Yes, but, with

Clear[f1, f11, fL]
f1[x_] := -(ArcTan[(1 + 2 x)/Sqrt[3]]/Sqrt[3]) - 1/3 Log[-1 + x] + 1/6 Log[1 + x + x^2]
f11[x_] := -(ArcTan[(1 + 2 x)/Sqrt[3]]/Sqrt[3]) - 1/3 Log[1 - x] + 1/6 Log[1 + x + x^2]
fL[x_] := -(ArcTan[(1 + 2 x)/Sqrt[3]]/Sqrt[3]) - 1/3 Log[Abs[1 - x]] + 1/6 Log[1 + x + x^2]

the integral table solution fL pretends to deliver the area under the real valued function (in the sense of Riemann) for $x\in[0,2]$

In[38]:= fL[2] - fL[0] // N
Out[38]= -0.0877506

In[40]:= f11[2] - f11[0] // N
Out[40]= -0.0877506 - 1.0472 I

In[41]:= f1[2] - f1[0] // N
Out[41]= -0.0877506 + 1.0472 I

whereas the correct anti-derivatives do not. Because there is a singularity at x = 1 let's show that fL does the job right. One rotates the function for $x\in[1,2]$ by 180° and considers the difference with the original function for $x\in[0,1]$.

Clear[f, fleft, fright]
f[x_] := x/(1 - x^3)
fleft[x_ /; 0 <= x <= 1] := f[x]
fright[x_ /; 0 <= x <= 1] :=(* Last[RotationTransform[\[Pi],{1,0}][{t,t/(1-t^3)}]]//.t\\[Rule]2-x *)
                            -((2 - x)/(1 - (2 - x)^3))

This way one has

In[137]:= Integrate[x/(1 - x^3) + (2 - x)/(1 - (2 - x)^3), {x, 0, 1}]
Out[137]= 1/18 (Sqrt[3] \[Pi] - 6 Sqrt[3] ArcTan[5/Sqrt[3]] + Log[343])

with other words

In[138]:= Integrate[x/(1 - x^3) + (2 - x)/(1 - (2 - x)^3), {x, 0, 1.}]
Out[138]= -0.0877506

as fL said.

Because you intended the argument being of no influence to this question,

Clear[f1, f11, fL]
    f1[x_] := -(ArcTan[(1 + 2 x)/Sqrt[3]]/Sqrt[3]) - 1/3 Log[-1 + x] + 
      1/6 Log[1 + x + x^2]
    f11[x_] := -(ArcTan[(1 + 2 x)/Sqrt[3]]/Sqrt[3]) - 1/3 Log[1 - x] + 
      1/6 Log[1 + x + x^2]
    fL[x_] := -(ArcTan[(1 + 2 x)/Sqrt[3]]/Sqrt[3]) - 
      1/3 Log[Abs[1 - x]] + 1/6 Log[1 + x + x^2]

you could have said

In[31]:= Simplify[PowerExpand[f1[z + 1] - f11[z + 1], Assumptions -> {}]]
Out[31]= \[Piecewise]   -((I \[Pi])/3)    Arg[z]>0
                          (I \[Pi])/3     True