# What is int xln(x)^2?

Mar 20, 2016

Supposing you mean $\ln {\left(x\right)}^{2} = {\left(\ln x\right)}^{2}$

You have to integrate by parts twice. Answer is:

${x}^{2} / 2 \left(\ln {\left(x\right)}^{2} - \ln x + \frac{1}{2}\right) + c$

Supposing you mean $\ln {\left(x\right)}^{2} = \ln \left({x}^{2}\right)$

You have to integrate by parts once. Answer is:

${x}^{2} \left(\ln x - \frac{1}{2}\right) + c$

#### Explanation:

Supposing you mean $\ln {\left(x\right)}^{2} = {\left(\ln x\right)}^{2}$

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

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

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

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

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

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

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

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

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

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

$= {x}^{2} / 2 \ln {\left(x\right)}^{2} - \left({x}^{2} / 2 \ln x - {x}^{2} / 4\right) + c =$

$= {x}^{2} / 2 \ln {\left(x\right)}^{2} - {x}^{2} / 2 \ln x + {x}^{2} / 4 + c =$

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

Supposing you mean $\ln {\left(x\right)}^{2} = \ln \left({x}^{2}\right)$

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

$2 \int x \ln x \mathrm{dx} =$

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

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

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

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

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

$= \cancel{2} \cdot {x}^{2} / \left(\cancel{2}\right) \left(\ln x - \frac{1}{2}\right) + c =$

$= {x}^{2} \left(\ln x - \frac{1}{2}\right) + c$