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

int_1^(e^2)t^2ln(t)dt∫e21t2ln(t)dt

3 Answers
Øko
Jan 24, 2018

1/9(5*e^6+1)19(5⋅e6+1)

Explanation:

We want to integrate

int_1^(e^2)t^2*ln(t)dt∫e21t2⋅ln(t)dt

Use integration by parts intudv=uv-intvdu∫udv=uv−∫vdu

Let u=ln(t)u=ln(t) and dv=t^2dtdv=t2dt

Then du=1/tdtdu=1tdt and v=1/3t^3v=13t3

[1/3ln(t)*t^3]_1^(e^2)-1/3int_1^(e^2) t^2dt[13ln(t)⋅t3]e21−13∫e21t2dt

[1/3ln(t)*t^3]_1^(e^2)-[1/9t^3]_1^(e^2)[13ln(t)⋅t3]e21−[19t3]e21

1/3ln(e^2)*(e^2)^3-1/9(e^2)^3+1/9*1^313ln(e2)⋅(e2)3−19(e2)3+19⋅13

6/9*e^6-1/9e^6+1/969⋅e6−19e6+19

1/9(5*e^6+1)19(5⋅e6+1)

Narad T.
Jan 24, 2018

The answer is =224.2=224.2

Explanation:

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

Anthony R.
Jan 24, 2018

int_1^(e^2)t^2ln(t)dt=(5e^6+1)/9~~224.23822

Explanation:

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

Related questions
Impact of this question
9345 views around the world
You can reuse this answer
Creative Commons License