Determine all p $\in \mathbb{R}$ such that the following integral converges The integral is: 
$$\int_0^{+\infty} \frac{(\cos(x) - 1)x^2}{x^p + (x + 1)^6} dx$$
Now, what I did was this: 
$$\frac{(\cos(x) - 1)x^2}{x^p + (x + 1)^6} \leq \frac{\cos(x)x^2}{x^p + (x + 1)^6} \leq \frac{\cos(x)x^2}{x^p + x^6} \leq \frac{x^2}{x^p + x^6} \leq \frac{x^2}{x^p}$$
And $\int_1^{+\infty} \frac{x^2}{x^p} dx$ converges when $p > 3$ because it is a $p$-integral. Is this approach right? 
Edit: I only have to know what to do with $\int_0^1$

Since
$$ \left|\int_0^{\infty}f(x)\ dx\right| \le \int_0^{\infty} |f(x)|\ dx $$
and $|\cos x-1| = 1-\cos x$, we consider
$$ \int_0^\infty \frac{(1-\cos x)x^2}{x^p + (x+1)^6}\ dx \le \int_0^\infty \frac{2x^2}{x^p + (1+x)^6}\ dx \le \int_0^1 2x^2 + \int_1^\infty \frac{2x^2}{x^p + x^6} $$
The first term clearly converges, so let's look at the other one
If $p > 6$
$$ \int_1^\infty \frac{2x^2}{x^p+x^6} \le \int_1^\infty \frac{x^2}{x^p}\ dx $$
which always converges
If $p \le 6$
$$ \int_1^\infty \frac{2x^2}{x^p+x^6} \le \int_1^\infty \frac{x^2}{x^6}\ dx $$
which always converges
So the integral converges for $\forall p$
