Improper integral $\int_{0}^{\infty}\left(\frac{1}{\sqrt{x^2+4}}-\frac{k}{x+2}\right)\text dx$ 
Given improper integral $$\int \limits_{0}^{\infty}\left(\frac{1}{\sqrt{x^2+4}}-\frac{k}{x+2}\right)\text dx \, ,$$
  there exists $k$ that makes this integral convergent.
  Find its integration value.  

Choices are $\ln 2$, $\ln 3$, $\ln 4$, and $\ln 5$.  

I've written every information from the problem.
Yet I'm not sure whether I should find the integration value from the given integral or $k$.  
What I've tried so far is,
$\int_{0}^{\infty} \frac{1}{\sqrt{x^2+4}} \, \text dx= \left[\sinh^{-1}{\frac{x}{2}}\right]_{0}^{\infty}$ 
How should I proceed?  

We have that for $x\to \infty$
$$\frac{1}{\sqrt{x^2+4}}=\frac1x(1+4/x^2)^{-1/2}\sim \frac1x-\frac2{x^3}$$
and
$$\frac k {x+2}\sim \frac k x$$
therefore in order to have convergence we need $k=1$ in such way that the $\frac1x$ term cancels out.
Then we need to solve and evaluate
$$\int_{0}^{\infty}\left (\frac{1}{\sqrt{x^2+4}}-\frac{1}{x+2}\right)\text dx=\left[\sinh^{-1}\frac x 2 -\log (x+2)\right]_{0}^{\infty}$$
and to evaluate the value at $\infty$ recall that
$$\sinh^{-1}\frac x 2=\log \left(\frac x 2 + \sqrt{\frac{x^2}{4}+1}\right) \sim \log x$$
