use the substitution u=2+ln(x) to show that int(e,1(ln(x)/x(2+ln(x)^2))dx)=p+ln(q) , where p and q are rational numbers.

So u=2+lnx, therefore du/dx=1/x , we can work out the new upper and new lower limit by substitute in e and 1 into 2+lnx , and we get 2+ln(e)=3 , 2+ln(1)=2Rearrange the differential we get dx=xdu , substitute u and dx and the equation becomes xln(x)/x(u)^2 , top and bottom xs cancel and with the top being ln(x) and u=2+ln(x) , we can also substitute the top with u-2 .we can now intergrate this as (u-2)/(u^2)with the limits being 3 and 2.we get a result of lnu+(2/u) , substitute in 3 and 2 and the final result is -1/3+ln(3/2).