[math]x^2\sqrt{x}[\math]
[math] \left(\begin{array}{cc}1\\1\\1\end{array}\right) = \left(\begin{array}{cc}2x\\-2y\\0\end{array}\right)\lambda + \left(\begin{array}{cc}2\\0\\1\end{array}\right)\m u [\math]
$\displaystyle \left(\begin{array}{cc}1\\1\\1\end{array}\right)$ = $\displaystyle \left(\begin{array}{cc}2x\\-2y\\0\end{array}\right)\lambda$ $\displaystyle +$ $\displaystyle \left(\begin{array}{cc}2\\0\\1\end{array}\right)$$\displaystyle \mu$
$\displaystyle so\ \mu=1$
$\displaystyle \left(\begin{array}{cc}2x\\-2y\\0\end{array}\right)\lambda$ = $\displaystyle \left(\begin{array}{cc}-1\\-1\\0\end{array}\right)$
$\displaystyle So\ \lambda \neq 0$
$\displaystyle \frac{1}{\lambda} = -2x$
$\displaystyle \frac{1}{\lambda} = -2y$
$\displaystyle -2x = -2y, and\ x = y$
$\displaystyle BUT x^2 - y^2 = 1???$
$\displaystyle u = -x^2$
$\displaystyle du = -2xdx$
$\displaystyle \frac{-1}{2}du = xdx$
Now, substitute in $\displaystyle \frac{-1}{2}du\ for\ xdx$.
Pull that 5 out to get $\displaystyle \frac{-5}{2}$ outside the integral.
It's cake from there...