Determine whether the improper integral $\int_{0}^{\infty}\frac{x^2}{1+x^4}\,dx$ exists

While doing an exercise I need to prove that $\frac{x^2}{1+x^4}$ is integrable.

So I have to see if $\int_{0}^{\infty} |\frac{x^2}{1+x^4}| dx < \infty$. I tried to divide it in two integrals but I don't know how to continue...

$\int_{0}^{\infty} |\frac{x^2}{1+x^4}| dx = \int_{0}^{\infty} \frac{x^2}{1+x^4} dx =\int_{0}^{1} \frac{x^2}{1+x^4} dx + \int_{1}^{\infty} \frac{x^2}{1+x^4} dx \leq \frac{1}{2}\int_{0}^{1} \frac{2x}{1+(x^2)^2} dx + \int_{1}^{\infty} \frac{x^2}{1+x^4} dx$

So the first integral is the arcotangent and it is finite but the second one?

I don't know if that's the best way to do this...

Could anyone help me please?

In $[1,\infty)$,$$\frac{x^2}{1+x^4}<\frac{x^2}{x^4}=\frac1{x^2},$$and$$\int_1^\infty\frac1{x^2}\,\mathrm dx=1.$$