Answer
$$\int \frac{1}{x\sqrt{4x^{2}+1}}dx=ln\left | \frac{\sqrt{4x^{2}+1}}{2x} -\frac{1}{2x} \right |+C$$
Work Step by Step
$$let\ x=\frac{1}{2}tan\,t,\,dx=\frac{1}{2}sec^{2}t\,dt$$
$$\int \frac{1}{x\sqrt{4x^{2}+1}}dx=\int \frac{sec^{2}}{tant\sqrt{tan^{2}t+1}}dt$$
$$=\int \frac{sect}{tant}dt=\int csc\,t\,dt$$
$$=ln\left | csc\,t-cot\,t \right |+C$$
$$=ln\left | \frac{\sqrt{4x^{2}+1}}{2x} -\frac{1}{2x} \right |+C$$
