Calculus of a Single Variable, Chapter 5, 5.5, Section 5.5, Problem 75

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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