Integrate $2x(2x-3)^\frac{1}{2} dx$. I am fairly new to integration.
I need to find the integral of
$$2x(2x-3)^\frac{1}{2} dx.$$
I will be using substitution, right? I tried using $u = 2x - 3$, but I'm not sure what to do with the $2x$. If I find $\frac{du}{dx}$, it turns out to be $2$ (so $du = 2dx$). There's that $x$ left over.

Substitute $(2x-3)=u.$Hence $dx = \frac {du}{2}.$ Your integral becomes -
$$\frac {1}{2}\int \sqrt {u}(u+3)du = \frac {1}{3}(u+3)\cdot u^{\frac {3}{2}} - \frac {1}{2}\int u^{\frac {3}{2}} du = \frac {(u+3)u^{\frac {3}{2}}}{3} - \frac {1}{5}u^{\frac {5}{2}}$$
