Answer
$$\frac{1}{\sqrt{6}} \ln \left|2 \sqrt{6}\left(\sqrt{6} x+\frac{1}{2 \sqrt{6}}\right)+\sqrt{24\left(6 x^{2}+x+\frac{1}{24}\right)-1}\right|+C$$
Work Step by Step
Given $$\int \frac{d x}{\sqrt{x+6 x^{2}}}$$
Since
\begin{align*}
x+6 x^{2}&= 6[x^2+\frac{1}{6}x]\\
&=6[(x+1/12)^2-\frac{1}{144}]\\
&=6(x+1/12)^2-\frac{6}{144}
\end{align*}
Let $$ \sqrt{6}(x+1/12)=\frac{\sqrt{6}}{12}\sec u\ \ \ \to \ \ \ \ \sqrt{6}dx=\frac{\sqrt{6}}{12}\sec u\tan udu$$
Then
\begin{align*}
\int \frac{d x}{\sqrt{x+6 x^{2}}}&=\int \frac{d x}{\sqrt{6(x+1/12)^2-\frac{6}{144}}}\\
&=\frac{1}{12}\int \frac{\sec u\tan udu}{\sqrt{\frac{6}{144}\sec^2u-\frac{6}{144}}}\\
&=\frac{1}{\sqrt{6}}\int \frac{\sec u\tan udu}{\tan u}\\
& = \frac{1}{\sqrt{6}}\int \sec u du \\
&= \frac{1}{\sqrt{6}}\ln| \sec u+\tan u|+C\\
&= \frac{1}{\sqrt{6}} \ln |2 u \sqrt{6}+\sqrt{24 u^{2}-1}|+C\\
&=\frac{1}{\sqrt{6}} \ln \left|2 \sqrt{6}\left(\sqrt{6} x+\frac{1}{2 \sqrt{6}}\right)+\sqrt{24\left(6 x^{2}+x+\frac{1}{24}\right)-1}\right|+C
\end{align*}
