How to Integrate $1/\left(x\sqrt{1-4x^2}\right)$?

Question

How do I integrate:

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

I am thinking about (Integrals of Inverse Hyperbolic Function):

$$\frac{-1}{a} \operatorname{arcsech}\left(\frac xa\right)+C$$

But do I need to use substitution for $2x$?

Answer 1

HINT:

$$\int\left(\frac{1}{x\sqrt{1-4x^2}}\right)\text{d}x=$$

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Substitute $u=\sqrt{1-4x^2}$ and $\text{d}u=-\frac{4x}{\sqrt{1-4x^2}}\text{d}x$:

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$$\int\left(\frac{1}{u^2-1}\right)\text{d}u=$$ $$-\int\left(\frac{1}{1-u^2}\right)\text{d}u=$$ $$-\tanh^{-1}(u)+C$$

Substitute everything back and you got the final answer!

Answer 2

HINT:

let $2x=\sin \theta\implies 2dx=\cos\theta d\theta$ $$\int \frac{1}{x\sqrt{1-4x^2}}\ dx$$ $$=\frac{2}{2}\int \frac{\cos\theta\ d\theta}{\sin\theta\sqrt{1-\sin^2\theta}}$$ $$=\int \frac{\cos\theta\ d\theta}{\sin\theta|\cos\theta|}$$