Integration by substitution using $u^2=3-x$ The question says to integrate $\frac x{\sqrt{3-x}}$ using the substitution $u^2=3-x$. 
What I did was first rearrange $u^2=3-x$ into $ 3 - u^2 =x$ and then subbed it into the integral .
I then rearranged $u^2=3-x$ to $u=\sqrt{3-x} $ and then also subbed that in.
Next I found the derivative of $u$ which I believe to be $du/dx = -1/(2\sqrt{3-x}$) I then sub in u and then rearrange to get $-2udu = dx $ 
I then also sub that in then have $y = -2(3-u^2)$ and integrate it. 
However when i integrate it i get $-6\sqrt{3-x} + (2(3-x)^3)/3 $
But the answer is $(-2/3)(x+6)\sqrt{3-x} $.
I tried to rearrange my answer to get that, but I was unsuccessful. Is there instead a mistake in my working out? 
Side note: I tried to put things to the power of a 1/2 but it did not work out I'm not sure how to do it 
thank you 

Using $u^2 = 3-x$, we have the following
$$\int \dfrac{x}{\sqrt{3-x}} dx = \int \dfrac{3-u^2}{\sqrt{u^2}} dx = \int \dfrac{3-u^2}{u} dx$$
Now, like you did, solve the transformation for $u$ such that $u=(3-x)^\frac{1}{2}$. Then, $du = -\dfrac{1}{2}(3-x)^{-\frac{1}{2}} dx = -\dfrac{1}{2}(u^2)^{-\frac{1}{2}}dx = -\dfrac{1}{2u}dx$ such that $-2u du = dx$. By substitution, we have 
$$\int \dfrac{x}{\sqrt{3-x}} dx = -\int \dfrac{(2u)(3-u^2)}{u} du=-\int (6-2u^2) du$$
Thus,
$$\int \dfrac{x}{\sqrt{3-x}} dx = -6u+\dfrac{2}{3}u^3 + C$$
Upon substitution of $u=\sqrt{3-x}$, we have
$$\begin{align}
\int \dfrac{x}{\sqrt{3-x}} dx &= -6\sqrt{3-x}+\dfrac{2}{3}(3-x)^{\frac{3}{2}} + C
\\
&=-6\sqrt{3-x}+\dfrac{2}{3}(3-x)\sqrt{3-x} + C
\\
&=\sqrt{3-x}\Bigg(-6+\dfrac{2}{3}(3-x)\Bigg)+C
\\
&=-\dfrac{2}{3}\sqrt{3-x}\Bigg(9-(3-x)\Bigg)+C
\\
&=-\dfrac{2}{3}\sqrt{3-x}(6+x)+C
\end{align}$$
