78. $\\int\\frac{dx}{x\\sqrt{4x^{2}-1}}$

Explanation:

Step1: Substitution

Let (u = 2x), then (du=2dx), and (dx=\frac{1}{2}du). The integral becomes (\int\frac{\frac{1}{2}du}{\frac{u}{2}\sqrt{u^{2}-1}}).

Step2: Simplify the integral

Simplify (\int\frac{\frac{1}{2}du}{\frac{u}{2}\sqrt{u^{2}-1}}) to (\int\frac{du}{u\sqrt{u^{2}-1}}). We know that the formula for (\int\frac{du}{u\sqrt{u^{2}-1}}=\text{arcsec}|u| + C).

Step3: Back - substitution

Since (u = 2x), the integral (\int\frac{dx}{x\sqrt{4x^{2}-1}}=\text{arcsec}|2x|+C).

Answer:

(\text{arcsec}|2x| + C)