#### Answer

It is not appropriate to substitute $u=x^2,\quad x=\sqrt{u},\quad dx=du/(2/\sqrt{u})$.

#### Work Step by Step

This is not appropriate because if we take $u=x^2$ then $x=\pm\sqrt{u}$. Only for nonnegative $x$ we could write $x=\sqrt{u}$ and this substitution would work only if the region of integration he nonnegative part of the real line. Since within the given bounds, $-1$ and $1$, $x$ takes both positive and negative values, this substitution won't work.