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Given the integral$\int_{0}^{+\infty} \frac{\lbrace \cos(x)-1 \rbrace x^2}{x^p + (x+1)^6}$, I need to find the values of $p \in \mathbb{R}$ such that it converges.

I started by trying to bound it, so then I could apply the comparison principle. But I am stuck in this step:

$\left|\frac{\lbrace \cos(x)-1 \rbrace x^2}{x^p + (x+1)^6}\right| \leq \frac{2(x+1)^2}{\left|x^p + (x+1)^6\right|}$

Am I on the right track? If so, how could I carry on with that?

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    Hint: It converges for all $p$ as $x\to\infty$. The only problem is checking convergence as $x\to0^+$. – 2017-02-26
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    How did you do to notice that? Also, why $x \rightarrow 0^+$? It is not discontinuous at $x=0$. – 2017-02-26

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Here's a better comparison for $x>1$.

$$\left|\frac{(\cos(x)-1)x^2}{x^p+(x+1)^6}\right|\le\frac{x^2}{x^6}=\frac1{x^4}$$

Thus, it converges for all $p$ as $x\to\infty$.

For $x<1$, apply this comparison:

$$\left|\frac{(\cos(x)-1)x^2}{x^p+(x+1)^6}\right|\sim_0\frac{x^4+\mathcal O(x^6)}{x^p}=\frac1{x^{p-4}}+\mathcal O\left(\frac1{x^{p-6}}\right)$$

where we used $\cos(x)=1-\frac12x^2+R$ where $|R|\le|\frac1{24}\cos(1)x^4|$

Thus, it converges for $p<4$ as $x\to0$, so your integral converges for $p<4$ and diverges elsewhere.