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.
 A: 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.
A: Hint: First note that by Comparison the integral does not converge if $C\le 0$. So we may assume that $C$  is positive. 
Bring to a common denominator, and then rationalize the numerator, by multiplying top and bottom by $x+2+C\sqrt{x^2+4}$. We get 
$$\frac{(x+2)^2-C^2(x^2+4)}{(x+2+C\sqrt{x^2+4})(x+2)(\sqrt{x^2+4})}.\tag{1}$$
Argue by Comparison that  the only positive value of $C$ for which the integral from $1$ to $\infty$ of (1) converges is given by $C=1$. 
Remark: We did algebraic manipulation, but in fact most of the answer was, at the informal level, immediate. The long run behaviour of $\frac{1}{\sqrt{x^2+4}}$ and of $\frac{1}{x+2}$ is effectively the same, and it is the same as that of $\frac{1}{x}$. Thus if $C\ne 1$, our difference will behave like $\frac{1-C}{x}$, and the integral must diverge. If $C=1$, there will be cancellation, and the combined term $\frac{1}{\sqrt{x^2+4}}-\frac{1}{x+2}$ has a chance to decay fast enough for the integral to converge. A calculation now shows that it does. 
