Given \(\int_0^\infty {\left( {\frac{1}{{\sqrt {{x^2} + 4} }} - \frac{C}{{x + 2}}} \right)} dx\)
\(\begin{aligned}{l}\int_0^\infty {\left( {\frac{1}{{\sqrt {{x^2} + 4} }} - \frac{C}{{x + 2}}} \right)} dx &= \mathop {\lim }\limits_{t \to \infty } \int_0^t {\left( {\frac{1}{{\sqrt {{x^2} + 4} }} - \frac{C}{{x + 2}}} \right)dx} \\ &\Rightarrow \mathop {\lim }\limits_{t \to \infty } \int_0^t {\left( {\frac{1}{{\sqrt {{x^2} + 4} }} - \frac{C}{{x + 2}}} \right)dx} \end{aligned}\)
and
\(\begin{aligned}{c} &\Rightarrow \mathop {\lim }\limits_{t \to \infty } \left( {\ln \left| {x + \sqrt {{x^2} + 4} } \right| - C\ln (x + 2)} \right)_0^t\\ &\Rightarrow \mathop {\lim }\limits_{t \to \infty } \left( {\ln \left| {x + \sqrt {{x^2} + 4} } \right| - \ln {{(x + 2)}^c}} \right)_0^t\\ &\Rightarrow \mathop {\lim }\limits_{t \to \infty } \left( {\ln \frac{{\left| {x + \sqrt {{x^2} + 4} } \right|}}{{{{(x + 2)}^c}}}} \right)_0^t\end{aligned}\)
If here we put \(t = \infty \), then the integral will be divergent \(\int_0^\infty {\left( {\frac{1}{{\sqrt {{x^2} + 4} }} - \frac{C}{{x + 2}}} \right)} dx = \infty \)
So, for making it convergent, we will take x common and try to eliminate it
\( &\Rightarrow \) &\( \Rightarrow \mathop {\lim }\limits_{t \to \infty } \left( {\log \left| {\frac{{x(1 + \sqrt {1 + \frac{4}{{{x^2}}}} }}{{{x^c}{{(1 + \frac{2}{x})}^c}}}} \right|} \right)_0^t\)
In above equation \(x\)should be equal to \({x^c}\) for convergence of the given integral
\(\begin{aligned}{c} \Rightarrow x = {x^c}\\ \Rightarrow C = 1\end{aligned}\)
Hence the value of C is 1
