Studying convergence

[ddiogo]

Hi, I'm trying to study the absolute convergence of this integral,

[math]\int^\infty_0 \frac{(x+1) \cos(2x)}{x^2-x+5}[/math]
but I'm getting a lot of dificulty, it seams that I cannot find a good function to compare... Please help me!

 

[AvgStudent]

How about this?
[math]-1\le \cos(2x)\le1\\ \frac{-(x+1)}{x^2-x+5}\le \frac{(x+1)\cos(2x)}{x^2-x+5} \le\frac{(x+1)}{x^2-x+5}[/math]

 

[blamocur]

How about this?
[math]-1\le \cos(2x)\le1\\ \frac{-(x+1)}{x^2-x+5}\le \frac{(x+1)\cos(2x)}{x^2-x+5} \le\frac{(x+1)}{x^2-x+5}[/math]

But both the left and the right bounds diverge, while the integral itself most likely converges, albeit not absolutely.
I would try bounding [imath]\int_{n\pi}^{(n+0.5)\pi} \frac{(x+1)\cos 2x}{x^2-x+5}[/imath] and [imath]\int_{(n+0.5)\pi}^{(n+1)\pi} \frac{(x+1)\cos 2x}{x^2-x+5}[/imath] by exploiting anti-symmetry of [imath]\cos 2x[/imath] relative to [imath](n+0.25)\pi[/imath] and [imath](n+0.75)\pi[/imath] respectively.