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.**

To find the integral 6x/(3x^2+4)

We put t = 3x^2+4.

We differentiate t = 3x^2+4.

di = 6xdx.

Therefore Int f(x) dx = Int (6xdx/(3x^2+4)

Int (6xdx/(3x^2+4) = Int dt//t

Int (6xdx/(3x^2+4) = lnt +C...(1)

Now we put t= 3x^2+4 in (1).

Therefore (6xdx/(3x^2+4) = ln(3x^2+4) + C, where C is the constant of integration.

We notice that the numerator is the derivative of denominator. So, we'll note the denominator by u:

3x^2 + 4 = u

3x^2 = u - 4

x^2 = (u-4)/3

x = sqrt[(u-4)/3]

We'll differentiate both sides:

6xdx = du

We'll re-write the integral in u:

Int 6xdx / (3x^2 + 4) = Int du/u

Int du/u = ln |u| + C

**Int 6xdx / (3x^2 + 4) = ln (3x^2 + 4) + C**

Since 3x^2 + 4 > 0, we will not take the absolute value of 3x^2 + 4.