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.