Question

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

`int_1^(e^2)t^2ln(t)dt`

Answer 1

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

Explanation:

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)`

Answer 2

The answer is `=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`

Answer 3

`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`