Calculus 2: Integration by parts?

Question

How do you solve ∫2x√(2x-3) dx using integration by parts.

Answer 1

Apply integration by parts  ∫u v dx = u∫v dx −∫u' (∫v dx) dx

Let u = 2x  , v =√(2x-3) ⇒ u' = 2

Now let t = 2x-3 ⇒ dt = 2 dx , dx = dt / 2

Now  let y = 2x - 3 ⇒ dy = 2 dx , dx = dy / 2