You need to find the indefinite integral of the function such that:

`int f(x)dx = int (4-x)/(2+sqrt x) dx`

Notice that `4-x` is a difference of squares, hence, you may convert the difference into the following special product such that:

`4 - x = 2^2 - (sqrt x)^2 = (2 - sqrt x)(2 + sqrt x)`

`int f(x)dx = int ((2 - sqrt x)(2 + sqrt x))/(2+sqrt x) dx`

Reducing the factor `(2 + sqrt x)` yields:

`int f(x)dx = int (2 - sqrt x) dx`

Using the property of linearityt of integrals yields:

`int f(x)dx = int2 dx - int sqrt x dx`

`int f(x)dx = 2x - (x^(1/2+1))/(1/2+1) + c`

`int f(x)dx = 2x - (x^(3/2))/(3/2) + c`

`int f(x)dx = 2x - (2/3)x*sqrt x + c`

**Hence, evaluating the antiderivative of the given function yields `int (4-x)/(2+sqrt x) dx = 2x - (2/3)x*sqrt x + c` .**