Tricky integral using substitution

**The Problem**

Use the substitution $\displaystyle u = 2x - 1$ to find the integral of $\displaystyle f(x) = x\sqrt{2x - 1}$

**Attempt**

$\displaystyle \frac{du}{dx} = 2x$, so $\displaystyle dx = \frac{1}{2} du$.

Therefore, this integral simplifies to $\displaystyle \frac{1}{2}(Int\ x\sqrt{u}\ du)$.

Rearranging, we get x = $\displaystyle \frac{u + 1}{2}$, so this becomes $\displaystyle \frac{1}{2}$ x $\displaystyle \frac{1}{2}\ Int (\sqrt{u}\ (u+1))\ du$.

Expanding, this becomes $\displaystyle \frac{1}{4}\ Int\ (u^\frac{3}{2}\ + \sqrt{u})\ du$.

As everything is now a power of $\displaystyle u$, we can integrate as follows:

$\displaystyle \frac{1}{4}(\frac{u^\frac{5}{2}}{\frac{5}{2}}\ + \ \frac{u^\frac{3}{2}}{\frac{3}{2}}) $.

Multiplying out the fractions:

$\displaystyle \frac{1}{4}(\frac{2u^\frac{5}{2}}{5}\ +\ \frac{2u^\frac{3}{2}}{3}}) $

Multiplying out the brackets:

$\displaystyle \frac{2u^\frac{5}{2}}{20}\ +\ \frac{2u^\frac{3}{2}}{12}$

Cancelling:

$\displaystyle \frac{u^\frac{5}{2}}{10}\ + \ \frac{u^\frac{3}{2}}{6}$

And as $\displaystyle u = 2x - 1$, we get:

$\displaystyle \frac{1}{10}(2x\ -\ 1)^\frac{5}{2} + \frac{1}{6}(2x\ -\ 1)^\frac{3}{2}+ C$.

I've checked this using 'Show Steps' after inputting the integral into WolframAlpha, and it appears to be correct.

However, the answer given in both the textbook and WolframAlpha is:

$\displaystyle \frac{1}{15}(2x\ -\ 1)^\frac{3}{2}(3x\ +\ 1)+C$.

I would like a hint as to how I would progress from my answer to the answer given.

OK, so what do I do next?

OK, so I took out the common factor to get:

$\displaystyle (2x-1)^\frac{3}{2}(\frac{1}{10}(2x\ - 1)\ + \frac{1}{6})$, then I could take out the factor of $\displaystyle \frac{1}{2}$ to get:

$\displaystyle \frac{1}{2}(2x\ -1)^\frac{3}{2}(\frac{1}{5}(2x\ -1)\ + \frac{1}{3})$