How do you find the indefinite integral of int x(5^(-x^2))?

1 Answer
Dec 12, 2016

The answer is =-5^(-x^2)/(2ln5)+C

Explanation:

We do the integral by substitution

Let u=x^2, then, du=2xdx

intx(5^(-x^2))dx=1/2int(5^(-u))du

Let v=5^(-u)

lnv=ln5^(-u)=-u ln5

v=e^(-u ln5)

Therefore,

intx(5^(-x^2))dx=1/2int(5^(-u))du=1/2inte^(-u ln5)du

=-e^(-u ln5)/(2ln5) =-5^(-u)/(2ln5)=-5^(-x^2)/(2ln5)+C