Recall that the indefinite integral is denoted as:

`int f(x) dx =F(x)+C`

There properties and basic formulas of integration we can apply to simply certain function.

For the problem `int (12)/(1+9x^2)dx`

we apply the `int cf(x)dx = c int f(x)dx ` to become:

`12 int 1/(1+9x^2)dx`

Then apply the basic inverse trigonometric function formula:

`int (du)/(a^2+u^2) = 1/a arctan(u/a)+C`

By comparison with the basic formula and the given problem, we can let:

`a^2 =1`

`u^2=9x^2 or (3x)^2`

then `du = 3 dx`

To satisfy the given formula, we need to multiply the integral by `3/3` to

be able to match ` du = 3 dx` .

The integral value will note change since multiplying by 3/3 is the same as multiplying by 1. Note: `3/3= 1 ` and` 3/3 = 3*(1/3)`

Then `12 int 1/(1+9x^2)dx * 3/3`

`= 12 int 1/(1+9x^2)dx * 3 * 1/3`

`= 12 (1/3)int 1*3/(1+9x^2)dx `

`=4 int (3 dx)/(1+9x^2)`

The` int (3 dx)/(1+9x^2) ` is now similar to `int (du)/(a^2+u^2) ` where:

`du =3dx` ,` a^2 =1` and `u^2 = 9x^2 or (3x)^2`

then `a=1 ` and `u =3x` .

Plug-in `a=1` and` u = 3x` in `1/a arctan(u/a)+C` , we get:

`4* int (3 dx)/(1+9x^2) = 4* 1/1 arctan((3x)/1)+C`

`=4 arctan(3x)+C`