Question

Integration of $x\log x$.

Answer 1

Step 1: Separate the expression

The given expression is $x\log x$.

Let its integration be $I$.

$\therefore$$I=\int x\log xdx$

Let $f\left(x\right)=\log x$ and $g\left(x\right)=x$.

Step 2: Apply Integration By parts

Using integration by parts where $\int u.vdx=u\int vdx-\int u\text{'}\left(\int vdx\right)dx$ we get,

$I=\log x.\frac{x^{1+1}}{1+1}-\int \frac{1}{x}.\frac{x^{1+1}}{1+1}.dx$

$=\log x\frac{x^2}{2}-\int \frac{1}{x}.\frac{x^2}{2}dx$

$=\log x.\frac{x^2}{2}-\frac{1}{2}\int xdx$

$=\log x.\frac{x^2}{2}-\frac{1}{2}.\frac{x^{1+1}}{1+1}$

$=\log x.\frac{x^2}{2}-\frac{x^2}{4}+c$ where $c$ is the constant of integration.

Hence, when $x\log x$ is integrated we get $\log x.\frac{x^2}{2}-\frac{x^2}{4}+c$.