How to Solve the Integral of (x^2+1)/(x^4+1)?

[totentanz]

Can anyone help me solving x²+1/x⁴+1...thanks

 

[micromass]

Just apply linearity and you'll get:

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

This are elementary integrals, so it is easily solvable.

 

[totentanz]

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

 

[micromass]

What don't you understand about it? You just apply the formula

$$\int{(f(x)+g(x))dx}=\int{f(x)dx}+\int{g(x)dx}$$

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

$$\int{x^ndx}=\frac{x^{n+1}}{n+1}+C$$ for $$n\neq -1$$.

 

[Caramon]

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

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

 

[micromass]

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: Parantheses in mathematics are EXTREMELY important! Don't add too few of them.

 

[Caramon]

So, which question is it? (Before I spend a year doing partial fractions, completing the square, substitution, and god knows what else)

 

[totentanz]

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

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

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

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

 

[ThomasEgense]

This is the solution, however getting there is not clear :)

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

 

[totentanz]

This is the solution, however getting there is not clear :)

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

 

[lurflurf]

-decompose into partial fractions (with completed squares)
-obvious
inverse tangent

 

[HallsofIvy]

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

That is, the denominator $x^4+ 1$ factors as
$$(x^2+ \sqrt{2}x+ 1)(x^2- \sqrt{2}x+ 1)$$
and you can use partial fractions with those.

 

[totentanz]

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

That is, the denominator $x^4+ 1$ factors as
$$(x^2+ \sqrt{2}x+ 1)(x^2- \sqrt{2}x+ 1)$$
and you can use partial fractions with those.

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

 

[totentanz]

sorry my friend but it is not working

 

[dextercioby]

Back in my high school days I knew tricks for these. For an anti-derivative sought in a subset of R excluding 0, we can make the following tricks:

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

 

[totentanz]

Back in my high school days I knew tricks for these. For an anti-derivative sought in a subset of R excluding 0, we can make the following tricks:

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

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

 

[HallsofIvy]

You have been given a number of different suggestions (dextercioby's making the problem almost trivial) but keep saying you cannot do it- and showing no work at all. Have you tried what dextercioby suggested? What did you do and where did you run into trouble.

 

[totentanz]

You have been given a number of different suggestions (dextercioby's making the problem almost trivial) but keep saying you cannot do it- and showing no work at all. Have you tried what 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

 

[totentanz]

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