∫ 2/x^1/2 dx = ?

Answer 1

4x^(1/2)

Answer 2

I interpret that being the same as ∫ 2x^(-1/2) dx
So you can bring the 2 out front to get 2 ∫x^(-1/2) dx which should be easier to integrate.

The integral of x^(-1/2) would be 2x^(1/2)
Because [2x^(1/2)]' = x^(-1/2)

So the answer is 4x^(1/2) or 4√x ...
Oh, and don't forget the + C at the very end since this is an indefinite integral. :)

So final answer ::: 4√x + C