Finding invert for complicated function 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?
 A: 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.
