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

 Method (no overloads)

Summary

An integrand that is a function of ln(x) and of nothing else, integrated by the
substitution u = ln(x), under which dx is e^u du.

Remarks

sin(ln(x)) had no antiderivative, and it is one substitution away from one that
does: under u = ln(x) it becomes e^u sin(u), which is the cyclic
by-parts integral and is closed by
SolveAPolynomialTimesAnExponentialAndATrigonometric(AngouriMath.Entity,AngouriMath.Entity.Variable). The same for
ln(x)^n beside anything, and for e^(1/ln(x)) and its kin.
The substitution is an inverse one, which is what makes it different from every
other substitution here and why the general one does not find it. The others divide by
du/dx and ask what is left; this one goes the other way — x = e^u, so
dx is e^u du, and the exponential is introduced rather than
cancelled. That only pays because the integrator answers exponentials times almost
anything.
The test is the rewrite itself: replace every ln(x) and see whether an
x survives. ln(x) + x keeps one and is declined, which is right — it is
not a function of the logarithm alone. A different base is read through first, since
log(b, x) is ln(x)/ln(b) and declining it for the spelling would be the
defect this file has had repeatedly.
Where the answer holds.u = ln(x) is a bijection from the positive reals
to the whole line, so the answer is an antiderivative for x > 0 — which is
where an integrand built from ln(x) is real in the first place. Nothing is
assumed that the integrand did not already assume.
 https://github.com/asc-community/AngouriMath/issues/718

























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