Calculate indefinite integral using Fourier transform

[Incand]

Homework Statement

Use the Fourier transform to compute
\int_{-\infty}^\infty \frac{(x^2+2)^2}{(x^4+4)^2}dx

Homework Equations

The Plancherel Theorem
$||f||^2=\frac{1}{2\pi}||\hat f ||^2$
for all $f \in L^2$.

We also have a table with the Fourier transform of some function, the ones of possible use may be
$(x^2+a^2)^{-1} \to (\pi/a)a^{-\xi^2/(2a)}$
and
$f(x)g(x) \to \frac{1}{2\pi} (\hat f * \hat g)(\xi )$

The Attempt at a Solution

Not really sure where to start here, every other similar exercise use Plancherel's theorem so I assume I should use that one here. The problem is then that I would need to computer the Fourier transform of
$\frac{x^2+2}{x^4+4}$ which doesn't seem to easy at all. I tried writing this as
$\frac{x^2+2}{(x^2+2i)(x^2-i2)}$ and somehow do partial fractions but doesn't seem to get me anywhere. Another similar approach that perhaps could work suggested by our professor was
$\int_{-\infty}^\infty \frac{(x^2+2)^2}{(x^4+4)^2} = \int_{-\infty}^\infty \left| \frac{(x^2+2)}{(x^2+2i)^2} \right|^2$ but this doesn't seem to work either.
Any ideas on how to procede?

Edit: I should mention that it's possible to find the Fourier transform in some tables, although not the one in our book. If I use that one however I would need to proove it first.

 

[ehild]

You can do the integral without Fourier transform. Factorize the denominator: it is the product of (x-x1)(x-x2)(x-x3)(x-x4) where x_k-s are the complex roots of x^4 = -4, x_k=√2 e ^{i(π/4+kπ/2)}. The products (x-x0)(x-x3) and (x-x12)(x-x2) are real so you write x^4+1 with two real factors. You can proceed with partial fractions then. It is lengthy, but solvable :)

 

[Incand]

You can do the integral without Fourier transform. Factorize the denominator: it is the product of (x-x1)(x-x2)(x-x3)(x-x4) where x_k-s are the complex roots of x^4 = -4, x_k=√2 e ^{i(π/4+kπ/2)}. The products (x-x0)(x-x3) and (x-x12)(x-x2) are real so you write x^4+1 with two real factors. You can proceed with partial fractions then. It is lengthy, but solvable :)

Yes I know I could use solve it that way, the problem is the exercise state I should use the Fourier transform in a course on Fourier analysis so I should make use of that somehow. But thanks for trying to help!
It's grouped together with 4 other question originally where the other 4 are really easy to solve with plancherel's theorem. I thought perhaps I should use partial fractions halfway and then the Fourier transform but doesn't seem to give me anything of use. Perhaps there is some mistake in the exercise not being easier with the Fourier transform. When I asked our professor about the question he couldn't come up with the solution either at the time.

 

[ehild]

$\frac{x^2+2}{x^4+4}=\frac{0.5}{(x+1)^2+1}+\frac{0.5}{(x-1)^2+1}$ You can find the Fourier transform of that form?

 

[Incand]

$\frac{x^2+2}{x^4+4}=\frac{0.5}{(x+1)^2+1}+\frac{0.5}{(x-1)^2+1}$ You can find the Fourier transform of that form?

Thanks! This must be what I was supposed to be doing!

Posting the rest of the solution:
Using the formula above with the shift $-1$ and $1$ we get
F.T. $0.5\left( \frac{1}{(x+1)^2+1}+\frac{1}{(x-1)^2+1}\right) \to 0.5\left(e^{i\xi }e^{-|\xi|}+e^{-i\xi}e^{-|xi|}\right)= e^{-|\xi|}\cos \xi$
Using Plancherel's theorem (and realising that both part of the functions is symmetric around zero)
$\frac{1}{2\pi} \int_{-\infty}^\infty e^{-2|\xi|}\cos^2 \xi d\xi = \frac{1}{4\pi} \int_0^\infty e^{-2\xi}\left(e^{-2i\xi}+e^{2i\xi} +2\right)d\xi =\\ \frac{1}{4\pi} \int_0^\infty \left(e^{(-2-2i)\xi}+e^{(-2+2i)\xi} +2e^{-2\xi}\right) = \frac{1}{4\pi} \left( \frac{1}{2(1+i)}+\frac{1}{2(1-i)}+1\right) = \frac{3}{8\pi}$.

 

[ehild]

The same result without Fourier transform :) http://www.wolframalpha.com/input/?i=int_{-infinity}^{înfinity}((x^2+2)^2/(x^4+4)^2dx)