`int dx/(x(sqrt(4x^2-1)))` Find the indefinite integral

We have to evaluate the integral : ```\int \frac{dx}{x\sqrt{4x^2-1}}`

Let `\sqrt{4x^2-1}=u`

So, `\frac{1}{2\sqrt{4x^2-1}}.8x dx=du`

      `\frac{4xdx}{\sqrt{4x^2-1}}=du`

       `\frac{dx}{\sqrt{4x^2-1}}=\frac{du}{4x}`

Hence we have,

`\int \frac{dx}{x\sqrt{4x^2-1}}=\int \frac{du}{4x^2}`

                 `=\int \frac{du}{u^2+1}`

                  `=tan^{-1}(u)+C`

                   `=tan^{-1}(\sqrt{4x^2-1})+C` where C is a constant

We have to evaluate the integral : ```\int \frac{dx}{x\sqrt{4x^2-1}}`

Let `\sqrt{4x^2-1}=u`

So, `\frac{1}{2\sqrt{4x^2-1}}.8x dx=du`

      `\frac{4xdx}{\sqrt{4x^2-1}}=du`

       `\frac{dx}{\sqrt{4x^2-1}}=\frac{du}{4x}`

Hence we have,

`\int \frac{dx}{x\sqrt{4x^2-1}}=\int \frac{du}{4x^2}`

                 `=\int \frac{du}{u^2+1}`

                  `=tan^{-1}(u)+C`

                   `=tan^{-1}(\sqrt{4x^2-1})+C` where C is a constant