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.

There is no need to do a substitution here. Do it by parts:\begin{align}\int2x\sqrt{2x-3}\,\mathrm dx&=\frac23x(2x-3)^{3/2}-\int\frac23(2x-3)^{3/2}\\&=\frac23x(2x-3)^{3/2}-\frac2{15}(2x-3)^{5/2}\\&=\frac23\left(x-\frac15(2x-3)\right)(2x-3)^{3/2}\\&=\frac25(x+1)(2x-3)^{3/2}.\end{align}
