Find convergence of integral and compute it Preparing for my calculus exam I found this exercise and I want to see if I'm on the right track
Find out the convergence of the following series and compute it.
$$\int_0^\infty \frac {\,dx} {(x+1)\sqrt {x^2 + x + 1}}$$
To find if the integral is convergent or divergent I'm thinking to compute this limit:
$$\lim_{t\rightarrow\infty} \int_0^t \frac {\,dx} {(x+1)\sqrt {x^2 + x + 1}}$$
And if that limit is a constant then my integral is convergent if not then the integral is divergent and I should not compute it.
Now, to compute the integral so I can compute that limit I'm thinking to apply integration by parts. 
That $\frac {1}{x+1}$ from the integral is $\frac {d}{\,dx}\ln(x+1)$.
Any tips and/or corrections?
 A: Let's try an elementary approach and make the substitution:
$$\sqrt{x^2+x+1}=x+t$$
and get 
$$x=\frac{t^2-1}{1-2t}$$
$$dx=-2 \cdot \frac{t^2-t+1}{(1-2t)^2}\ dt$$
Therefore, our integral gets reduced to 
$$2\int_{1}^{\frac{1}{2}}\frac{1}{t(t-2)} \ dt=[\log(2-t)-\log(t)]_{1}^{\frac{1}{2}}= \ln 3.$$
Note that for the integration limits, particularly for the case when $x$ tends to $\infty$, I used a celebre limit whose limit is well known, namely:
$$\lim_{x\to\infty} \sqrt[k]{x^k+x^{k-1}+\cdots + 1}-x = \frac{1}{k}, \space k\geq 2$$
that is easily deduced by using Taylor expansion.
In our case we dealt with 
$$\lim_{x\to\infty} \sqrt{x^2+x+1}-x = \frac{1}{2}$$
If you are not used to Taylor expansion, use the substitution $\displaystyle x=\frac{1}{u}$ and pay attention that u tends to $0$.  Then you may nicely finish it by using L'Hôpital's rule.
Finally, if you don't like either way, just use the conjugates. 
Q.E.D. (Chris)
A: 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.
$$\begin{align*}
\int_{\frac{\pi}{3}}^{\frac{2\pi}{3}} \frac{du}{\sin u}
&= \int_{\frac{\pi}{3}}^{\frac{2\pi}{3}} \frac{\sin u}{\sin^2 u} \; du
= \int_{\frac{\pi}{3}}^{\frac{2\pi}{3}} \frac{\sin u}{1 - \cos^2 u} \; du \\
&= \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{ds}{1 - s^2} \qquad (s = \cos u) \\
&= \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{1}{2}\left( \frac{1}{1-s} + \frac{1}{1+s} \right) \; ds \\
&= \frac{1}{2}\left[ \log(1+s) - \log(1-s) \right]_{-\frac{1}{2}}^{\frac{1}{2}}
= \log 3.
\end{align*}$$
