How do you integrate #(t^2)e^(4t)#?

1 Answer
Mar 9, 2018

Answer:

#(t^2e^(4t))/4-1/2((te^(4t))/4-e^(4t)/16)+C#

Explanation:

We have #intt^2e^(4t)dt#.
According to Integration by Parts, #intf(t)g(t)dt=f(t)intg(t)dt-intf'(x)(intg(t)dt)dt#

Here, #f(t)=t^2# and #g(t)=e^(4t)#. So we input:

#t^2inte^(4t)dt-int(t^2)'(inte^(4t)dt)dt#

A logical route to take is to find #inte^(4t)dt#

According to Integration by Substitution, #intf(g(t))g'(t)dt=intf(u)du#, where #u=g(t)#. We can write the above as:

#1/4inte^(4t)4dt#

#1/4inte^(u)du#

#1/4e^u#

#e^(4t)/4#. So we input:

#(t^2e^(4t))/4-1/2intte^(4t)dt#

Apply integration by parts for the integral:

#tinte^(4t)dt-intt'(inte^(4t)dt)dt#

Since we know that #inte^(4t)dt# is:

#(te^(4t))/4-inte^(4t)/4dt#

#(te^(4t))/4-1/4*1/4e^(4t)#

#(te^(4t))/4-e^(4t)/16#

We can input this into our eariler calculations:

#(t^2e^(4t))/4-1/2((te^(4t))/4-e^(4t)/16)#

Add the constant of integration:

#(t^2e^(4t))/4-1/2((te^(4t))/4-e^(4t)/16)+C#