How do you integrate (t^2)ln(t) using integration by parts?

We want to integrate

#int_1^(e^2)t^2*ln(t)dt#

Use integration by parts #intudv=uv-intvdu#

Let #u=ln(t)# and #dv=t^2dt#

Then #du=1/tdt# and #v=1/3t^3#

#[1/3ln(t)*t^3]_1^(e^2)-1/3int_1^(e^2) t^2dt#

#[1/3ln(t)*t^3]_1^(e^2)-[1/9t^3]_1^(e^2)#

#1/3ln(e^2)*(e^2)^3-1/9(e^2)^3+1/9*1^3#

#6/9*e^6-1/9e^6+1/9#

#1/9(5*e^6+1)#

Integration by parts is :

#intu'v=uv-intuv'#

Here,

#intt^2lntdt=?#

Let #v=lnt#, #=>#, #v'=1/t#

#u'=t^2#, #=>#, #u=t^3/3#

Therefore,

#intt^2lntdt=t^3/3lnt-int1/t*t^3/3dt#

#=t^3/3lnt-1/3intt^2dt#

#=t^3/3lnt-1/9t^3+C#

Then, the definite integral is

#int_1^(e^2)t^2lntdt=[t^3/3lnt-1/9t^3]_1^ (e^(2))#

#=(e^6/3ln(e^2)-e^6/9)-(1/3ln(1)-1/9)#

#=5/9e^6+1/9#

#=224.2#

First solve for the indefinite integral

Using integration by parts

#intudv=uv-intvdu#

Let

#u=lnt=>du=1/tdt#

#dv=t^2=>v=t^3/3larr# (See Proof Below)

Use #color(blue)(intt^adt=(t^(a+1))/(a+1)#

So #color(red)(intt^2dt=t^(2+1)/(2+1)=t^3/3#

Substituting in for the formula:

#I=(ln(t))*(t^3/3)-int(t^3/3)(1/tdt)#

Simplify (Take out the constant #1/3#):

#I=(t^3ln(t))/3-1/3intt^3/tdt#

#I=(t^3ln(t))/3-1/3intt^2dt#

Now we solve for #intt^2dt# but actually solved this earlier at the start of the problem:

So...

#I=(t^3ln(t))/3-1/3[t^3/3]#

Simplify

#I=[(t^3ln(t))/3-t^3/9]_1^(e^2)#

Now we evaluate the upper/lower limits

#I=[(color(red)((e^2))^3ln(color(red)(e^2)))/3-(color(red)(e^2))^3/9]-[(color(red)1^3ln(color(red)1))/3-color(red)1^3/9]#

#I=(5e^6)/9-(-1/9)#

#I=(5e^6)/9+1/9#

#I=(5e^6+1)/9~~224.23822#