I'm trying to prove a certain integral converges, and for that I need to find the inverse of $f(x) =\dfrac{\ln(x)}{1-x^2}$. I've gotten this far:
$x = \dfrac{\ln(y)}{1-y^2} \rightarrow e^x = e^{\left( \frac{\ln(y)}{1-y^2} \right)} = y^{\frac{1}{1-y^2}}$
But I'm stuck here. I know the function should have an inverse but I just can't find it. Any help will be appreciated, thanks in advance!
I am not an expert. But, I would go something like the following :
$$\begin{align}&\ln x-y+yx^2=0\\ \iff &\ln x^2-2y+2yx^2=0\end{align}$$
Let $\thinspace x^2=e^u$ and $2y=z$, then :
$$ \begin{align}u-z+ze^{u}&=0\\ u+\ln z-z+e^{u+\ln z}&=\ln z\\ \end{align} $$
Then let $\thinspace u+\ln z=t$, you have :
$$ \begin{align}&t+e^t=z+\ln z\\ \implies &e^{t+e^t}=e^{z+\ln z}\\ \implies &e^te^{e^t}=e^{\ln \left(ze^z\right)}\\ \implies &e^t=W\left(ze^z\right)\\ \implies &u=\ln W\left(ze^z\right) -\ln z\\ \implies &u=\ln \frac {W\left(2ye^{2y}\right)}{2y}\end{align} $$
Finally, the inverse function becomes : $x=±\sqrt {e^u}$
$$ \begin{align}±\sqrt {\frac {W\left(2xe^{2x}\right)}{2x}}\thinspace . \end{align} $$