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Can anyone help me solving x

^{2}+1/x^{4}+1.........thanksYou are using an out of date browser. It may not display this or other websites correctly.

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- Thread starter totentanz
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- #1

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Can anyone help me solving x^{2}+1/x^{4}+1.........thanks

- #2

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[tex]\int{x^2dx}+\int{\frac{1}{x^4}dx}+\int{dx}[/tex].

This are elementary integrals, so it is easily solvable.

- #3

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thank you...but can you explain it more?

- #4

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[tex]\int{(f(x)+g(x))dx}=\int{f(x)dx}+\int{g(x)dx}[/tex]

Then you have brought the integral to three basic integrals, which you can easily solve with the formula

[tex]\int{x^ndx}=\frac{x^{n+1}}{n+1}+C[/tex] for [tex]n\neq -1[/tex].

- #5

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Did you use partial fractions to get that result? I think he may be referring to:

[tex]

\int{\frac{{x^2}+1}{{x^4}+1}dx}

[/tex]

[tex]

\int{\frac{{x^2}+1}{{x^4}+1}dx}

[/tex]

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- #6

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Did you use partial fractions to get that result? I think he may be referring to the integral of ((x^2 + 1)/(x^4 + 1))dx

Ah yes, that may make more sense...

To the OP:

- #7

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- #8

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Did you use partial fractions to get that result? I think he may be referring to:

[tex]

\int{\frac{{x^2}+1}{{x^4}+1}dx}

[/tex]

Yes this is what I mean...but when you decompose the fraction I get

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

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- #9

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http://integrals.wolfram.com/index.jsp?expr=(x*x+1)/(x*x*x*x+1)&

- #10

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http://integrals.wolfram.com/index.jsp?expr=(x*x+1)/(x*x*x*x+1)&

Thank you...but it has a method or just a rule

- #11

lurflurf

Homework Helper

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-decompose into partial fractions (with completed squares)

-obvious

inverse tangent

-obvious

inverse tangent

- #12

HallsofIvy

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[itex]x^4+ 1= 0[/itex] has no real roots but we can write [itex]x^4= -1[/itex] so that [itex]x^2= \pm i[/itex] and then get [itex]x= \frac{\sqrt{2}}{2}(1+ i)[/itex], [itex]x= \frac{\sqrt{2}}{2}(1- i)[/itex], [itex]x= \frac{\sqrt{2}}{2}(-1+ i)[/itex], and [itex]x= \frac{\sqrt{2}}{2}(-1- i)[/itex] as the four roots. We can pair those by cojugates:

[tex]\left(x- \frac{\sqrt{2}}{2}(1+ i)\right)\left(x- \frac{\sqrt{2}}{2}(1- i)\right)[/tex][tex]= \left((x- \frac{\sqrt{2}}{2})- i\frac{\sqrt{2}}{2}\right)\left((x- \frac{\sqrt{2}}{2})+ i\frac{\sqrt{2}}{2}\right)[/tex][tex]= (x- \sqrt{2}{2})^2+ \frac{1}{2}= x^2- \sqrt{2}x+ 1[/tex]

and

[tex]\left(x- \frac{\sqrt{2}}{2}(-1+ i)\right)\left(x- \frac{\sqrt{2}}{2}(-1- i)\right)[/tex][tex]= x^2+ \sqrt{2}x+ 1[/tex]

That is, the denominator [itex]x^4+ 1[/itex] factors as

[tex](x^2+ \sqrt{2}x+ 1)(x^2- \sqrt{2}x+ 1)[/tex]

and you can use partial fractions with those.

[tex]\left(x- \frac{\sqrt{2}}{2}(1+ i)\right)\left(x- \frac{\sqrt{2}}{2}(1- i)\right)[/tex][tex]= \left((x- \frac{\sqrt{2}}{2})- i\frac{\sqrt{2}}{2}\right)\left((x- \frac{\sqrt{2}}{2})+ i\frac{\sqrt{2}}{2}\right)[/tex][tex]= (x- \sqrt{2}{2})^2+ \frac{1}{2}= x^2- \sqrt{2}x+ 1[/tex]

and

[tex]\left(x- \frac{\sqrt{2}}{2}(-1+ i)\right)\left(x- \frac{\sqrt{2}}{2}(-1- i)\right)[/tex][tex]= x^2+ \sqrt{2}x+ 1[/tex]

That is, the denominator [itex]x^4+ 1[/itex] factors as

[tex](x^2+ \sqrt{2}x+ 1)(x^2- \sqrt{2}x+ 1)[/tex]

and you can use partial fractions with those.

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- #13

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[itex]x^4+ 1= 0[/itex] has no real roots but we can write [itex]x^4= -1[/itex] so that [itex]x^2= \pm i[/itex] and then get [itex]x= \sqrt{2}{2}(1+ i)[/itex], [itex]x= \sqrt{2}{2}(1- i)[/itex], [itex]x= \sqrt{2}{2}(-1+ i)[/itex], and [itex]x= \sqrt{2}{2}(-1- i)[/itex] as the four roots. We can pair those by cojugates:

[tex]\left(x- \frac{\sqrt{2}}{2}(1+ i)\right)\left(x- \frac{\sqrt{2}}{2}(1- i)\right)[/tex][tex]= \left((x- \frac{\sqrt{2}}{2})- i\frac{\sqrt{2}}{2}\right)\left((x- \frac{\sqrt{2}}{2})+ i\frac{\sqrt{2}}{2}\right)[/tex][tex]= (x- \sqrt{2}{2})^2+ \frac{1}{2}= x^2- \sqrt{2}x+ 1[/tex]

and

[tex]\left(x- \frac{\sqrt{2}}{2}(-1+ i)\right)\left(x- \frac{\sqrt{2}}{2}(-1- i)\right)[/tex][tex]= x^2+ \sqrt{2}x+ 1[/tex]

That is, the denominator [itex]x^4+ 1[/itex] factors as

[tex](x^2+ \sqrt{2}x+ 1)(x^2- \sqrt{2}x+ 1)[/tex]

and you can use partial fractions with those.

Thanks bro...this is what I was looking for...I hope it works

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- #14

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sorry my friend but it is not working

- #15

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[tex] \int \frac{x^2 +1}{x^4 +1} \, dx = \int \frac{1+\frac{1}{x^2}}{x^2 + \frac{1}{x^2}} \, dx = \int \frac{d\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^2 + \left(\sqrt{2}\right)^2} = ... [/tex]

- #16

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[tex] \int \frac{x^2 +1}{x^4 +1} \, dx = \int \frac{1+\frac{1}{x^2}}{x^2 + \frac{1}{x^2}} \, dx = \int \frac{d\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^2 + \left(\sqrt{2}\right)^2} = ... [/tex]

Thanks my friend,this is great...because I am stuck in this exo,and I can not go on

- #17

HallsofIvy

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- #18

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no workat all. Have youtriedwhat dextercioby suggested? What did you do and where did you run into trouble.

I am working on solving it all week,and I can not find a complete answer except what Mr.dextercioby show...and I already solve this "normalization of the wavefunction",thank you for your care

- #19

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I am working on solving it all week,and I can not find a complete answer except what Mr.dextercioby show...and I already solve this "normalization of the wavefunction",thank you for your care

Ok...I solve it

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