int 12/(1+9x^2) dx Find the indefinite integral

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