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 Contacts

← Back to list of members

SolveByExponentialSubstitution​(AngouriMath.​Entity,​AngouriMath.​Entity.​Variable,​System.​Boolean)

 Method (no overloads)

Summary

A rational function of e^(k x), turned into a rational function of one variable
by u = e^(k x).

Remarks

With u = e^(k x) we have dx = du/(k u), and every exponential in the
integrand becomes a whole power of u, so a quotient built from them is a
quotient of polynomials — which the rational integrator answers.
What was missing.1/(1 + e^x) had no antiderivative, and neither did
1/(e^x + e^(-x)). e^x/(1 + e^x) did, which is the shape of the gap: the
general substitution answers an exponential integrand only when the numerator happens
to be the derivative of something in it, and declines the rest for want of anything to
substitute for.
It is also how the hyperbolic functions get integrated, since they are not nodes
here — tanh(x) is built as (e^(2x) - 1)/(e^(2x) + 1) and sech(x) as 2/(e^x + e^(-x)), so both are rational functions of an exponential, and
tanh, coth, sech and csch had no antiderivative at all.
 The answers come out in the exponential rather than as ln(cosh(x)) or
2 arctan(e^x), and unfolded — the rational integrator writes a logarithm as
ln((2ax + b - D)/(2ax + b + D)) and leaves the arithmetic in the coefficients
standing. That is its shape on plain rational integrands too, not something this
rewrite introduces; 1/(x(x+1)) comes out as
ln((2x + 1 - 1)/(2x + 1 + 1)) with nothing exponential in sight.
One k for the whole integrand. The exponents are read as linear in the
variable and their slopes taken together by greatest common divisor, so e^x beside e^(2x) gives u and u^2 under one substitution, and
e^(x/2) beside e^x gives u = e^(x/2). Symbolic slopes are taken
where they are rational multiples of one another. An exponent that is not linear —
e^(x^2), whose integral is not elementary at all — is declined.
No condition is owed.e^(k x) is positive for every real x and
never zero, so the substitution is invertible on the whole line and introduces no
interval of its own; u is a genuine change of variable rather than a branch.
 https://github.com/asc-community/AngouriMath/issues/718

























Angouri © 2019-2026 · Project's repo · Site's repo · Octicons · Transparency · 5057 pages online