Finding invert for complicated function

Question

I'm looking for an invert for this function (when x > 0): wolframalpha

Just wondering if there is any chance it exists..

Maybe some numerical methods could help finding it?

Answer 1

Your function is $f(x) =log((x*x+9)/(x*x))*sqrt((3*x-x*x*atan(3/x))/atan(3/x))/(-2*x*atan(3/x)+6)\\ =\log\left(\dfrac{x^2+9}{x^2}\right)\dfrac{\sqrt{\dfrac{3x-x^2\arctan(3/x)}{\arctan(3/x)}}}{-2x\arctan(3/x)+6}\\ =\log\left(1+\dfrac{9}{x^2}\right)\dfrac{\sqrt{\dfrac{3x}{\arctan(3/x)}-x^2}}{-2x\arctan(3/x)+6}\\ =\log\left(1+(3/x)^2\right)\dfrac{\sqrt{\dfrac{3}{x\arctan(3/x)}-1}}{-2\arctan(3/x)+6/x}\\ =\frac12\log\left(1+(3/x)^2\right)\dfrac{\sqrt{\dfrac{3}{x\arctan(3/x)}-1}}{-\arctan(3/x)+3/x}\\ $

Letting $3/x=y$, so $x = 3/y$,

$f(3/y) =\frac12\log\left(1+y^2\right)\dfrac{\sqrt{\dfrac{y}{\arctan(y)}-1}}{-\arctan(y)+y}\\ =\frac12\log\left(1+y^2\right)\dfrac{\sqrt{\dfrac{y-\arctan(y)}{\arctan(y)}}}{y-\arctan(y)}\\ =\frac12\log\left(1+y^2\right)\sqrt{\dfrac{1}{\arctan(y)(y-\arctan(y))}}\\ =\dfrac{\log\left(1+y^2\right)}{2\sqrt{\arctan(y)(y-\arctan(y))}}\\ $

Letting $y = \tan(z)$,

$f(3/\tan(z)) =\dfrac{\log\left(1+\tan^2(z)\right)}{2\sqrt{z(\tan(z)-z)}}\\ $

Considering that $\tan(z)-z$ can't be inverted (at least I don't think it can), there is, in my opinion, no hope of inverting this.