Official solution
\[I = \int\limits_1^3 { - 3dx + \int\limits_1^3 {\left[ {{{(x - 1)}^2}} \right]dx} } \]
Put x - 1 = t ; dx = dt
\[I = ( - 6) + \int\limits_0^2 {\left[ {{t^2}} \right]} dt\]
\[I = - 6 + \int\limits_0^1 {0dt} + \int\limits_1^{\sqrt 2 } {1dt} + \int\limits_{\sqrt 2 }^{\sqrt 3 } {2dt} + \int\limits_{\sqrt 3 }^2 {3dt} \]
\[I = - 6 + \left( {\sqrt 2 - 1} \right) + 2\sqrt 3 - 2\sqrt 2 + 6 - 3\sqrt 3 \]
\[I = - 1 - \sqrt 2 - \sqrt 3 \]