AngouriMath
TryReadOverSquareTimesRoot(AngouriMath.Entity,AngouriMath.Entity.Variable,AngouriMath.Entity@,AngouriMath.Entity@,AngouriMath.Entity@)
Method (no overloads)
Summary
The antiderivative of sqrt(a x^2 + b x + c) :
(2ax + b) sqrt(Q) / (4a) + ((4ac - b^2) / (8a)) times the integral of
1/sqrt(Q) -- integration by parts once, leaving the reciprocal form that
IntegrateOverRootOfQuadratic(AngouriMath.Entity,AngouriMath.Entity,AngouriMath.Entity,AngouriMath.Entity,AngouriMath.Entity,AngouriMath.Entity.Variable) already knows.
IntegrateOverRootOfQuadratic(AngouriMath.Entity,AngouriMath.Entity,AngouriMath.Entity,AngouriMath.Entity,AngouriMath.Entity,AngouriMath.Entity.Variable) already knows.
Remarks
Only where the leading coefficient is a number other than zero. With a = 0 this
is the square root of something linear, which the ordinary power rule already
integrates, and dividing by a would not be allowed anyway.
is the square root of something linear, which the ordinary power rule already
integrates, and dividing by a would not be allowed anyway.
Summary
Reads k / (x^2 * sqrt(ax^2 + c)) , giving back k, the radicand and its
constant term. The radicand has to be a quadratic in x with no linear term and
with neither of its two coefficients zero: a zero constant makes the formula
below divide by it, and with a zero a there is no root of x left to speak of.
constant term. The radicand has to be a quadratic in x with no linear term and
with neither of its two coefficients zero: a zero constant makes the formula
below divide by it, and with a zero a there is no root of x left to speak of.
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