integration - 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 $\ln 5$.

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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?

Answer

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

$$\sinh^{-1}\frac x 2=\log \left(\frac x 2 + \sqrt{\frac{x^2}{4}+1}\right) \sim \log x$$