Integral of (xdx) / [sqrt(1-x^4)]

Answer 1

Let u = x². Then du = 2xdx. So xdx = (1/2)du

The integral then becomes (1/2)∫[1 / √(1 - u²)]du = (1/2)Arcsinu + C = (1/2)Arcsin(x²) + C

Answer 2

You might be tempted to try trig sub but its not quite in the right form for that because of the x^4 term under the radical. However can you think of a u-sub you could do first to transform the integral into something you can use trig sub on?