# 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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### 1 Answer

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

`F=ln^2x+1`

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

Using integration by substitution:

`u=lnx` and `du = dx/x`

`intdx/(xln^2x+x)=int(dx/x)(1/(ln^2x+1))=int(du)/(1+u^2)`

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

`int(du)/(1+u^2)=tan^-1(u)+C`

Substituting `u=lnx` :

`intdx/(xln^2x+x)=tan^-1(lnx)+C`

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

`tan^-1(lnx)=tan^-1(lne)=tan^-1(1)=pi/4`

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.

`lim_(x->1^-)tan^-1(lnx)=tan^-1(ln1)=0`

Therefore:

`int^e_1dx/(xln^2x+x)=pi/4-0=pi/4`

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