>
> Keep in mind that indefinite integrals are only defined up to a
> constant.
>
My math teacher just reminded me to definitely not disagree on that ;)
Let me pls. follow up in a different way. We can do
a,b = sp.symbols('a b')
r1=u.subs({x:a,y:sp.Integer(2),G:sp.Integer(1)}) -
u.subs({x:b,y:sp.Integer(2),G:sp.Integer(1)})
r2=v.subs({x:a})-v.subs({x:b})
With u and v defined as in my original post.
Mathematically r1 and r2 are the same.
Do you know what kind of simplification I would have to apply to either of
the two expressions r1,r2, to make SymPy actually also tell me that?
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