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

Sep 5, 2016

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

#### Explanation:

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

$\int {u}^{5} \ln u \setminus \mathrm{du}$

$= \int \left({u}^{6} / 6\right) ' \setminus \ln u \setminus \mathrm{du}$

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

$= {u}^{6} / 6 \setminus \ln u - \int {u}^{6} / 6 \setminus \left(\ln u\right) ' \setminus \mathrm{du}$

$= {u}^{6} / 6 \setminus \ln u - \int {u}^{6} / 6 \setminus \left(\frac{1}{u}\right) \setminus \mathrm{du}$ !!

$= {u}^{6} / 6 \setminus \ln u - \int {u}^{5} / 6 \setminus \mathrm{du}$

$= {u}^{6} / 6 \setminus \ln u - {u}^{6} / 36 + C$

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