Calculus: Early Transcendentals, Chapter 7, Chapter Review, Section Review, Problem 22

Let us first rewrite the integral a bit.
int te^sqrt t dt=int(sqrt t ^3 e^sqrt t)/(sqrt t)dt=
Now we use substitution x=sqrt t,\ dx=dt/(2sqrt t)
2int x^3e^x dx=
Now we use partial integration |[u=x^3,dv=e^xdx],[du=3x^2dx,v=e^x]|
2(x^3e^x-3int x^2e^x dx)=
Again we use partial integration |[u=x^2,dv=e^xdx],[du=2xdx,v=e^x]|
2x^3e^x-6(x^2e^x-2int xe^xdx)=
Partial integration once more |[u=x,dv=e^xdx],[du=dx,v=e^x]|
2x^3e^x-6x^2e^x+12(xe^x-int e^xdx)=
2x^3e^x-6x^2e^x+12xe^x-12e^x+C=
Now we return the substitution.
2sqrt t ^3e^sqrt t-6te^sqrt t+12sqrt t e^sqrt t-12e^sqrt t+C=
e^sqrt t(2sqrt t^3-6t+12sqrt t-12)+C