Answer
$$\int \frac{x}{\sqrt{6 x+1}} d x =\frac{1}{18}\left(\frac{1}{3}(6x+1)^{3/2}-\sqrt{6x+1}\right)+c$$
Work Step by Step
Since $$\int \frac{x}{\sqrt{6 x+1}} d x$$
Let $$ u^2=6x+1 \ \ \to\ \ 2udu=6dx $$
Then
\begin{align*}
\int \frac{x}{\sqrt{6 x+1}} d x&=\int \frac{u^2-1}{\sqrt{u^2}} \frac{udu}{18}\\
&=\frac{1}{18}\int (u^2-1)du\\
&=\frac{1}{18}\left(\frac{1}{3}u^3-u\right)+c\\
&=\frac{1}{18}\left(\frac{1}{3}(6x+1)^{3/2}-\sqrt{6x+1}\right)+c
\end{align*}
