
Dayv O. answered 05/02/22
Caring Super Enthusiastic Knowledgeable Pre-Calculus Tutor
I(x)=∫ x4e-x^3dx
let u=-x2/3,,,,,,,,,,,,,,,,,dv=-3x2e-x^3dx
du=-(2/3)xdx,,,,,,,,,,,,,,,v=e-x^3
I(x)=-(1/3)x2e-x^3+ ∫ (2/3)xe-x^3dx
that is as far as integration by parts will go
in another possibility is saying x4e-x^3 = sum of x4(1-x-3+x-6/2!-x-9/3! +/-...)
and integrate.
Dayv O.
Gamma(z)=integral, 0 to infinity of,,,, t^(z-1)e^(-t)dt. I do not see this as a gamma problem.05/02/22