If the problem is to examine the convergence alone, just observe that for some constant $C > 0$ we have $ \frac{1}{(x+1)\sqrt{x^2+x+1}} \leq C \min\left\{ 1, \frac{1}{x^2}\right\}. \tag{1}$ Indeed, on $[0, 1]$, continuity of the integrand shows that it is bounded by some constant $C_1$. On $[1, \infty)$, we have $ \frac{1}{(x+1)\sqrt{x^2+x+1}} \leq \frac{1}{x \sqrt{x^2}} = \frac{1}{x^2}.$ Thus for $C = \max \{ C_1, 1 \}$ we have $(1)$. Therefore $ \begin{align*} \int_{0}^{\infty} \frac{dx}{(x+1)\sqrt{x^2+x+1}} & \leq C \int_{0}^{\infty} \min\left\{ 1, \frac{1}{x^2}\right\} \; dx \\ & = C \left( \int_{0}^{1} dx + \int_{1}^{\infty} \frac{1}{x^2} \; dx \right) = 2C < \infty. \end{align*} $
This method is quite general for convergence analysis. When applying this method, we first list all the singularities (the point where the integrand blows up) of the integrand, including points $\pm \infty$ at infinity. For example, if we are dealing with
$ \int_{0}^{\infty} \frac{dx}{x^{3/2}\sqrt{x + 1}} $
instead, then our list of singularity will be $\{0, \infty\}$. Then examine the behavior of the integrand near each singularity point $x_0$, by approximating it to familiar functions such as $(x - x_0)^{r}$. In many cases, this information solely determines the convergence behavior of the integral. In our example above, near $x = 0$ the function is approximately $x^{-3/2}$, whose integral near $x = 0$ diverges to infinity. This proves the divergence of the integral above. Rigorous justification of this estimation would be to find a suitable estimation for the integral. In this example, we may argue by
$ \frac{1}{x^{3/2}\sqrt{x + 1}} \geq \frac{1}{x^{3/2}\sqrt{2}} \quad \text{on} \quad (0, 1]. $
But in this case, we are asked to find its value. When we succeed in finding its value, then convergence also follows.
We make the substitution $x + \frac{1}{2} = \frac{\sqrt{3}}{2} \tan t$. As $x$ ranges from $0$ to $\infty$, $t$ ranges from $\frac{\pi}{6}$ to $\frac{\pi}{2}$. Also differentiating both sides, we have $ dx = \frac{\sqrt{3}}{2} \sec^2 t \; dt$. Thus
$ \begin{align*} \int \frac{dx}{(x+1)\sqrt{x^2 + x + 1}} &= \int \frac{1}{\left( \frac{1}{2}+\frac{\sqrt{3}}{2} \tan t \right) \left( \frac{\sqrt{3}}{2} \sec t \right)} \cdot \frac{\sqrt{3}}{2} \sec^2 t \; dt \\ &= \int \frac{dt}{\frac{1}{2}\cos t + \frac{\sqrt{3}}{2} \sin t} \\ &= \int \frac{dt}{\sin\left(t+\frac{\pi}{6}\right)}. \end{align*}$
This shows that
$ \int_{0}^{\infty} \frac{dx}{(x+1)\sqrt{x^2 + x + 1}} = \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{dt}{\sin\left(t+\frac{\pi}{6}\right)} = \int_{\frac{\pi}{3}}^{\frac{2\pi}{3}} \frac{du}{\sin u} $
for $u = t + \frac{\pi}{6}$. Already it is clear that this improper integral converges, for the integrand is bounded. To find its value, we proceed the calculation.
$\begin{align*} \int_{\frac{\pi}{3}}^{\frac{2\pi}{3}} \frac{du}{\sin u} &= \int_{\frac{\pi}{3}}^{\frac{2\pi}{3}} \frac{\sin u}{\sin^2 u} \; du = \int_{\frac{\pi}{3}}^{\frac{2\pi}{3}} \frac{\sin u}{1 - \cos^2 u} \; du \\ &= \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{ds}{1 - s^2} \qquad (s = \cos u) \\ &= \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{1}{2}\left( \frac{1}{1-s} + \frac{1}{1+s} \right) \; ds \\ &= \frac{1}{2}\left[ \log(1+s) - \log(1-s) \right]_{-\frac{1}{2}}^{\frac{1}{2}} = \log 3. \end{align*}$