Integrate $2x(2x-3)^frac{1}{2} dx$.

Substitute $(2x-3)=u.$Hence $dx = frac {du}{2}.$ Your integral becomes -

$$frac {1}{2}int sqrt {u}(u+3)du = frac {1}{3}(u+3)cdot u^{frac {3}{2}} - frac {1}{2}int u^{frac {3}{2}} du = frac {(u+3)u^{frac {3}{2}}}{3} - frac {1}{5}u^{frac {5}{2}}$$

There is no need to do a substitution here. Do it by parts:begin{align}int2xsqrt{2x-3},mathrm dx&=frac23x(2x-3)^{3/2}-intfrac23(2x-3)^{3/2}\&=frac23x(2x-3)^{3/2}-frac2{15}(2x-3)^{5/2}\&=frac23left(x-frac15(2x-3)right)(2x-3)^{3/2}\&=frac25(x+1)(2x-3)^{3/2}.end{align}

$$int2x(2x-3)^frac{1}{2} dx$$

This can be recognized to be in the form $int{x^m(1+x^n)^p}$ right away

Let $2x-3=t^2implies dx=tdt$

$$int(t^2+3)t^2 dt$$

Can you finish?

$2x=u+3$

So, integrand becomes $$(u+3)sqrt u frac{du}2$$

Can you do now?

Since $u=2x-3$, then $2x=(u+3)$ and $du=2dx$ so $dx=frac{1}{2}du$, we have $$int2x(2x-3)^frac{1}{2} dx=int (u+3)u^{frac{1}{2}}frac{du}{2}$$ $$=frac{1}{2}int[u^{frac{3}{2}}+3u^{frac{1}{2}}]du$$

Can you end it?