Given f(x)= 6x/ (3x^2 + 4)

We need to find the integral of f(x).

Let us assume that y= 3x^2 + 4

==> dy = 6x dx

Let us substitute:

intg f(x) = intg ( 6x/ (3x^2 + 4) dx

            = intg ( dy / y)

             = ln y + C

==> Now we will substitute with y= 3x^2 + 4

==> intg f(x) = ln ( 3x^2 + 4)  + C

It is given that f(x) = 6x / (3x^3 + 4)

Let t = 3x^2 + 4

dt / dx = 6x

=> dt = 6x dx

Int [6x / (3x^3 + 4) dx]

=> Int [dt / t]

=> ln t + C

replace t with 3x^2 + 4

=> ln (3x^2 + 4) + C

Therefore the integral of f(x) = 6x / (3x^2 + 4) is ln (3x^2 + 4) + C.