$\int\limits_0^∞[\frac{3}{x^2+1}]dx$ is equal to ([.] denotes the greatest integer function)

$\frac{3}{\sqrt{2}}$

Here $0<\frac{3}{x^2+1}≤3$,$\frac{3}{x^2+1}=2$

$⇒x=\frac{1}{\sqrt{2}}$ and $\frac{3}{x^2+1}=1⇒x=\sqrt{2}$

$⇒\int\limits_0^{1/\sqrt{2}}2dx+\int\limits_{1/\sqrt{2}}^{\sqrt{2}}1.dx+\int\limits_{\sqrt{2}}^∞0dx=\sqrt{2}+\sqrt{2}-\frac{1}{\sqrt{2}}$

$=2\sqrt{2}-\frac{1}{\sqrt{2}}=\frac{3}{\sqrt{2}}$

Hence (C) is the correct answer.