# How do you find the integral of f(x)=x^5lnx using integration by parts?

Oct 31, 2015

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

#### Explanation:

$\ln x = u \implies \frac{\mathrm{dx}}{x} = \mathrm{du}$

${x}^{5} \mathrm{dx} = \mathrm{dv} \implies v = {x}^{6} / 6$

$\int u \mathrm{dv} = u v - \int v \mathrm{du}$

$I = \int {x}^{5} \ln x \mathrm{dx} = {x}^{6} / 6 \ln x - \int {x}^{6} / 6 \frac{\mathrm{dx}}{x} = {x}^{6} / 6 \ln x - \frac{1}{6} \int {x}^{5} \mathrm{dx}$

$I = {x}^{6} / 6 \ln x - \frac{1}{6} {x}^{6} / 6 + C$

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