Finding the indefinite integral of a root function I'm stuck on a particular integral problem.  The problem is stated as:
$$\int x \sqrt{2x+1} dx$$
My working thus far:
$$\int x \sqrt{2x+1} dx = \frac{1}{2}x^2\frac{2}{3}(2x+1)^\frac{3}{2}$$
$$\frac{1}{3}x^2(2x+1)\frac{3}{2}$$
...
I think I am totally wrong here and completely on another course.  Will somebody please point out where I'm screwing up, and possibly guide me to the correct answer?
Thanks!

It looks like to integrate the product, you found the product of the integration of each factor. We can't do that! 
If $\int f(x) \,dx = F(x)$ and $\int g(x)\,dx = G(x)$

$$\int f(x)\cdot g(x) \,dx \neq F(x)G(x) + C$$


Let's start over.
Note that since the integrand is defined only for $2x+1\geq0$, we put
$$\underbrace{u^2 = 2x+1}_{u = \sqrt{2x+1}} \implies 2u\,du = 2\,dx\iff u\,du = dx$$  
Note also that we also have $2x = u^2 -1 \iff x = \frac 12( u^2 - 1)$.
Then $$I = \int x \sqrt{2x+1} dx = \frac 12\int (u^2 -1)u\cdot u \,du =\frac 12\int(u^4 - u^2)\,du$$
$$\begin{align} I & = \frac 12\left( \frac {u^5}{5} - \frac{u^3}{3}\right) + C \\
& = \frac 12 \left(\frac{(2x + 1)^{5/2}}{5} - \frac{(2x+1)^{3/2}}{3}\right) + C\end{align}$$
