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`
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