`int x(5^(-x^2))dx` Find the indefinite integral

Indefinite integral are written in the form of` int f(x) dx = F(x) +C`

 where: `f(x)` as the integrand

         ` F(x) ` as the anti-derivative function 

          `C`  as the arbitrary constant known as constant of integration

For the given problem `int x(5^(-x^2)) dx` has an integrand in a form of exponential function.

 To evaluate this, we may let:

`u = -x^2` then  ` du= -2x dx or (-1/2)(du)= x dx` .

Applying u-substitution, we get:

`int x(5^(-x^2)) dx =int (5^(-x^2)) * x dx`

                            `=int (5^(u)) *(-1/2du)`

                           ` =-1/2int (5^(u) du)`

The integral part resembles the basic integration formula:

`int a^u du = a^u/(ln(a))+C`

Applying it to the problem: 

`-1/2int (5^(u) du)=-1/2 * 5^(u)/ln(5) +C`

Pug-in `u =-x^2` , we get the definite integral:

`-1/2 * 5^(-x^2)/ln(5) +C`

 or

`- 5^(-x^2)/(2ln(5)) +C`