Evaluate: `intxlogx\ dx`

Explanation

We have given `intxlogx\ dx`

`I=int xlogx\ dx`

By Integrating by parts,

`I=(x^2)/2logx−int(x^2)/2.(1/x)dx`

`I=(x^2)/2logx-1/2int xdx`

`I=(x^2)/2logx-1/2(x^2)/2+c`

`I=(x^2)/2logx-1/4x^2+c`