Looking for a cleaner/quicker way to evaluate the integral of a quadratic Looking for a clean/quick way to evaluate
$$\int_{2-\sqrt{3}}^{2+\sqrt{3}}{ \left(x^2-4x+1\right)\textrm{d}x}=\left.\frac{x^3}{3}-2x^2+x \right|_{2-\sqrt{3}}^{2+\sqrt{3}}$$
So I evaluated all of this out leaving this expression as it was and there was some pretty nice cancelation that led me to the answer $-\sqrt{3}$. However, this was a question from a contest so I'm expecting that if we factor out the $x$ and then evaluate at the bounds there is something we can do to make the algebra nicer.
$$=\left.x(\frac{x^2}{3}-2x+1) \right|_{2-\sqrt{3}}^{2+\sqrt{3}}$$
Anyone see any quick way to do this? Thanks!

Notice that $x^2 - 4x + 1 = (x - 2)^2 - 3$, so a substitution $y = x - 2$ yields
\begin{align*}
\int_{2-\sqrt{3}}^{2+\sqrt{3}} (x^2-4x+1)dx = \int_{-\sqrt{3}}^{\sqrt{3}}(y^2-3)dy = \left[\frac{y^3}{3} - 3y\right]_{-\sqrt{3}}^{\sqrt{3}} = 2\sqrt{3}-6\sqrt{3} = -4\sqrt{3}
\end{align*}
