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

2 Answers
Apr 3, 2018

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

Apr 3, 2018

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