How do you integrate $\int u^{5} ln u$ $int u^5lnu$ by integration by parts method?

Question

1607 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-05T13:09:20

$= \frac{u^{6}}{6} (ln u - \frac{1}{6}) + C$ $= u^6/6 ( lnu - 1/6) + C$

tactically, the $ln u$ $ln u$ is the bit you want to differentiate cos that just turns it into another $u$ $u$ term. So we approach the IBP as

$int u^5lnu \ du$

$= int (u^6/6)' \ lnu \ du$

and so by IBP we get to differentiate it in the next line

$= u^6/6 \ lnu - int u^6/6 \ ( lnu)' \ du$

$= u^6/6 \ lnu - int u^6/6 \ ( 1/u) \ du$ !!

$= u^6/6 \ lnu - int u^5/6 \ du$

$= u^6/6 \ lnu - u^6/36 + C$

$= u^6/6 ( lnu - 1/6) + C$

How do you integrate int u^5lnu∫u5lnuint u^5lnu by integration by parts method?