Answer
$$\int \frac{dx}{x^{2}\sqrt{4x^{2}-1}}=\frac{\sqrt{4x^{2}-1}}{x}+C$$
Work Step by Step
$$let\ x=\frac{1}{2}sec\,t,dx=\frac{1}{2}tan\,t\,sec\,t\,dt,sin\,t=\frac{\sqrt{4x^{2}-1}}{2x}$$
$$\int \frac{dx}{x^{2}\sqrt{4x^{2}-1}}=\int \frac{\frac{1}{2}sec\,t\,tan\,t\,dt}{\frac{1}{4}sec^{2}t\sqrt{sec^{2}t-1}}$$
$$=2\int \frac{sec\,t\,tan\,t\,dt}{sec^{2}t\,tan\,t}=2\int cos\,t\,dt$$
$$=2sin\,t+C=\frac{\sqrt{4x^{2}-1}}{x}+C$$
