<s>[INST] Find $C$ for which improper integral is convergent Find $C$ for which $\displaystyle I=\int\limits_{1}^{\infty} \left(\dfrac{1}{\sqrt{x^2+4}}-\dfrac{C}{x+2}\right)dx$  is convergent.
 [/INST] The only problem is at $+\infty$.
We have
$$\dfrac{1}{\sqrt{x^2+4}}-\dfrac{C}{x+2}=\frac{1}{x}\left(\left(1+\frac{4}{x^2}\right)^{-1/2}-C\left(1+\frac{2}{x}\right)^{-1}\right)=\frac{1}{x}\left(1-C+\frac{2C}{x}+O\left(\frac{1}{x^2}\right)\right)$$
so if $C=1$ we have
$$\dfrac{1}{\sqrt{x^2+4}}-\dfrac{C}{x+2}\sim_\infty\frac{2}{x^2}$$
and the integral is convergent.
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