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.
 A: $$\int2x(2x-3)^\frac{1}{2} dx$$
This can be recognized to be in the form $\int{x^m(1+x^n)^p}$ right away
Let $2x-3=t^2\implies dx=tdt$
$$\int(t^2+3)t^2 dt$$
Can you finish?
A: 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?
A: $2x=u+3$
So, integrand becomes $$(u+3)\sqrt u \frac{du}2$$
Can you do now?
A: 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}
A: Substitute $(2x-3)=u.$Hence $dx = \frac {du}{2}.$ Your integral becomes -
$$\frac {1}{2}\int \sqrt {u}(u+3)du = \frac {1}{3}(u+3)\cdot u^{\frac {3}{2}} - \frac {1}{2}\int u^{\frac {3}{2}} du = \frac {(u+3)u^{\frac {3}{2}}}{3} - \frac {1}{5}u^{\frac {5}{2}}$$
