`((x+1)(x+logx)^(2))/(x)`

Explanation

`int((x+1)(x+logx)^(2))/(x)dx" माना "x+logx=t`
`=intt^(2)dt=(1)/(3)t^(3)+c" "rArr (1+(1)/(x))dx=dt`
`=(1)/(3)(x+logx)^(3)+c" "rArr ((x+1)/(x))dx=dt`