Question

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

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:

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