AngouriMath

Symbolics for .NET


Table of contents

AngouriMath

Navigation
Home Quick start Wiki Almanac Research What's new Try Use Angouri.org GitHub Report a bug Donate Contacts

← Back to list of members

SolveALinearBesideTheRootOfAQuadratic​(AngouriMath.​Entity,​AngouriMath.​Entity.​Variable)

 Method (no overloads)

Summary

A constant over a linear beside the square root of a quadratic,
K/((x - p) sqrt(Q)), by the reciprocal of the linear: with t = 1/(x - p) the root becomes one of a quadratic in t alone, and the table answers that.

Remarks

Q(p + 1/t) = (Q(p) t^2 + Q'(p) t + a)/t^2, so sqrt(Q) = sqrt(R(t))/|t| with R(t) = Q(p) t^2 + Q'(p) t + a, and dx = -dt/t^2: the integrand is
-K sgn(t) dt/sqrt(R(t)), whose integral is the table's, an arcsine or a
logarithm by the sign of Q(p) -- a piecewise where that sign is a symbol's.
The sign of t is the sign of x - p, and the antiderivative is
-K sgn(x - p) G(1/(x - p)) on both sides of p, exactly.
 This is the shape the Euler substitution answers at length, as a partial-fraction
decomposition in t, and declines for a leading coefficient that is a
symbol of the wrong sign: Hearn's 1/(r sqrt(-alpha^2 - epsilon^2 + 2h r^2 - 2k r^4)) is 1/(2u sqrt(-alpha^2 - epsilon^2 + 2h u - 2k u^2)) under u = r^2, with
-2k in front and -alpha^2 - epsilon^2 behind, and neither Euler's first
nor second substitution has a real radical to take. Here Q(0) is
-alpha^2 - epsilon^2, the arcsine arm, and the answer is Rubi's. Closed, and
volunteered at any depth for it: the reciprocal substitution and u = x^2 both
hand it what they make.
 https://github.com/asc-community/AngouriMath/issues/718

























Angouri © 2019-2023 · Project's repo · Site's repo · Octicons · Transparency · 4378 pages online