How do we integrate [2(x)^3*e^(-x^2) dx] ?
By observation, we may note that,
d(-x^2.e^(-x^2))/ dx = 2x^3.e^(-x^2) - 2x.e^(-x^2)
Integrating the above,
-x^2.e^(-x^2) = int{2x^3.e^(-x^2)} - 2int{x.e^(-x^2)}
-x^2.e^(-x^2) = int{2x^3.e^(-x^2)} - 2(-1/2.e^(-x^2)
int{2x^3.e^(-x^2)} = e^(-x^2)(-1 - x^2)
Just noticed, you asked for a substitution.. Try u = x^2, with du = 2x.dx