Answer
$$\displaystyle\int{\frac{1}{x\sqrt{4x+1}}dx}=\displaystyle2\arctan \left( \sqrt{4x+1} \right) +C$$
Work Step by Step
$I=\displaystyle\int{\frac{1}{x\sqrt{4x+1}}dx}$
$\left[\begin{matrix}
w=\sqrt{4x+1}& dx=\displaystyle\frac{\sqrt{4x+1}}{2}dw& x=\displaystyle\frac{w^2-1}{4}
\end{matrix}\right]$
$I=\displaystyle\int{\frac{1}{x\sqrt{4x+1}}\times \frac{\sqrt{4x+1}dw}{2}}\\
I=\displaystyle\frac{1}{2}\int{\frac{1}{x}dw}\\
I=\displaystyle2\int{\frac{1}{w^2-1}dw}\\
I=\displaystyle2\arctan w+C\\
I=\displaystyle2\arctan \left( \sqrt{4x+1} \right) +C
$
