We carry out the details. In a comment at the end, we show a somewhat simpler way.
Let $u=\sqrt{1-x}$. Then $\dfrac{du}{dx}=-\dfrac{1}{2\sqrt{1-x}}$.
Thus $du=-\dfrac{1}{2\sqrt{1-x}}\,dx$. You left out the $dx$, which may be part of the reason you are puzzled.
So $dx=-2\sqrt{1-x} \,du=-2u\,du$.
Also, since $u^2=1-x$, we have $x=1-u^2$, and therefore $x^2=(1-u^2)^2$.
Expressing everything in terms of $u$, we get
$$\int (1-u^2)^2 (u)(-2u)\,du.$$
Note that by everything, we include $dx$.
Now expand the $(1-u^2)^2$, multiply through by $2u^2$, and integrate term by term.
Remark: I would prefer to do the same substitution in the form $u^2=1-x$. Then $2u\,du=-dx$, no unpleasant square roots, less algebra. Try it, you will like it.
Best Answer
Try completing the square to get
\begin{equation*} \int x\sqrt{(\sqrt{7}x+3\sqrt{7})^2-4)}\,\mathrm dx \end{equation*}
& use the substitution $u=\sqrt{7}x+3\sqrt{7},~du=\sqrt{7}dx.$