Match the inner derivative in a radical integral

$d(1+6x^2)/dx=12x$. Thus $\int x\sqrt{1+6x^2}\,dx=\tfrac1{12}\cdot\tfrac23(1+6x^2)^{3/2}=\tfrac1{18}(1+6x^2)^{3/2}+C$.

$\int_0^a x\sqrt{1+6x^2}\,dx=\tfrac1{18}[(1+6a^2)^{3/2}-1]$. This holds for real $a$; when $a<0$ the reversal of bounds compensates for the negative integrand on the negative x-axis.