The answer is:
#I=-e^-t(t^5+5t^4+20t^3+60t^2+120t+120)+c#
We have to use for five times (!) the integration by parts, that says:
#intf(t)g'(t)dt=f(t)g(t)-intg(t)f'(t)dt#
We can assume that #f(t)# is, step by step, the polynomial function and #g'(t)# is the exponential function
First step:
#f(t)=t^5# and #g'(t)=e^-x#
so:
#f'(t)=5t^4# and #g'(t)=-e^-x#.
#I=intt^5e^-tdt=-t^5e^-t-int5t^4(-e^-t)dt=#
#=-t^5e^-t+5intt^4e^-tdt#.
Second step:
#f(t)=t^4# and #g'(t)=e^-x#
so:
#f'(t)=4t^3# and #g'(t)=-e^-x#.
#I=-t^5e^-t+5[-t^4e^-t-int4t^3(-e^-t)dt]=#
#=-t^5e^-t-5t^4e^-t+20intt^3e^-tdt#.
Third step:
#f(t)=t^3# and #g'(t)=e^-x#
so:
#f'(t)=3t^2# and #g'(t)=-e^-x#.
#I=-t^5e^-t-5t^4e^-t+20[-t^3e^-t-int3t^2(-e^-t)dt]=#
#=-t^5e^-t-5t^4e^-t-20t^3e^-t+60intt^2e^-tdt#.
Fourth step:
#f(t)=t^2# and #g'(t)=e^-x#
so:
#f'(t)=2t# and #g'(t)=-e^-x#.
#I=-t^5e^-t-5t^4e^-t-20t^3e^-t+60[-t^2e^-t-int2t(-e^-t)dt]=#
#=-t^5e^-t-5t^4e^-t-20t^3e^-t-60t^2e^-t+120intte^-tdt#.
Fifth step:
#f(t)=t# and #g'(t)=e^-x#
so:
#f'(t)=1# and #g'(t)=-e^-x#.
#I=-t^5e^-t-5t^4e^-t-20t^3e^-t-60t^2e^-t+120[-te^-t-int1(-e^-t)dt]=#
#=-t^5e^-t-5t^4e^-t-20t^3e^-t-60t^2e^-t-120te^t+120inte^-tdt=#
#=-t^5e^-t-5t^4e^-t-20t^3e^-t-60t^2e^-t-120te^t-120e^-t+c=#
#=-e^-t(t^5+5t^4+20t^3+60t^2+120t+120)+c=#