MATH EXPERTS NEEDED! How do you solve this CALCULUS Integral question?

Question

Evaluate this integral:

∫ (4y dy) / √(2y^2 + 1)

Answer 1

Just use a simple substitution:

let u = (2y^2+1)

then du = 4y dy (quite convinient)

now just substitute back:

∫ du / √(u)

∫ u^(-1/2) du

u^(1/2) / (1/2)

2 u ^ (1/2)

now just replace u

2 (2y^2+1)^(1/2)

2 * √(2y^2+1)

Answer 2

I take it you really meant INT 4y /√(2y^2 + 1) dy – at least I hope you did!
The way I look at it is this:
if v = (2y^2 + 1)^ 0.5
then v’ = -0.5 (2y^2 + 1)^ -0.5 x 4y ( which is nearly the same as the question)
So ….
PS I would love to know how you get integral sign - I just got a box when I tried