# 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.