Let $$f(x) = \frac{x+2}{\sqrt{x}\left(x^2 + x + 1\right)^4}$$ Discuss the convergence of $\displaystyle\int_0^1f(x)\,\mathrm dx$ and $\displaystyle\int_1^{+\infty}f(x)\,\mathrm dx$.
I encountered this problem in a test about improper integrals, so I expect that there is a solution which does not require evaluating the integral.
My attempt
Substitute $\sqrt{x} = t$ so that the bounds remain the same and the integral becomes:
$$\int\frac{2t^2 + 4}{\left(t^4 + t^2 + 1\right)^4}\mathrm dt$$
Since $\displaystyle\lim_{t \to 0} f(t) = 4$ and $f(t)$ is decreasing, the function is bounded on $(0, 1]$ and the integral converges.
What about the other one? I was not able to conclude anything useful, other than that it goes to $0$ towards $+\infty$.
$$f(x)\sim_0\frac{2}{\sqrt x}$$ and the integral $$\int_0^1\frac{dx}{\sqrt x}$$ is convergent hence the integral $$\int_0^1f(x)dx$$ is also convergent.
$$f(x)\sim_\infty\frac1{x^7\sqrt x}$$ and since $$\int_1^\infty\frac{dx}{x^7\sqrt x}$$ is convergent then $$\int_1^\infty f(x)dx$$ is also convergent.