do you know it is solvable?

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

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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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quote:

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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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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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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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quote:

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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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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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I don't have enough time to actually to solve it atm, but I think I know how I would attempt it.

int*(x + 1/x) * int*((x^2 + 1/x^2)^(1/2))

As I'm sure someone pointed out earlier.

1/x^2 indeed is 1/x. Try it with a number and put it in your calculator. Insttead of agruing about the square root part, why don't just fuse the expressions under the root to one? Might be more complicated, but at least you can't have different opinions then.

x^2 + 1/x^2 = (x^2 + 1) / x^2

Now of course, we will have the fun of realising there are no x outside the parenthesis, so we will have to divide and get some funky number, which I'm certaninly not arsed to do atm.

EDIT: As I wasn't sure about the thing you are arguing on, I decided just to comment it. Shouldn't it be possible just to take the square root of x^2, which is x, and the root of 1/x^2, which is 1/x, and then write it as x + 1/x and then int* that part?

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Originally posted by Vilkku
int*(x + 1/x) * int*((x^2 + 1/x^2)^(1/2))

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quote:

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Originally posted by Vilkku
EDIT: As I wasn't sure about the thing you are arguing on, I decided just to comment it. Shouldn't it be possible just to take the square root of x^2, which is x, and the root of 1/x^2, which is 1/x, and then write it as x + 1/x and then int* that part?

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Originally posted by Chrono

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Originally posted by Vilkku
int*(x + 1/x) * int*((x^2 + 1/x^2)^(1/2))

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you cant do that, lol

counterexample:
int x^2 =(X^3)/3
int x * int x = (x^4)/4

i tried to solve it earlier but didnt manage to get to an answer i used to be good at those, but it's been a long time since i had to integrate a difficult expression

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Originally posted by Rubber Stamp

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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 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)
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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You sure the question wasn't (x + 1/x)(x^2 + 2 + 1/(x^2))^(1/2) ?

if that was the case then it would be simply the integral of (x + 1/x)^2 but without that +s under the quareroot it makes it almightily difficult.  Or even if (x + 1/x) was replaced with the derivative of (x^2 + 1/(x^2)).  I'll keep trying my best to get an answer, I got something at home but it's a bit ugly and I have to check with it's derivative just to make sure I did it right but I didn't want to just yet because of it's ugliness

I tried and I thought I had the good idea, but I came this far (see attachement). Now I'm stuck again...it is really a difficult one!

haha i did the exact same thing, that was actually my first try .

The file's nice, but you did way too many steps. You could have done it in three lines. it's too tedious to read all that

I did the same thing in about as many steps as chrono.  Will post my steps from there.


EDIT:  Here are my steps to follow on...

code:

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dy/dx = int( (x^4 + 2x^2 + 2 + 2/x^2 + 1/x^4)^1/2 )   what you had

y = (x^4 + 2x^2 + 2 + 2/x^2 + 1/x^4)^3/2              reversal of the chain rule
     3/2 * (4x^3 + 4x - 4/x^3 - 4/x^4)

y = (x^4 + 2x^2 + 2 + 2/x^2 + 1/x^4)^3/2             simplified somewhat
     6 * (x^3 + x - 1/x^3 - 1/x^4

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quote:

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Originally posted by Chrono
The file's nice, but you did way too many steps. You could have done it in three lines. it's too tedious to read all that

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Wow!  It appears that everyone is getting something different.  Here is what I did.  I used substitution and realized that the 1/x^2 can be rewritten as 1x^-2.

Integration: (x + 1/x) (sqrroot(x^2+x^-2)) dx

u = x^2+x^-2 dx
du = 2x - 2x^-1 dx
(1/2)du = x - x^-1
x- x^-1 = x + 1/x

[color=black]

Substitute (x + 1/x) with 1/2 and x^2+x^-2 with "u"

Integrate from here: 1/2 (u^(1/2))

1/2 (u^(3/2)) / (3/2)


      2u^(3/2)
        2 * 3

this equals: 2(x^2 + 1/x^2)^(3/2)
                                 6

THIS WAS CONFIRMED BY A TI-89 Titanium.  Check it yourself.  No integration by parts neccessary!

quote:

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Originally posted by Buttah N Bred
Wow!  It appears that everyone is getting something different.  Here is what I did.  I used substitution and realized that the 1/x^2 can be rewritten as 1x^-2.

Integration: (x + 1/x) (sqrroot(x^2+x^-2)) dx

u = x^2+x^-2 dx
du = 2x - 2x^-1 dx
(1/2)du = x - x^-1
x- x^-1 = x + 1/x

[color=black]

Substitute (x + 1/x) with 1/2 and x^2+x^-2 with "u"

Integrate from here: 1/2 (u^(1/2))

1/2 (u^(3/2)) / (3/2)


      2u^(3/2)
        2 * 3

this equals: 2(x^2 + 1/x^2)^(3/2)
                                 6

THIS WAS CONFIRMED BY A TI-89 Titanium.  Check it yourself.  No integration by parts neccessary!

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no,  I wasn't focused on the date that it was posted.  this was for anybody else who may have had this problem.