Calculus: Early Transcendentals, Chapter 7, 7.1, Section 7.1, Problem 19

intz^3e^zdz
If f(x) and g(x) are differentiable functions, then
intf(x)g'(x)=f(x)g(x)-intf'(x)g(x)dx
If we write f(x)=u and g'(x)=v, then
intuvdx=uintvdx-int(u'intvdx)dx
Using the above integration by parts,
Let u=z^3 , u'=3z^2
and let v=e^z, v'=e^z
intz^3e^z=z^3inte^zdz-int(3z^2inte^zdz)dz
=z^3e^z-int(3z^2e^z)dz
=z^3e^z-3intz^2e^zdz
again applying integration by parts,
=z^3e^z-3(z^2inte^zdz-int(d/dz(z^2)inte^zdz)dz
=z^3e^z-3(z^2e^z-int(2ze^z)dz
=z^3e^z-3z^2e^z+6intze^zdz
again applying integration by parts,
=z^3e^z-3z^2e^z+6(zinte^zdz-int(d/dz(z)inte^zdz)dz)
=z^3e^z-3z^2e^z+6(ze^z-int(1*e^z)dz)
=z^3e^z-3z^2e^z+6(ze^z-e^z)
adding constant to the solution,
=z^3e^z-3z^2e^z+6ze^z-6e^z+C