It often pays to consider an equivalent problem for which the algebra looks a bit easier. At least it provides an alternative check on doing things the hard way.
Here the length of the curve $y = 1 - \sqrt{x}$ on $[0,1]$ is by vertical translation and reflection in the $x$-axis the same as for $y = \sqrt{x}$ on the unit interval.
Now exchange the roles of $x$ and $y$, which amounts to reflection in $y = x$, and we would have the same length for $y = x^2$ on $[0,1]$.
Then applying the arclength formula that Chandru has nicely formatted:
\begin{align*} \ell = \int\limits_{0}^{1} \sqrt{1+4x^2} \ dx \end{align*}
one gets a proper integral that yields to trigonometric substitution:
$ x = \frac{1}{2} \tan{\theta}$
$ dx = \frac{1}{2} \sec^{2}{\theta} d\theta$
with respective limits of integration for $\theta \in [0,\tan^{-1}(2)]$.
Thus: \begin{align*} \ell = \int\limits_{0}^{1} \sqrt{1+4x^2} \ dx = \frac{1}{2} \int\limits_{0}^{\tan^{-1}(2)} \sec^3{\theta} \ d\theta \end{align*}
Consulting a table of trigonometric identities, I find $\sec^{3}{\theta}$ has antiderivative:
$ \frac{1}{2} ( \sec{\theta} \tan{\theta} + \ln{| \sec{\theta} + \tan{\theta}|} ) + C$
Fortunately the nonconstant terms are zero when $\theta = 0$, so we only have the value at $\theta = \tan^{-1}(2)$ to simplify:
$ \sec(\tan^{-1}(2)) = \sqrt{5}$
$ \ell = \frac{1}{2} (\sqrt{5} + \frac{1}{2} \ln( 2 + \sqrt{5}) )$
Numerically this gives $\ell = 1.47894...$ or slightly more than the straightline distance $\sqrt{2}$, which seems plausible.