How to tell if this integral converges? $$\int_{0}^{\infty} \frac{\sqrt{1-x+x^2}}{1-x^2+x^4} dx$$
What's the method for determining if this integral converges or diverges? The integral seems to converge if I put it into Wolfram Alpha.
But do we assume it's similar to $$\int_{0}^{\infty} \frac{1}{x^4} dx$$
Because if so I can't get that one to converge since it's $-\frac{1}{3x^3}$ where $x=\infty$ and $x=0$ (I don't know how to write the "right bar" notation for integrals) which goes to $-0 + \infty$ which is divergent. So I am not sure what is right.

To prove convergence of $\int_0^{\infty}f(x)dx$  for a continuous $f(x)$ it often helps to find a crude but effective upper bound for $|f(x)|$ for large $x$. 
When $x>1$ we have $\sqrt {1-x+x^2}\; <\sqrt {1+2x+x^2}=x+1<2x .$ 
And when $x>2$ we have $x^4-x^2+1>x^4-x^2>x^4/2.$ 
So when $x>2$ we have $$0<\frac {\sqrt {1-x+x^2}}{1-x^2+x^4}<\frac {2x}{x^4/2}=\frac {4}{x^3}$$ so the integral converges.
We can also observe that when $x>0$ we have $\sqrt {1-x+x^2}=x(1+g(x))$ and $1-x^2+x^4=x^4(1+h(x))$ where $g(x)$ and $h(x)$ converge to $0$ as $x\to \infty.$ 
So for all sufficiently large $x$ we have $|1+g(x)|<3/2$ and $|1+h(x)|>1/2.$ Thus $$0<\left|\frac {x(1+g(x)}{x^4(1+h(x)}\right|<\frac {x(3/2)}{x^4(1/2)}=\frac {3}{x^3}$$ for all sufficiently large $x.$
