Question

Integral `int_(1/e)^e ln(x)^2018/(x+1) dx` ?

Answer 1

`F(e)-F(1/e)= 2/2019`

Explanation:

`F(x)= int_(1/e)^eln^2018x/(x+1)dx=ln^2019x/2019`

Evaluate `F(e)-F(1/e)`

`F(e)=ln^2019e/2019=1^2019/2019=1/2019`

`F(1/e)=ln^2019(1/e)/2019=(-1)^2019/2019=-1/2019`

Finally: `F(e)-F(1/e)=1/2019-(-1/2019)=2/2019`

Answer 2

`int_(1/e)^(e) (Lnx)^2018/(x+1)*dx=1/2019`

Explanation:

`I=int_(1/e)^(e) (Lnx)^2018/(x+1)*dx`

After using `x=1/y` and `dx=-dy/y^2` transform,

`I=int_(e)^(1/e) (Ln(1/y))^2018/(1/y+1)*(-dy/y^2)`

=`int_(e)^(1/e) (-Lny)^2018/((y+1)/y)*(-dy/y^2)`

=`int_(e)^(1/e) (Lny)^2018/(y^2+y)*(-dy)`

=`int_(1/e)^(e) (Lny)^2018/(y^2+y)*dy`

=`int_(1/e)^(e) (Lnx)^2018/(x^2+x)*dx`

After summing 2 integrals,

`2I=int_(1/e)^(e) (Lnx)^2018/(x+1)*dx`+`int_(1/e)^(e) (Lnx)^2018/(x^2+x)*dx`

=`int_(1/e)^(e) (x*(Lnx)^2018)/(x^2+x)*dx`+`int_(1/e)^(e) (Lnx)^2018/(x^2+x)*dx`

=`int_(1/e)^(e) ((x+1)*(Lnx)^2018*dx)/(x*(x+1))`

=`int_(1/e)^(e) ((Lnx)^2018*dx)/x`

=`[(Lnx)^2019/2019]_(1/e)^(e)`

=`2/2019`

Thus,

`I=int_(1/e)^(e) (Lnx)^2018/(x+1)*dx=1/2019`