Integral of a rational function over the positive real axis

[muppet]

Hi everyone, this isn't really a homework question- this integral has come up during project work- but this seems like a sensible place to ask it nontheless.

Homework Statement

Compute
$$I(t)=\int^{\infty}_{0} dx \frac{x^5}{(t+x^2)(1+(t+x^2)^4)}$$
where t is real and positive.

Homework Equations

If $$f(z)=h(z)/g(z)$$, and $$g$$ has a simple pole at $$z=z_0$$, then

$$Res(f,z_0) = \frac{h(z_0)}{g'(z_0)}$$

$$\oint_ Cf(z)dz=2\pi i \sum_{j}Res(f,z_j)+\pi i \sum_{k} Res(f,b_k)$$
where C is a contour enclosing poles at $$z=z_j$$ that "infinitesmally semi-circles around" simple poles at $$z=b_k$$

The Attempt at a Solution

As my integral only runs over the positive real axis, I can't integrate around the "usual" semi-circle of infinite radius in the upper half-plane. However, as the integrand is odd in x (or, changing language to the conventions of complex analysis, in z), I think I can take the real part of an integral around a quarter-circle of infinite radius along the positive real axis and down the negative imaginary axis. This gives me a simple pole on the boundary at $$z=i\sqrt{t}$$ and a simple pole at $$z=\sqrt{e^{i \pi/4}-t}$$. Plugging these in and taking the real part gives me
$$I(t)=Re \left[ \frac{\pi i}{4}e^{-i \pi /2}(e^{i \pi /4}-t)^{2} \right] =\frac{\pi}{4}(t^2-t\sqrt{2})$$

This is unfortunately wrong- mathematica gives me a decreasing function of t, but in a horribly complicated form, whilst I happen to know that a much simpler form exists. Any help would be greatly appreciated.

 

[jbunniii]

The integrand is not odd in $x$, because of the $(t + x)$ term.

 

[muppet]

Sorry, TeX error! Thanks for that.