`int_1^3 r^3 ln(r) dr` Evaluate the integral

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`int_1^3 r^3 ln(r) dr`

To evaluate, apply integration by parts `int udv = uv - vdu` .

So let

`u = ln r`

and 

`dv = r^3 dr`

Then, differentiate u and integrate dv.

`u=1/r dr`

and

`v= int r^3 dr=r^4/4`

Plug-in them to the formula. So the integral becomes:

`int r^3 ln(r) dr`

`= ln (r)* r^4/4 - int r^4/4 * 1/rdr`

`= (r^4 ln(r))/4 - 1/4 int r^3 dr`

`= (r^4 ln(r))/4 - 1/4*r^4/4 `

`=(r^4 ln(r))/4 - r^4/16`

And, substitute the limits of the integral.

`int_1^3 r^3 ln(r) dr`

`= ((r^4ln(r))/4 - r^4/16) |_1^3`

`= ( (3^4ln(3))/4 - 3^4/16) - ((1^4ln(1))/4-1^4/16)`

`= (3^4 ln(3))/4-3^4/16 +1/16`

`= (81ln(3))/4-81/16+1/16`

`=(81ln(3))/4-80/16`

`=(81ln(3))/4-5`

 

Therefore,  `int_1^3 r^3 ln(r) dr = (81ln(3))/4-5` .

Approved by eNotes Editorial Team
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