# How to do this integration?

1. Dec 13, 2008

### KFC

1. The problem statement, all variables and given/known data
There is a integration need to be done

$$I = \int_{-\infty}^\infty \frac{1+x^2}{1+x^4}dx$$

2. The attempt at a solution
I use the following substitution

$$x=\tan \theta$$

such that

$$dx = \frac{d\theta}{\cos^2\theta}$$

Now the integration becomes
$$I = \int_{-\pi/2}^{\pi/2} \frac{1}{1-0.5\sin^2(2\theta)}d\theta$$

But I still stuck with the simplified integration. Any other way to do that?

2. Dec 13, 2008

### tiny-tim

Hi KFC!

Hae you tried factorising 1 + x4 and using partial fractions?

3. Dec 13, 2008

### KFC

Thanks a lot for your reply. I try to do something like that

$$1+x^4 = (1+x^2)(Ax^2 + Bx + C)$$

and figure out A, B and C by comparing it to 1+x^4, but it will gives a complex terms.

4. Dec 13, 2008

### Dick

There are two quadratic real polynomials whose product is (x^4+1). To my mind the easiest what to find them is to factor (x^4+1)=(x-c1)(x-c2)(x-c3)(x-c4) over the complex numbers and then pick the pairs where ci and cj are complex conjugates and multiply them to get the quadratic factors. Can you do that? There might be tricks you can use to stay in the real numbers, but I don't know them offhand.

5. Dec 14, 2008

### tiny-tim

Nooo … 1+x2 is not a factor of 1+x4.

As Dick says, roots come in complex conjugate pairs,

so x4 + 1 = (x2 + Ax + B)(x2 + Cx + D), with A B C and D real.

6. Dec 14, 2008

### TD

As Dick mentioned, there's also a little trick possible if you notice that

x4+1 = (x2+1)2-2x2

Now you can just use the factorization of a difference of two squares.

7. Dec 14, 2008

### dextercioby

The factoring is simpler to get if you add and subtract $2x^2$ to the denominator and you those 6th grade formulas for expanding the square of a sum and the difference of two squares...

8. Dec 14, 2008

### KFC

Thanks all of you guys, it helps. I know how to do it now. :)

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