Integration with square root in denominator I am honestly embarrassed to ask this because i feel like i should know how to do this but:
$
\int \frac{x}{\sqrt{2x-1}}dx
$
Try to use u-substitution please

Another substitution you could use is $u=\sqrt{2x-1}$, so $x=\frac{1}{2}(u^2+1)$ and $dx=u du$.
This gives $\displaystyle\int\frac{\frac{1}{2}(u^2+1)}{u} u du=\frac{1}{2}\int(u^2+1) du=\frac{1}{2}\left[\frac{1}{3}u^3+u\right]+C$
$\displaystyle\hspace{.5 in}=\frac{1}{6}(2x-1)^{3/2}+\frac{1}{2}(2x-1)^{1/2}+C$
