antiderivative [(3x)/((2x^2+1)^1/2)]

what i did:

let u=2x^2+1

du/dx=4x

(3/4)du=(3/4)(4x)(dx)

3/4 du=(3x)(dx)

x=[(u-1)/2)]^1/2

antiderivative [(u^1/2)(3[(u-1)/2]^1/2)(du)]

= antiderivative [(3/u^1/2)[(u-1)/2]^1/2

from here i get stuck..i need to multiply du in as well but im not sure how to do that since i only found 3/4du=3xdx

how would i do this? or am i doing something wrong?