how to solve integral?

Question

how do you solve this? substitution?

Integral of x / Sqrt( 2+3(x^2) )

Answer 1

I = ∫ x / (2 + 3 x ²)^(1/2) dx
Let u = 2 + 3 x ²
du = 6x dx
du / 6 = x dx
I = (1/6) ∫ du / u^(1/2)
I = (1/6) ∫ u^(-1/2) du
I = (1/6) u^(1/2) / (1/2) + C
I = (1/3) (2 + 3 x ²)^(1/2) + C

Answer 2

Integral of x / Sqrt( 2+3(x^2) )
integral of ( x*(2+3x^2)^(-1/2))
let u= 2+3x^2
du=6x
1/6 Integral of 6x / Sqrt( 2+3(x^2) )
1/6*[(2+3x^2)^(-1/2+1)]/(-1/2+1)
1/6*[(2+3x^2)^(1/2)]/(1/2)
1/3*(2+3x^2)^(1/2)

Answer 3

If you differentiate 2 + 3(x^2) you get 6x, whcih is close enough to x to let us know we are probably on the right track.
Here is the first line:
Let u = 2+ 3(x^2)