Evaluate the integral $\int\frac{dx}{x^2\sqrt{4x^2-1}}$

Question

I solved the question, but Wolfram Alpha and Symbolab both give me two completely different answers.

Here's my work:

Let $u = 2x$ and $a = 1$

Then $du = 2dx$ and $dx = \frac{du}{2}$

Then $\int\frac{dx}{x^2\sqrt{4x^2-1}}$ = $\int\frac{1}{u\sqrt{u^2-a^2}}$

$\therefore \space\int\frac{dx}{x^2\sqrt{4x^2-1}} = sec^{-1}(2x) + C$

Can anyone verify this solution?

Answer 1

Hint: Let $2x=\cosh u$ and then $I=\dfrac{\sqrt{4x^2-1}}{x}$.

Answer 2

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

take $x=\dfrac{\sec \theta}{2}$, then $dx=\dfrac{\sec \theta\tan \theta}{2}d\theta$

Now $\sin\theta=\pm\dfrac{\sqrt{4x^2-1}}{2x}$

When $x>\dfrac{1}{2}$, then $\sqrt{4x^2-1}=\tan \theta$

$$\int\dfrac{dx}{x^2\sqrt{4x^2-1}}=2\int \dfrac{\sec \theta\tan \theta}{\sec^2\theta\sqrt{\sec^2\theta-1}}d\theta=2\int \dfrac{\sin\theta}{\tan \theta}d\theta\\=2\int \cos\theta d\theta=2\sin\theta+C=2\sin\Big(\cos^{-1}\frac{1}{2x}\Big)+C$$

Similarly, When $x<-\dfrac{1}{2}$, then $\sqrt{4x^2-1}=-\tan \theta$

$$\int\dfrac{dx}{x^2\sqrt{4x^2-1}}=-2\sin\Big(\cos^{-1}\frac{1}{2x}\Big)+C$$