Integral question

1. Jan 9, 2008

transgalactic

((x^2)+1)/((x^4)+1)

i tride to solve it in every way possible
1.by splitting it into two polinomials
2.by subtsituting x^4=t
3.by puttinh x=tan t

nothing works

how do i solve it

2. Jan 9, 2008

Mystic998

As far as I can tell, after some cheating, you're supposed to somehow figure out that $$x^4 + 1 = (x^2 + \sqrt{2}x + 1)(x^2 - \sqrt{2}x + 1)$$. I'm not sure exactly how one would easily recognize that though (assuming I didn't make a mistake).

3. Jan 9, 2008

Defennder

Well let's first put it in LaTeX for the benefit of the homework helpers:

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

How exactly did you cheat Mystic?

4. Jan 9, 2008

Mystic998

Wolfram's Integrator.

5. Jan 9, 2008

rocomath

Buy Gelfand's Algebra book and you'll be able to factor almost any polynomial :p

Last edited: Jan 9, 2008
6. Jan 9, 2008

Mystic998

It's somewhat depressing getting recommendations for high school algebra books while sitting in my crappy three-person TA office.

Anyway, out of curiosity, how would one proceed in getting that factorization of x^4 + 1. I've honestly got no clue, and that would probably be handy information.

7. Jan 9, 2008

rocomath

Gelfand may be intended for HS students, but his books are hard. I just finished Calc 2 and I still struggle on a lot of his problems.

$$x^4+2x^2+1-2x^2$$

$$(x^2+1)^2-(\sqrt{2}x)^2$$

$$(x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1)$$

8. Jan 9, 2008

Mystic998

Wow, I feel silly now. Thanks.

9. Jan 9, 2008

rocomath

Exactly how I felt before learning a few things in his book :-x

10. Jan 9, 2008

rocomath

After splitting it into 2, and using Mystic's advice ...

$$\int\frac{x^2}{(x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1)}dx+\int\frac{1}{(x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1)}dx$$

Partial Fraction?

Last edited: Jan 9, 2008
11. Jan 9, 2008

transgalactic

so how do i solve it after this step

12. Jan 9, 2008

rocomath

You have to use Partial Fractions, have you learned that method yet?

13. Jan 9, 2008

transgalactic

yes i did but it looks very scary
will it work?

14. Jan 9, 2008

rocomath

Yes and it's pretty much the only method I can think of. After that, you may want to mess around with completing the square, I did that in order to use arctan.

15. Jan 12, 2008

epenguin

Have you done it? Note I think it saves some work if you do not split it as rocophysics suggests but just notice that $$\frac{A}{(x^2+\sqrt{2}x+1)}+\frac{B}{(x^2-\sqrt{2}x+1)}$$
is capable of giving the form $$\frac{(x^2 + 1)}{(x^4 + 1)}$$ and that part of it is quite simple to get.

Edit: I mean easy to see or find what A and B must be.

Last edited: Jan 12, 2008
16. Jan 12, 2008

dextercioby

Since $$x^{4}+1 =(x^2 +i) (x^2 -i)$$, partial fraction decomposition is simply

$$\frac{x^{2}+1}{x^{4}+1} =\frac{1+i}{2}\frac{1}{x^2 +i} +\frac{1-i}{2}\frac{1}{x^2 -i }$$ and then the integration would be easy.

17. Jan 12, 2008

coomast

The remark of epenguin is correct, you can llok for A and B to make the integrals more easy to solve. However if you notice the following:

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

And also that:

$$x^4+1 = \left[\left(x+\frac{\sqrt{2}}{2}\right)^2+\left(\frac{\sqrt{2}}{2}\right)^2\right] \cdot \left[\left(x-\frac{\sqrt{2}}{2}\right)^2+\left(\frac{\sqrt{2}}{2}\right)^2\right]$$

You get two integrals involving arctan.