Question:

$\int \sqrt{x^2 - 3x^6} dx = \int \sqrt{1 - 3x^4} \cdot x dx$ [Put $\sqrt{3}x^2 = t \Rightarrow 2\sqrt{3}xdx = dt$] <br> $= \int \sqrt{1 - t^2} \frac{dt}{2\sqrt{3}} = \frac{1}{4\sqrt{3}} [t\sqrt{1 - t^2} + \sin^{-1} t] + C$ <br> $= \frac{1}{4\sqrt{3}} [\sqrt{3}x^2 \sqrt{1 - 3x^4} + \sin^{-1} (\sqrt{3}x^2)] + C$