Where have I made my mistake in my integrating $\int x \sqrt{2x - 1} \, dx$? $$\int x \sqrt{2x - 1} \,dx$$
Let $u = 2x - 1$
$$\int x \sqrt{u}\, dx$$
$$\frac{du}{dx} = 2 \implies \frac{1}{2}du = dx.$$
So the integral is written as
$$\int \frac{1}{2} u^{\frac{1}{2}} \, du$$
$$ = \frac{1}{2} \left(\frac{2}{3} u^{\frac{3}{2}} \right)$$
$$ = \frac{1}{3} (2x - 1)^{\frac{3}{2}} + c$$
But apparently this is wrong.
Wolfram says it is something completely different. Where have I made the mistake?
EDIT: OOOPS!! Sorry. Silly mistake.
 A: $$\int x \sqrt{u} dx\tag{1}$$
$$\int \frac{1}{2} u^{\frac{1}{2}} du\tag{2?}$$
What happened to the factor of $x$ going from $(1) \to (2)$?
We have that $u = 2x-1$, so $x = \dfrac{u+1}{2},\;$ and as you know, $dx =  \dfrac 12 du.\,$ This gives us:
$$\int x \sqrt{2x - 1} \,dx \quad = \quad \frac{1}{2} \int \frac{(u+1)}{2}  u^{1/2} \, du\tag{2} \quad =\quad \frac 14 \int \left(u^{3/2} + u^{1/2}\right) \,du$$
A: If $u = 2x-1$ then $x = \frac{u+1}{2}$ and so
$$\int  x \sqrt{2x-1}dx = \int \frac{1}{2}\frac{u+1}{2}\sqrt{u}du$$
A: I'd do it like this:
$$
u = \sqrt{2x-1}
$$
$$
u^2 = 2x-1
$$
$$
2u\,du = 2\,dx
$$
$$
u\,du = dx
$$
$$
x = \frac{u^2+1}{2}
$$
$$
\int x\sqrt{2x-1}\,dx = \int\frac{u^2+1}{2}\, u \,\Big(u\,du\Big)
$$
Then you're integrating a polynomial function.  When you've done that, change it back to a function of $x$ by putting $\sqrt{2x-1}$ in place of $u$
A: You dropped your $x$ when you went from $$\int x\sqrt{u}\,dx \implies \int\frac{1}{2}\sqrt{u}\,du $$
Since $x=\frac{1}{2}(u+1)$, the correct expression would be $$\int\frac{1}{4}(u+1)u^{1/2}\,du$$
