`inte^x/x(1+xInx)dx`

Explanation

`inte^x/x(1+xInx)dx`
=`inte^xdx+inte^xInxdx`
[Integrating by parts taking `e^x` as first and 1/x as second function.]
=`e^x.Inx-inte^x.Inxdx+inte^xInxdx+C`
=`e^x.Inx+C`