# How do you integrate int x^2 ln ^2x dx  using integration by parts?

Mar 8, 2017

$\int {x}^{2} {\ln}^{2} x \mathrm{dx} = {x}^{3} / 27 \left(9 {\ln}^{2} x - 6 \ln x + 2\right) + C$

#### Explanation:

We can integrate by parts using the logarithm as integral part, so that in the resulting integral we have a rational function:

$\int {x}^{2} {\ln}^{2} x \mathrm{dx} = \int {\ln}^{2} x d \left({x}^{3} / 3\right)$

$\int {x}^{2} {\ln}^{2} x \mathrm{dx} = \frac{{x}^{3} {\ln}^{2} x}{3} - \frac{1}{3} \int {x}^{3} d \left({\ln}^{2} x\right)$

$\int {x}^{2} {\ln}^{2} x \mathrm{dx} = \frac{{x}^{3} {\ln}^{2} x}{3} - \frac{2}{3} \int {x}^{3} \ln \frac{x}{x} \mathrm{dx}$

$\int {x}^{2} {\ln}^{2} x \mathrm{dx} = \frac{{x}^{3} {\ln}^{2} x}{3} - \frac{2}{3} \int {x}^{2} \ln x \mathrm{dx}$

Solve the resulting integral by parts again:

$\int {x}^{2} \ln x \mathrm{dx} = \int \ln x d \left({x}^{3} / 3\right)$

$\int {x}^{2} \ln x \mathrm{dx} = \frac{{x}^{3} \ln x}{3} - \frac{1}{3} \int {x}^{3} d \left(\ln x\right)$

$\int {x}^{2} \ln x \mathrm{dx} = \frac{{x}^{3} \ln x}{3} - \frac{1}{3} \int {x}^{3} \frac{\mathrm{dx}}{x}$

$\int {x}^{2} \ln x \mathrm{dx} = \frac{{x}^{3} \ln x}{3} - \frac{1}{3} \int {x}^{2} \mathrm{dx}$

$\int {x}^{2} \ln x \mathrm{dx} = \frac{{x}^{3} \ln x}{3} - \frac{1}{9} {x}^{3} + C$

Substituting in the first expression:

$\int {x}^{2} {\ln}^{2} x \mathrm{dx} = \frac{{x}^{3} {\ln}^{2} x}{3} - \frac{2}{3} \left(\frac{{x}^{3} \ln x}{3} - \frac{1}{9} {x}^{3}\right) + C$

and simplifying:

$\int {x}^{2} {\ln}^{2} x \mathrm{dx} = \frac{{x}^{3} {\ln}^{2} x}{3} - \frac{2}{9} \left({x}^{3} \ln x\right) + \frac{2}{27} {x}^{3} + C$

$\int {x}^{2} {\ln}^{2} x \mathrm{dx} = {x}^{3} / 27 \left(9 {\ln}^{2} x - 6 \ln x + 2\right) + C$