` int_(1)^(e^6)[ (log x)/(3)] dx=e^(a)-e^(b )` then `(a-b)=? `

int_(1)^(e) log (x) dx=

int_(1)^(e )((log x)^(3))/(x)dx=

int_(e^(-1))^(e^(2))|(ln x)/(x)|dx

The value of int_(1)^(e^(6)) [(log x)/(3)] dx (where [.] denotes the greatest integer function) is (e^(a)-e^(b)) then the value of (a)/(b) is

int_(1//e)^(e) (dx)/(x(log x)^(1//3))

int_(e )^(e^(2))log x dx =

int e^(x) (e^(log x)+1) dx

int _(1) ^(e ) (dx )/(ln (x ^(x) e ^(x)))

int_(1)^(x)log_(e)[x]dx

int(e^(x log a)+e^(a log x))dx