Evaluate $$ \int_{1}^{\infty} \left(\frac{1}{\,\sqrt{\, x^{2} + 4\,}\,} - \frac{C}{x + 2}\right)\,\mathrm{d}x $$
I am stuck at this question where I should find the value of C for which the integral converges and evaluate the integral for this value of C. Any hints or answers ??.
On
Note that $(x+2)^2\geqslant x^2+4$ for $x>0$ therefore $$\int^{\infty}_1\frac{1}{\sqrt{x^2+4}}-\frac{C}{x+2}\,dx\geqslant \int^{\infty}_{1}\frac{1}{\sqrt{(x+2)^2}}-\frac{C}{x+2}\,dx=\int^{\infty}_1\frac{1-C}{x+2}\,dx$$ The left side is convergent only if the right side is. That happens only if $C=1$.
$\frac{1}{\sqrt{x^2+4}}$ behaves like $\frac{1}{x}$ in a left neighbourhood of $+\infty$ and $\frac{1}{x}$ is not an integrable function over there.
So, in order to have a convergent integral, we need $C=1$ to compensate such singular behaviour.
$\frac{1}{\sqrt{x^2+4}}-\frac{1}{x+2}$ behaves like $\frac{K}{x^2}$ for $x\to +\infty$ and we are fine. In explicit terms
$$ \int_{1}^{+\infty}\left(\frac{1}{\sqrt{x^2+4}}-\frac{1}{x+2}\right)\,dx = \log\left(\tfrac{3}{\varphi}\right)\approx \varphi-1 $$ where $\varphi=\frac{1+\sqrt{5}}{2}$ is the golden ratio.