How do you find the indefinite integral of int x(5^(-x^2))? Calculus Introduction to Integration Integrals of Exponential Functions 1 Answer Narad T. ยท Steve M 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 Answer link Related questions How do you evaluate the integral inte^(4x) dx? How do you evaluate the integral inte^(-x) dx? How do you evaluate the integral int3^(x) dx? How do you evaluate the integral int3e^(x)-5e^(2x) dx? How do you evaluate the integral int10^(-x) dx? What is the integral of e^(x^3)? What is the integral of e^(0.5x)? What is the integral of e^(2x)? What is the integral of e^(7x)? What is the integral of 2e^(2x)? See all questions in Integrals of Exponential Functions Impact of this question 1725 views around the world You can reuse this answer Creative Commons License