$Version
(*"12.0.0 for Microsoft Windows (64-bit) (April 6, 2019)"*)

? = -(? + 1/?); 
q = x^2 + (1 - x)^2 - x (1 - x)*?; 
p0 = ?/q; 
r = (1 + ?^2)/(1 - ?^2);
func = -1/r*p0*Log[x]*Log[1 - x] // FullSimplify
sol=Integrate[func, {x, 0, 1}, Assumptions -> ? > 0]

(*for ? > 1 *)

(* (24*? - 12*EulerGamma*? - 2*Pi^2*? - 33*?^2 + 12*EulerGamma*?^2 - Pi^2*?^2 + 
  2*EulerGamma*Pi^2*?^2 + 9*?^3 - 15*?^2*Log[?] + 2*Pi^2*?^2*Log[?] - 
  18*?^2*Log[-1 + ?]*Log[?] + 12*EulerGamma*?^2*Log[-1 + ?]*Log[?] + 
  9*?^2*Log[?]^2 - 6*EulerGamma*?^2*Log[?]^2 + 12*EulerGamma*?^2*PolyLog[2, 1 - ?] + 
  18*?^2*PolyLog[2, ?^(-1)] - 12*EulerGamma*?^2*PolyLog[2, ?^(-1)] - 
  12*EulerGamma*?^2*PolyLog[2, (-1 + ?)/?] - 3*(-1 + ?)^2*?^2*
   Derivative[0, 0, 1, 0][Hypergeometric2F1][1, 2, 3, 1 - ?] + 
  12*(-1 + EulerGamma)*(-1 + ?)^2*Derivative[0, 0, 1, 0][Hypergeometric2F1Regularized][1, 
    2, 3, (-1 + ?)/?] + 6*?^2*Derivative[0, 0, 2, 0][Hypergeometric2F1][1, 2, 3, 
    1 - ?] - 12*?^3*Derivative[0, 0, 2, 0][Hypergeometric2F1][1, 2, 3, 1 - ?] + 
  6*?^4*Derivative[0, 0, 2, 0][Hypergeometric2F1][1, 2, 3, 1 - ?] - 
  12*Derivative[0, 0, 2, 0][Hypergeometric2F1Regularized][1, 2, 3, (-1 + ?)/?] + 
  24*?*Derivative[0, 0, 2, 0][Hypergeometric2F1Regularized][1, 2, 3, (-1 + ?)/?] - 
  12*?^2*Derivative[0, 0, 2, 0][Hypergeometric2F1Regularized][1, 2, 3, (-1 + ?)/?] + 
  6*?^2*Derivative[0, 1, 1, 0][Hypergeometric2F1][1, 2, 3, 1 - ?] - 
  12*?^3*Derivative[0, 1, 1, 0][Hypergeometric2F1][1, 2, 3, 1 - ?] + 
  6*?^4*Derivative[0, 1, 1, 0][Hypergeometric2F1][1, 2, 3, 1 - ?] - 
  12*Derivative[0, 1, 1, 0][Hypergeometric2F1Regularized][1, 2, 3, (-1 + ?)/?] + 
  24*?*Derivative[0, 1, 1, 0][Hypergeometric2F1Regularized][1, 2, 3, (-1 + ?)/?] - 
  12*?^2*Derivative[0, 1, 1, 0][Hypergeometric2F1Regularized][1, 2, 3, (-1 + ?)/?])/
 (12*?^2)*)

If we assume ?=2 then:

 sol[[1]]/. ? -> 2 // N
 (*  -0.486141 *)
 NIntegrate[func/.? -> 2, {x, 0, 1}]
 (* -0.486141 *)

Yours formula:

   4*PolyLog[3, 1 - ?] + 2*PolyLog[3, ?] - 
        4*PolyLog[2, 1 - ?]*Log[1 - ?] - 
        4*PolyLog[2, ?]*Log[1 - ?] - 
        4*Log[?]*Log[1 - ?]^2 + 
        Log[?]^2*Log[1 - ?] + 
        2*?^2*Log[1 - ?]/3 - 2 Zeta[3] /. ? -> 2 // N
  (*-0.486141 + 7.10543*10^-15 I*)

Simplify more:

   Log[1 - ?] Log[?]^2 + 4 PolyLog[3, 1 - ?] + 2 PolyLog[3, ?] - 2 Zeta[3] /. ? -> 2 // N
   (*-0.486141 + 2.66454*10^-15 I*)

You STILL did not follow the instructions about formatting Mathematica code!

The answer I am expecting is the following

I want an answer in this form

S2 = 4*PolyLog[3, 1 - \[Alpha]] + 2*PolyLog[3, \[Alpha]] - 
  4*PolyLog[2, 1 - \[Alpha]]*Log[1 - \[Alpha]] - 
  4*PolyLog[2, \[Alpha]]*Log[1 - \[Alpha]] - 
  4*Log[\[Alpha]]*(Log^2)[1 - \[Alpha]] + (Log^2)[\[Alpha]]*
   Log[1 - \[Alpha]] + 2*\[Pi]^2*Log[1 - \[Alpha]]/3 - 2 Zeta[3]

I am also not sure whether these two are the same or different but it's highly likely that these two are not the same.

\[Gamma] = -(\[Alpha] + 1/\[Alpha]);
q = x^2 + (1 - x)^2 - x (1 - x)*\[Gamma] ;
p0 = \[Gamma]/q ;
r = (1 + \[Alpha]^2)/(1 - \[Alpha]^2) ;   

Find the following

I6 = -1/r \!\(
\*SubsuperscriptBox[\(\[Integral]\), \(0\), \(1\)]\(p0*Log[x]*
    Log[1 - x] \(\(\[DifferentialD]\)\(x\)\(\ \)\)\)\)