Evaluate the following :
`int_(1)^(e)(logx)dx`

Explanation

`int_(1)^(2)(logx)^(2)dx=int_(1)^(e)(logx)^(2)(1)dx`
`=[(logx)^(2)int1dx]_(1)^(e)-int_(1)^(e)[(d)/(dx)(logx)^(2).int1 dx]dx`
`=[(logx)^(2).x]_(1)^(e)-int_(1)^(e)(2log x)/(x).xdx`
`=[x(logx)^(2)]_(1)^(e)-2int_(1)^(e)(logx)(1)dx`
`=2(loge)^(2)-(log1)^(2)-2{[(logx)int 1 dx]_(1)^(e)-int [(d)/(dx)(logx)int 1 dx]dx}`
`=e-2[x logx]_(1)^(e)+2 int_(1)^(e)(1)/(x).xdx`
`=e-2(elog e =log 1)+2 int_(1)^(e)1 dx`
`=e-2 (e)+2[x]_(1)^(e)=-e+2(e-1)" ... "[because log e= 1 and log 1=0]`
`=e-2.`