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?  

There are no integration issue in a right neighbourhood of the origin, but when $x\to +\infty$ we have that the integrand function behaves like $\frac{1-k}{x}+O\left(\frac{1}{x^2}\right)$, so a necessary and sufficient condition for the integrability is $k=1$. In such a case
$$\begin{eqnarray*} \int_{0}^{+\infty}\left[\frac{1}{\sqrt{x^2+4}}-\frac{1}{x+2}\right]\,\text dx &\stackrel{x\mapsto 2z}{=}& \int_{0}^{+\infty}\left[\frac{1}{\sqrt{z^2+1}}-\frac{1}{z+1}\right]\,\text dz\\[0.3cm]&=&\left[\text{arcsinh}(z)-\log(z+1)\right]_{0}^{+\infty}\\[0.3cm]&=&\lim_{z\to +\infty}\text{arcsinh}(z)-\log(z+1)\\&\stackrel{z\mapsto\sinh t}{=}&\lim_{t\to +\infty} \log\left(\frac{e^t}{\sinh t+1}\right)=\color{red}{\log 2}.\end{eqnarray*}$$
