Question

Integrate $\int x{\left(\log x\right)}^2\mathrm{d}x$

Answer 1

Using Integration by parts

Given, $\int x{\left(\log x\right)}^2\mathrm{d}x$

Let, $I=\int x{\left(\log x\right)}^2\mathrm{d}x$

From integration by parts, we know that

$\int f\left(x\right)·g\left(x\right)=f\left(x\right)\int g\left(x\right)\mathrm{d}x-\int \left(\frac{df\left(x\right)}{dx}\int g\left(x\right)\mathrm{d}x\right)\mathrm{d}x$

Here, take $f\left(x\right)={\left(\log x\right)}^2$ and $g\left(x\right)=x$

$\begin{array}{cc}\Rightarrow & I={\left(\log x\right)}^2\int x\mathrm{d}x-\int \left(\frac{d{\left(\log x\right)}^2}{dx}\int x\mathrm{d}x\right)dx\\ \Rightarrow & I={\left(\log x\right)}^2·\frac{x}{}\end{array}$

Apply integration by parts on $\int x\log \left(x\right)dx$ taking $f\left(x\right)=\log \left(x\right)$ and $g\left(x\right)=x$. Thus,

$\begin{array}{cc}\Rightarrow & I=\frac{{\left[x\log \left(x\right)\right]}^2}{2}-\left[\log \left(x\right)\int x\mathrm{d}x-\int \left(\frac{d\log \left(x\right)}{dx}\int xdx\right)dx\right]\\ \Rightarrow & I=\frac{{\left[x\log \left(x\right)\right]}^2}{2}-\left[\log \left(x\right)·\frac{x^2}{2}-\int \frac{1}{x}·\frac{x^2}{2}dx\right]\\ \Rightarrow & I=\frac{{\left[x\log \left(x\right)\right]}^2}{2}-\left[\log \left(x\right)·\frac{x^2}{2}-\frac{x^2}{4}\right]+C\\ \Rightarrow & I=\frac{{\left[x\log \left(x\right)\right]}^2}{2}-\frac{x^2\log \left(x\right)}{2}+\frac{x^2}{4}+C\end{array}$

Therefore, $\int x{\left(\log x\right)}^2\mathrm{d}x=\frac{{\left[x\log \left(x\right)\right]}^2}{2}-\frac{x^2\log \left(x\right)}{2}+\frac{x^2}{4}+C$