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`
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