# How do you evaluate the integral int x(lnx)^2dx?

May 23, 2017

Recursively use integration by parts

#### Explanation:

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

Let $u = {\left(\ln \left(x\right)\right)}^{2} \mathmr{and} \mathrm{dv} = x \mathrm{dx}$, then $\mathrm{du} = \frac{2 \ln \left(x\right)}{x} \mathrm{dx} \mathmr{and} v = {x}^{2} / 2$

$\int x {\left(\ln \left(x\right)\right)}^{2} \mathrm{dx} = {\left(x \ln \left(x\right)\right)}^{2} / 2 - \int x \ln \left(x\right) \mathrm{dx}$

Integration by parts for the second integral:

Let $u = \ln \left(x\right) \mathmr{and} \mathrm{dv} = x \mathrm{dx}$, then $\mathrm{du} = \frac{1}{x} \mathrm{dx} \mathmr{and} v = {x}^{2} / 2$

$\int x {\left(\ln \left(x\right)\right)}^{2} \mathrm{dx} = {\left(x \ln \left(x\right)\right)}^{2} / 2 - {x}^{2} / 2 \ln \left(x\right) + \frac{1}{2} \int x \mathrm{dx}$

$\int x {\left(\ln \left(x\right)\right)}^{2} \mathrm{dx} = {\left(x \ln \left(x\right)\right)}^{2} / 2 - {x}^{2} / 2 \ln \left(x\right) + \frac{1}{4} {x}^{2} + C$