Set $y\=1-3 x^2$ so that $\mathrm{dy}\=-6 x \mathrm{dx}$ and $x \mathrm{dx}\=-\frac{\mathrm{dy}}{6}$. Under this change of variable, the given indefinite integral becomes

 $\int \frac{\sqrt{y} \mathrm{dy}}{-6}\= -\frac{1}{6}\int \sqrt{y}\mathit{\ⅆ}y\= -\frac{1}{6}\frac{y^{3\/2}}{3\/2}\= -\frac{1}{9}{\left(1-3 x^2\right)}^{3\/2}$  

Of course, there are settings in which the addition of an arbitrary constant is deemed essential.

An alternate approach to making the substitution $y\=1-3 x^2$ starts with the recognition that $\mathrm{dy}\=-6 x \mathrm{dx}$, and that outside the square root there's an $x$, but not a factor of $-6$. So, insert this factor inside the integral, and compensate with a factor of $-1\/6$ outside the integral, to obtain $-\frac{1}{6}\int \sqrt{y}\mathit{\ⅆ}y$ immediately.

It is interesting to note that Maple might make a different substitution, as per Table 4.4.3(a) where the Context Panel's "Student Calculus1≻All Solution Steps" option is applied to the inert form of the indefinite integral.

Maple has made the substitution $u^2\=1-3 x^2$, so that $u\=\sqrt{1-3 x^2}$ and $2 u \mathrm{du}\=-6 x \mathrm{dx}$. Thus, $x \mathrm{dx}\=-\frac{2}{6}u \mathrm{du}$ and the new integrand is $-\frac{1}{3}{u}^2$. Of course, the final results for the two approaches will agree, but some of the intermediate steps will not.