Integrate the following w.r.t.x.
`(1)/(x[6(logx)^(2)+7(log x)+2])`

Explanation

`"Let I"=int (1)/(x[6(logx)^(2)+7(log x)+2])dx`
`"Put log x"=1 therefore (1)/(x)dx=dt`
`therefore I=int (1)/(6t^(2)+7t+2)dt`
`=(1)/(6)int (1)/(t^(2)+(7)/(6)+(2)/(6))dt`
`=(1)/(6)int (1)/((t^(2)+(7)/(6)t+(49)/(144))-(49)/(144)+(2)/(6))dt`
`=(1)/(6) int (1)/((t+(7)/(12))^(2)-((1)/(12))^(2))dt`
`=(1)/(6)xx(1)/(2xx(1)/(12)).log |(t+(7)/(12)-(1)/(12))/(t+(7)/(12)+(1)/(12))|+c`
`=log|(12t+6)/(12t+8)|+c`
`=log|(6logx+3)/(6log x+4)|+c`