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hey... can some on try this.it's a bit hard, but i will look for who solves. okey check this out:
integrate{1/(x^4+1)}

By Fairyofwind (Fairyofwind) on Sunday, August 10, 2003 - 11:52 am: Edit

(1/(4*Sqrt[2]))(2*ArcTan[1+x*Sqrt[2]]-2*ArcTan[1-x*Sqrt[2]]-Ln[(x*Sqrt[2]-x^2-1)/(x*Sqrt[2]+x^2+1)])+C.

By Sovietman (Sovietman) on Sunday, August 10, 2003 - 01:15 pm: Edit

is this calculus?

By Fairyofwind (Fairyofwind) on Sunday, August 10, 2003 - 01:48 pm: Edit

yea

By Quack (Quack) on Sunday, August 10, 2003 - 02:07 pm: Edit

•••• and im taking that! i'm going to die

By Serene (Serene) on Sunday, August 10, 2003 - 02:21 pm: Edit

Fairy: that looks complicated. Where did you use trig substitution? (ie please show steps *sigh*)

Edited: nevermind... I don't think I learned that much calculus yet

By Fairyofwind (Fairyofwind) on Sunday, August 10, 2003 - 03:26 pm: Edit

Using elementary methods, first decompose x^4+1 into partial fractions, x^4+1=(x^2+x*Sqrt[2]+1)(x^2-x*Sqrt[2]+1). Solving for a,b,c,d in (ax+b)/(x^2+Sqrt[2]x+1)+(cx+d)/(x^4-Sqrt[2]x+1) = 1/(x^4+1), yields equations, a+c=0, -Sqrt[2]a+Sqrt[2]c+b+d=0, a-Sqrt[2]b+c+Sqrt[2]d=0, b+d=1, yields, a=1/(2*sqrt[2]),b=1/2, c=-1/(2*Sqrt[2]), d=1/2. So Integral[1/(x^4+1),x]=Integral[(2+Sqrt[2]x)/(4x^2+4Sqrt[2]x+4) + (Sqrt[2]x-2)/(-4x^2+4Sqrt[2]x-4),x]=Integral[(2/(4x^2+4Sqrt[2]x+4))+(Sqrt[2]x)/(4x^2+4Sqrt[2]x+4)+(Sqrt[2]x)/(-4x^2+4Sqrt[2]x-4)-(2/(-4x^2+4Sqrt[2]x-4)),x]. Notice that we've reduced the problem to the integrals Integral[1/(x^2+bx+c),x] and Integral[x/(x^2+bx+c),x], since obviously Integral[k*f[x],x]=k*Integral[f[x],x]. Completing the square on x^2+bx+c=(x+(b/2))^2+(c-(b^2/4)), let k=Sqrt[c-(b^2/4)], so x^2+bx+c=(x+(b/2))^2+k^2. Knowing that Integral[1/(a^2+x^2),x]=(1/a)*ArcTan[x/a]+C, we can now integrate 2/(4x^2+4Sqrt[2]x+4) and -2/(-4x^2+4Sqrt[2]x-4). Yet, also knowing Integral[x/(a^2+x^2),x]=(1/2)Ln[a^2+x^2], we can integrate the remaining two terms. Now all that remains to do is simplify...

By Serene (Serene) on Sunday, August 10, 2003 - 03:34 pm: Edit

*speechless*
wow... (somewhere else I saw another way of integration which used some really really weird method but anyway...) hmm... interesting factorization there... wow... how did you ever think to factor x^4+1 like that... *full of admiration* =)

By Aoe2guy (Aoe2guy) on Sunday, August 10, 2003 - 03:37 pm: Edit

serene, is that a "lester approach?" haha

By Serene (Serene) on Sunday, August 10, 2003 - 03:43 pm: Edit

shush. *goes back to admiration*

fairy: do you by any chance go to AAST or some magnet school? I would hardly understand your solution when I was in sophomore year.

you know however cc is, at least i'm getting minor minor calc reviews here... I hope xaq posts these problems more often! =)

By Emma (Emma) on Sunday, August 10, 2003 - 04:32 pm: Edit

errrm i'm really confused, that looks like a normal simple integration question.... never seen all that stuff fairy has done before, what is it? hehe, surely one can just do (or did i miss the point of the question???):

{1/(x^4+1)} = x^-4 + 1

so when you intergrate:

(x^-3 /-3) + (1^2 / 2) + c

= (-1 / 3x^3) + (1/2) + c

Whats the answer Xaq?
Was i meant to do the crazy a,b,c, d thingys fairy did?

