Show that the improper integral is conditionally convergent 
Find out if the following improper integral converges absolutely, converges conditionally or diverges:
  $$\int_0^\infty \frac{\cos(x)}{\sqrt{x+1}+\sqrt{x}}\,dx.$$

I'm struggling to show that the integral converges conditionally (I already found that the original integral converges).
I tried to use comparison tests, but had no success. Here are a couple of my tries:  

1) $$\int_0^\infty \left|\frac{\cos(x)}{\sqrt{x+1}+\sqrt{x}}\right|\,dx=\int_0^\infty\frac{\left|\cos(x)\right|}{\sqrt{x+1}+\sqrt{x}}\,dx\le\int_0^\infty\frac{dx}{\sqrt{x+1}+\sqrt{x}}\,dx;$$
  this one didn't really helped me, since the last integral diverges.  
2) $$\int_0^\infty \left|\frac{\cos(x)}{\sqrt{x+1}+\sqrt{x}}\right|\,dx=\int_0^\infty\frac{\left|\cos(x)\right|}{\sqrt{x+1}+\sqrt{x}}\,dx\ge\int_0^\infty\frac{\cos^2(x)}{\sqrt{x+1}+\sqrt{x}}\,dx;$$
  this one seems that it might help, but I couldn't find a way to use it.

Any help would be appreciated,
Thanks
 A: The integral converges by Dirichlet test.
To tackle the absolute convergence, note that $|\cos(x)|\geq \cos(x)^2=\frac{1+\cos(2x)}2$ and use Dirichlet again to conclude that you have the sum of a convergent and a divergent integral. Hence the integral with absolute values diverges.
A: To show that the integral does not converge absolutely:
You are integrating over the interval
$$(0,\infty)=(0,\pi/4]\cup (\pi/4, 3\pi/4]\cup (3\pi/4, 5\pi/4]\cup (5\pi/4, 7\pi/4]\space\cup\space ...$$
Note that on the even-numbered intervals $(\pi/4,3\pi/4],(5\pi/4,7\pi/4],$ and so on, it holds that $|\cos(x)|>1/\sqrt{2}$. This means that
$$\begin{align}
\int_0^\infty \frac{|\cos(x)|}{\sqrt{x+1}+\sqrt{x}}dx &\ge \sum_{k=1}^\infty \int_{\pi k-\pi/4}^{\pi k+\pi/4} \frac{|\cos(x)|}{\sqrt{x+1}+\sqrt{x}}dx \\
&\ge \sum_{k=1}^\infty \int_{\pi k-\pi/4}^{\pi k+\pi/4} \frac{1/\sqrt{2}}{\sqrt{x+1}+\sqrt{x}}dx \\
&\ge \sum_{k=1}^\infty \int_{\pi k-\pi/4}^{\pi k+\pi/4} \frac{1/\sqrt{2}}{\sqrt{x}+\sqrt{x}}dx \\
&\ge \sum_{k=1}^\infty \int_{\pi k-\pi/4}^{\pi k+\pi/4} \frac{dx}{2\sqrt{2x}} \\
\end{align}$$
Then evaluate the integral in each term of the series and evaluate the convergence or divergence of the series. Can you take it from here?
A: Since $\int_0^M \frac{1}{\sqrt{x}} < \infty$, you can ignore some finite initial interval. Also, notice that the denominator is
$$
\sqrt{x} + \sqrt{x+1} = \sqrt{x} (2 + O(x^{-1})).
$$
Using your attempt on 2), if you use that
$$
\cos^2 x = \frac{\cos 2x + 1}{2},
$$
and consider $M$ large enough you get for some positive constant $c$ that
$$
\int_M^{\infty} \frac{|\cos x|}{\sqrt{x} + \sqrt{x+1}} \ dx
\geq c \left( \int_M^{\infty} \frac{\cos 2x}{\sqrt{x}} \ dx + \int_M^{\infty} \frac{1}{\sqrt{x}} \ dx \right).
$$
Since you already know that the first integral converges and the second integral diverges, you conclude that the original integral is not absolutely convergent.
