Given F=ax+b if x<1 F=(ln x)^2+1, x>1, calculate integral from 1 to e 1/(x*F)?



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Posted on (Answer #1)

On the interval (1,e] the function is defined as:


Therefore `1/(xF)=1/(xln^2x+x)`

Using integration by substitution:

`u=lnx`  and `du = dx/x`


The derivative of `tan^-1(x)=1/(1+x^2)` ; therefore:


Substituting `u=lnx` :


For the closed end of the interval, x=e:


For the open end of the interval` x-gt1^-` we must take the limit as x approaches 1 from the left.  We can simply substitute x=1 into the function because the function is defined and continuous at x=1.





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