By Jason817 (Jason817) on Sunday, August 10, 2003 - 04:36 pm: Edit

•••• and im taking that! i'm going to die

He's making it seem WAY too hard. You would never actually do all that stuff. It's much easier than it looks. Plus, with a ti-89, it's only a matter of typing them into your calculator

By Callmecollege (Callmecollege) on Sunday, August 10, 2003 - 04:53 pm: Edit

emma, u sorta did it wrong
you can't break it up into what you did x^-4 + 1, only if the question wuz (x^4+1)/1 would u be able to split it up into 2 fractions
and these sorta problems are NOT on calculus ab, and if they were on calculus bc, they'd be alot easier..so no worries out there calculus haters

By Emma (Emma) on Sunday, August 10, 2003 - 05:01 pm: Edit

o yea, been too long since school, my brain has melted, hottest day in London since the records began today.

By Serene (Serene) on Sunday, August 10, 2003 - 05:03 pm: Edit

It's not even on BC *_*

By Quack (Quack) on Sunday, August 10, 2003 - 05:04 pm: Edit

Thats great... except I have a TI-30X IIS! Is it completely necessary to have a graphing calculator for calculus (AB). They're pretty dang expensive. You'd think the price would go down by now.

By Serene (Serene) on Sunday, August 10, 2003 - 05:16 pm: Edit

I heard someone talking about finding calculators on ebay...

By Jason817 (Jason817) on Sunday, August 10, 2003 - 07:42 pm: Edit

A ti89 helps significantly but isnt necessary. I wouldnt have gotten as high on the AP test without it, thats for sure.

By Xag (Xag) on Monday, August 11, 2003 - 05:04 am: Edit

by the way it's not trig substitution serene.i know the solution.emma you confused. problem is correct. my request to every one is try to modify question withot any substitions at early stages and try.

By Serene (Serene) on Monday, August 11, 2003 - 11:29 am: Edit

I thought fairy had a good answer?

By Heatwave345 (Heatwave345) on Monday, August 11, 2003 - 02:05 pm: Edit

don't you just use trig substitution?

make a right triangle with sides 1, x^4+1, and sqrt(x^4+1). choose one of the azute angles and call it A

sinA= 1/sqrt(x^4+1)

so 1/(x^4+1)=(sinA)^2

and the integral of (sinA)^2 is A/2-sinAcosA/2

then find A in terms of x and plug it in. A=arctan(1/x^4)

By Fairyofwind (Fairyofwind) on Monday, August 11, 2003 - 02:07 pm: Edit

heatwave, what about dx instead of your dA... hehe... dx != dA.

By Heatwave345 (Heatwave345) on Monday, August 11, 2003 - 02:57 pm: Edit

oh man, summer has made me forget everything. disregard my last post.


thx for cathing that

By Jason817 (Jason817) on Monday, August 11, 2003 - 03:04 pm: Edit

Well apparently I forgot everything too because I don't understand what you did! Damn I'm screwed for BC.

By Emma (Emma) on Monday, August 11, 2003 - 05:05 pm: Edit

I don't know what i'm going to do next year, i seem to have forgotten everything!

