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

Explanation

`int(x^(2)+1)log x dx`
`=log x. int (x^(2)+1)dx`
`" "-int{(d)/(dx)logx int(x^(2)+1)dx}dx`
`=logx.((x^(3))/(3)+x)-int(1)/(x)((x^(3))/(3)+x)dx+C`
`=(x+(x^(3))/(3))logx-int((x^(2))/(3)+1)dx+C`
`=(x+(x^(3))/(3))logx-(x+(x^(3))/(9))+C`