Thursday, April 23, 2009

`int p^5 ln p dp` Evaluate the integral

`int p^5 ln(p) dp`


To evaluate, apply integration by parts intu dv = uv  -int vdu.


So let


`u= ln (p)`


and


`dv = p^5 dp`


Then, differentiate u and integrate dv.


`du=1/p dp`


and


`v = int p^5dp = p^6/6`


And, plug-in them to the formula. So the integral becomes:


`int p^5 ln(p) dp`


`= ln(p) *p^6/6 - int p^6/6*1/pdp`


`= (p^6ln(p))/6 - int p^5/6 dp`


`= (p^6 ln(p))/6 - 1/6 int p^5 dp`


`=(p^6 ln(p))/6 - 1/6*p^6/6 + C`


`=(p^6ln(p))/6 -p^6/36+C`



Therefore,  `int p^5 ln(p) dp = (p^6 ln(p))/6 - p^6/36 + C` .

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