By Xag (Xag) on Wednesday, August 13, 2003 - 08:14 am: Edit

i have a good answer. solution is
we can write given problem as
1/(x^2+1/x^2)*1/x^2
then write it as
1/[(x+1/x)^2-2]*1/x^2
take 1/x=t
then problem changes as
1/[(t+1/t)^2-(2^(1/2))^2]
which is in form of 1/(y^2-a^2)
if we integrate 1/(y^2-a^2) we get
(1/2*a)*log[(a+y)/(a-y)]
so easy(if you are aware of the formulae)

By Fairyofwind (Fairyofwind) on Wednesday, August 13, 2003 - 12:32 pm: Edit

Actually it is -1/[(t+1/t)^2-(2^(1/2))^2], but the thing is you can't just use that formula, because y=(t+(1/t)). dy? Now if y=t+3, that would be okay, but dy != dt, and if you do that, the new expression is no easier to integrate than the first.

By Zerg_Vvins (Zerg_Vvins) on Wednesday, August 13, 2003 - 12:49 pm: Edit

Fairyofwind: My question to you is "WHAT ARE YOU?"

J/K

Where do you go to school? Do you even need to go to school?

By Xag (Xag) on Thursday, August 14, 2003 - 04:36 am: Edit

yes, fariy what you said was correct.
i tried another solution out.
modify it as:
1/2[(x^2+1)-(x^2_1)]/(x^4+1)
you can divide it as partial fraction
1/2[(x^2+1)/(x^4+1)]-1/2[(x^2-1)/(x^4+1)]
then divide numerator and denominator by x^2
we get
1/2[(1+(1/x)^2)/(x^2+(1/x)^2]
-1/2[(1-(1/x)^2)/(x^2+(1/x)^2]
keep x-1/x=t in the first partial fraction and x+1/x=v
then we get
1/2[1/(t^2+{(2^(1/2)}^2)]-1/2[1/(v^2-{(2^(1/2)}^2]
dx changes to dt in first partial fraction and dx changes to dv in second partial fraction
after integrating them we get
1/[2*(2^(1/2))]*arc tan[t/(2^(1/2))]
-1/[4*(2^(1/2))]*log [(v-2^(1/2))/v+2^(1/2)]
then we substitute values of t and v to get final answer. to check whether it is correct or not differentiate the answer.
i hope this will be correct now.

By Xag (Xag) on Thursday, August 14, 2003 - 04:43 am: Edit

yes, fariy what you said was correct.
i tried another solution out.
modify it as:
1/2[(x^2+1)-(x^2_1)]/(x^4+1)
you can divide it as partial fraction
1/2[(x^2+1)/(x^4+1)]-1/2[(x^2-1)/(x^4+1)]
then divide numerator and denominator by x^2
we get
1/2[(1+(1/x)^2)/(x^2+(1/x)^2]
-1/2[(1-(1/x)^2)/(x^2+(1/x)^2]
keep x-1/x=t in the first partial fraction and x+1/x=v
then we get
1/2[1/(t^2+{(2^(1/2)}^2)]-1/2[1/(v^2-{(2^(1/2)}^2]
dx changes to dt in first partial fraction and dx changes to dv in second partial fraction
after integrating them we get
1/[2*(2^(1/2))]*arc tan[t/(2^(1/2))]
-1/[4*(2^(1/2))]*log [(v-2^(1/2))/v+2^(1/2)]
then we substitute values of t and v to get final answer. to check whether it is correct or not differentiate the answer.
i hope this will be correct now.

By Y17k (Y17k) on Thursday, August 14, 2003 - 06:39 am: Edit

sry for profaning but...

WAT THE ****


i dont even frigging compared with u guys T.T

By Fairyofwind (Fairyofwind) on Thursday, August 14, 2003 - 09:42 am: Edit

Xaq, this is correct. VERY impressive and elegant solution! This inspires me to provide a simplification of the final result of my original solution. 2(ArcTan[1+x*Sqrt[2]]-ArcTan[1-x*Sqrt[2]])=2(ArcTan[Tan[ArcTan[1+x*Sqrt[2]]-ArcTan[1-x*Sqrt[2]]]])=2(ArcTan[Tan[a-b]])=2(ArcTan[(Tan[a]-Tan[b])/(1+Tan[a]Tan[b])])=2(ArcTan[(x*Sqrt[2])/(1-x^2)]).