AngouriMath
SolveAsTheDerivativeOfAFunctionOverAPowerOfALogarithmOrAnInverse(AngouriMath.Entity,AngouriMath.Entity.Variable)
Method (no overloads)
Summary
The derivative of a quotient U/V^k for V a logarithm or an inverse
trigonometric function:(U/V^k)' = U'/V^k - k U V'/V^(k + 1) , so the terms of the
integrand withV^(k + 1) below the bar are -k U V' over it, and
U = -B/(k V') for their sum B . Rubi's
x^3/((1 + a^2 x^2) atan(a x)^3) - 3 x^2/(2 a atan(a x)^2) is
(-x^3/(2 a atan(a x)^2))' , and
1/((c + d x)(c - a + (d - b) x) ln(Q)) + ln(1 - Q)/((a + b x)(c + d x) ln(Q)^2) , for
Q = (a + b x)/(c + d x) , is (-ln(1 - Q)/((b c - a d) ln(Q)))' .
trigonometric function:
integrand with
Remarks
Neither term is elementary on its own, so every split loses it, and by parts on either
leaves the other's question again; both were searches past the budget. The rational part
ofB/(k V') is brought to lowest terms, where it is a constant or a polynomial for
these, and the answer isU/V^k only where its derivative is the whole integrand at
sampled points, the terms with lower powers ofV included: where it is not, the
integrand is no such derivative, and this declines.
https://github.com/asc-community/AngouriMath/issues/233
leaves the other's question again; both were searches past the budget. The rational part
of
these, and the answer is
sampled points, the terms with lower powers of
integrand is no such derivative, and this declines.
https://github.com/asc-community/AngouriMath/issues/233
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