Let's look at more general form of limits:
In[]:=
ClearAll[lftLim,rgtLim];​​lftLim[h_,n_]:=-h^n;​​rgtLim[h_,n_]:=h^n;
Integrated function
ClearAll[ourFunc];ourFunc[x_,h_]:=Abs
1+x^2
-
1+x
-
2
x
4*h
+h;
​Firstly for h  0 we see that ourFunc 
-
2
x
4*h
and integral over { -h^n, h^n} is roughly
h^(3n-1)
6
. So if n ≥ 1/3 (in example n = 2) minimum is exactly h=0 with minimum value 0. So it's interesting find a minimum for different n.​
In[]:=
ClearAll[goalNum];​​goalNum[h_,n_]:=NIntegrate[ourFunc[x,h],{x,lftLim[h,n],rgtLim[h,n]},​​Exclusions(h0),MinRecursion4]/;NumericQ[h];​​
Take a look at Plots:
In[]:=
nSamples={-1,-0.5,0.,0.1,0.33,1.,2.};​​Plot[goalNum[h,#]&/@nSamples,{h,0.0,3.0}]
Out[]=
0.5
1.0
1.5
2.0
2.5
3.0
5
10
15
Now we may try to find h value that minimize goalNum[h,n] for given n.
​
In[]:=
ClearAll[minForN];​​minForN[n_]:=NMinimize[{goalNum[h,n],0<=h<=5},{h},​​Method{"NelderMead","ShrinkRatio"0.85,"ContractRatio"0.85,"ReflectRatio"2}];​​
In[]:=
minForN[2]​​
Out[]=
{3.21451×
-24
10
,{h1.17137×
-8
10
}}
In[]:=
​​minForN[0.34]
Out[]=
{0.116691,{h0.0749159}}
In[]:=
​​Quiet@minForN[-0.5]
Out[]=
{1.72258,{h0.490799}}