AngouriMath
SolveAPolynomialOverAPowerOfXTimesAnOddHalfPowerOfAQuadratic(AngouriMath.Entity,AngouriMath.Entity.Variable)
Method (no overloads)
Summary
The same reduction with a power of x below: P Q^(m/2)/x^n is
R Q^(k + 1/2)/x^(n - 1) + K_1/sqrt(Q) + K_2/(x sqrt(Q)) , the two remainders the
table's and the linear-beside-the-root rule's.
table's and the linear-beside-the-root rule's.
Remarks
With s = k + 1/2 the derivative of R Q^s/x^(n - 1) is
Q^(s - 1) x^(-n) ((R' x - (n - 1) R) Q + s R Q' x) , and the two remainders
over the sameQ^(s - 1) x^(-n) are K_1 Q^(-k) x^n and
K_2 Q^(-k) x^(n - 1) , polynomials both; the identity is one equation per
power of x, linear in the coefficients ofR and in the two constants.
Stewart'ssqrt(x^2 - a^2)/x^4 and sqrt(a^2 - x^2)/x^2 had no
antiderivative, with the symbol in the radicand; the second is
-sqrt(a^2 - x^2)/x - arcsin(x/a) .
https://github.com/asc-community/AngouriMath/issues/718
over the same
power of x, linear in the coefficients of
Stewart's
antiderivative, with the symbol in the radicand; the second is
https://github.com/asc-community/AngouriMath/issues/718
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