(* This is formula 4.271.8 *)

f[n_]:=Integrate[(Log[x])^(2*n)/(1-x^2),{x,0,1},Assumptions->Element[n,Integers]&&(n>0)]

f[n]

(* This is the value in the table *)

f1[n_]:=(2^(2*n+1)-1)*(2*n)!*Zeta[2*n+1]/2^(2*n+1);

(* We check they are the same *)

FullSimplify[f[n]/f1[n]]

(* Correct *)

(* Now we test numerical values to test numerical integration *)

g[n_]:=N[f1[n]];

g1[n_]:=NIntegrate[(Log[x])^(2*n)/(1-x^2),{x,0,1}];

Table[N[g[n]/g1[n]],{n,1,8}]

(* Everything matches *)

(* Is n natural necessary? *)

fnew[n_]:=Integrate[(Abs[Log[x]])^(2*n)/(1-x^2),{x,0,1}];

g1new[n_]:=NIntegrate[(Abs[Log[x]])^(2*n)/(1-x^2),{x,0,1}];

fnew[1/2]

f1[1/2]

g1[1/2]

(* The formula fails *)

fnew[1/4]

N[fnew[1/4]]

g1new[1/4]

(* Correct after putting Log[x] in absolute value *)