Hello,
I think you want integrate #x^2 e^{-5x}#, not #x^2 e^{-5}x#...
Use two integrations by parts :
#\int_0^x u(t) v'(t) dt = [u(t)v(t)]_0^x - \int_0^x u'(t) v(t) dt#.
You can prove that if you write #(uv)' = u'v+uv'# and if you integrate between #0# and #x#.
Let #F# be the function define by #F(x) = \int_0^x t^2 e^{-5t} dt#.
Take #u(t) = t^2# and #v(t) =-\frac{1}{5}e^{-5t}#. So you have
#u'(t) = 2t# and #v'(t) = e^{-5t}#
Apply the integration by parts formula :
#F(x) = [-\frac{1}{5} t^2 e^{-5t} ]_0^x +\frac{2}{5} \int_0^x te^{-5t}dt#
Use integration by parts again to calculate #\int_0^x te^{-5t}dt#. Take
#u(t) =t# and #v(t) = -\frac{1}{5}e^{-5t}#. So you have
#u'(t) = 1# and #v'(t) = e^{-5t}#, and
#\int_0^x te^{-5t}dt = [-\frac{1}{5}t e^{-5t}]_0^x + \frac{1}{5}\int_0^x e^{-5t}dt = [-\frac{1}{5}t]_0^x + \frac{1}{5}[-\frac{1}{5}e^{-5t}]_0^x#
You can simplify :
#\int_0^x te^{-5t}dt =-\frac{1}{5}xe^{-5x} - \frac{1}{25}(e^{-5x}-1)#
Finally, plug that in the above expression :
#F(x) = -\frac{1}{5}x^2e^{-5x} - \frac{2}{5}(\frac{1}{5}xe^{-5x} + \frac{1}{25}(e^{-5x}-1))#
Primitives (or antiderivatives) of #x^2e^{-5x}# are
#-(\frac{1}{5}x^2 + \frac{2}{25}x + 2/125)e^{-5x} + c#
where #c \in RR#.
