AngouriMath
SolveARationalFunctionBesideTheRootOfAQuadratic(AngouriMath.Entity,AngouriMath.Entity.Variable)
Method (no overloads)
Summary
A polynomial over a product of distinct linears, beside the square root of a
quadratic above or below the bar, taken apart as the rational function it is over
that root:N sqrt(Q)/D is N Q/(D sqrt(Q)) , and P/D is a polynomial
plus a constant over each linear, so the integrand is a polynomial over the root
plus oneK/((x - p) sqrt(Q)) per linear -- the rule before this one's shape,
each. And with a quadratic below the bar, a power of a linear or a power of the root's
own quadratic, a linear over the quadratic or a polynomial over the power for each,
every one of them closed beside the root.
quadratic above or below the bar, taken apart as the rational function it is over
that root:
plus a constant over each linear, so the integrand is a polynomial over the root
plus one
each. And with a quadratic below the bar, a power of a linear or a power of the root's
own quadratic, a linear over the quadratic or a polynomial over the power for each,
every one of them closed beside the root.
Remarks
so this is
half over
the Euler substitution takes the generic first substitution for a symbolic leading
coefficient, with
smaller: a polynomial over the root alone, or one linear with nothing divided out,
is the rule before this one's or the chain's already, and asking again would be
asking the same question. The polynomial part goes to the chain, for
linears are closed.
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