calculus and analysis - Problem with evaluation of the integral

I am trying to evaluate the following integral:

Integrate[(1 + k^2)/(2 k Sqrt[1 + k^2 + k^4]) - 1/(2 k ), k]

which does not work. Mathematica is just unable to find the primitive function. However, if I separate this integral into two integrals

Integrate[(1 + k^2)/(2 k Sqrt[1 + k^2 + k^4]), k] + Integrate[-1/(2 k ), k]

Mathematica yields

1/4 (ArcSinh[(1 + 2 k^2)/Sqrt[3]] - Log[2 + k^2 + 2 Sqrt[1 + k^2 + k^4]])

I don't understand why the first formula does not work. Am I missing some conditions or something?

Answer

Here are a few ways that work. These give answers that differ by a constant.

Substitutions

ClearAll[trysub];
SetAttributes[trysub, HoldFirst];
trysub[Integrate[int_, x_], sub_Equal, u_] := Module[{sol, newint},

   sol = Solve[sub, x];                                        (* should check Solve *)
   newint = int*Dt[x] /. Last[sol] /. Dt[u] -> 1 // Simplify;
   Integrate[newint, u] /. Last@Solve[sub, u] // Simplify      (* should check Solve *)
   ];

Try the expression inside Sqrt:

Assuming[u > 1 && k ∈ Reals,
 int1 = trysub[
   Integrate[(1 + k^2)/(2 k Sqrt[1 + k^2 + k^4]) - 1/(2 k), k],
   1 + k^2 + k^4 == u, u]

 ]
(*  1/4 (ArcSinh[(1 + 2 k^2)/Sqrt[3]] - Log[4 + 2 k^2 + 4 Sqrt[1 + k^2 + k^4]])  *)

Try the Sqrt (squared):

Assuming[u > 1 && k ∈ Reals,
 int2 = trysub[
   Integrate[(1 + k^2)/(2 k Sqrt[1 + k^2 + k^4]) - 1/(2 k), k],
   1 + k^2 + k^4 == u^2, u]
 ]
(*

1/4 (-2 Log[1 + Sqrt[1 + k^2 + k^4]] + 
   Log[(1 + 2 k^2 + 2 Sqrt[1 + k^2 + k^4]) (2 - 2 k^2 + 4 Sqrt[1 + k^2 + k^4])])
*)

Try u = k^2:

Assuming[u > 0 && k ∈ Reals,
 int3 = trysub[
   Integrate[(1 + k^2)/(2 k Sqrt[1 + k^2 + k^4]) - 1/(2 k), k],
   k^2 == u, u]
 ]

(*  1/4 (ArcSinh[(1 + 2 k^2)/Sqrt[3]] - Log[2 + k^2 + 2 Sqrt[1 + k^2 + k^4]])  *)

Use the real Surd instead of the complex Sqrt.

int4 = Simplify[
  Integrate[(1 + k^2)/(2 k Surd[1 + k^2 + k^4, 2]) - 1/(2 k), k],
  k ∈ Reals]
(*  1/4 (ArcSinh[(1 + 2 k^2)/Sqrt[3]] - Log[2 + k^2 + 2 Sqrt[1 + k^2 + k^4]])  *)

I believe this works because 1/Sqrt[u] is rewritten as Power[u, -1/2], where as 1/Surd[pu, 2] is unaltered. For an unknown reason (to me), Mathematica fails to come up with a key transformation of the integral in the Sqrt case, but succeeds in the Surd case.