Solving equation with $\arctan(x)$ and $\ln(x)$ Last week my group had math exam and this questions popped up. I approximated $x$ using Newton-Rhapson method ($x\approx 0.341$). However, after the exam professor told us that we should have found an exact solution. I tried however; I couldn't make it with methods I know of, that's why I ask for your help.
Solve for $x$:
$$\frac{\arctan(x)}{x} + \frac{\ln(x)}{x^2+1} = 0$$
Thanks upfront for help.
 A: This equation is of a form that cannot be solved by only applying only a finite number of elementary functions (= elementary operations).
The left-hand side of the equation is the function term of an elementary function $f$. The elementary functions (according to Liouville and Ritt) are the functions of one complex variable that are generated by applying finite numbers of $\exp$, $\ln$ and/or algebraic functions.
Transforming the equation into its $\exp$-$\ln$-form yields an algebraic equation of $x$, $\ln(x)$ and $\ln(\frac{i+x}{i-x})$.
They don't have algebraic points in common.
Assuming the (until now unproven) Schanuel conjecture is true, $x$, $\ln(x)$ and $\ln(\frac{i+x}{i-x})$ are algebraically independent.
According to a theorem of Ritt (1925) that is proven also by Risch (1979), $f$ therefore cannot have an inverse that is an elementary function. Therefore you cannot solve the equation by applying the inverses of the elementary functions / operations you can read from the equation.
We can assume that your equation also cannot have solutions in the elementary numbers (Chow 1999).
MAPLE, MATHEMATICA and WolframAlpha couldn't find a solution with special functions either.
A: Only a partial solution:
I will suppose we are looking for $x$ in the interval $I=(0;+\infty)$.
It is at least easy to show that the system has a unique solution. Consider $f$ and $g$ defined by $f(x)=(x^2+1)\arctan(x)$ and $g(x)=-x\ln(x)$. Then the system is equivalent to $f(x)=g(x)$.
Furthermore, $f''(x)=2\arctan(x)+\dfrac{2x}{x^2+1}$ and $g''(x)=-\dfrac1x$ so that $f$ and $g$ are both respectively strictly convex and strictly concave on $I$. Hence, the equation has at most two solutions on $I$. But, as $\lim\limits_{x \to 0} g(x) = \lim\limits_{x \to 0} f(x) = 0$, there is in fact only one solution in I.
Now, considering $h$, defined on $I$ by $h(x)=\arctan(x)\ln(x)$. Then we see that: $$h'(x)=\dfrac{\arctan{x}}x+\dfrac{\ln{x}}{x^2+1}$$
Hence, the system is equivalent to $h'(x)=0$.
But as $\lim\limits_{x \to 0} h(x) = h(1) = 0$, by Rolle's theorem, we know that the equation has a solution in $(0;1)$.
Furthermore, as $h$ is strictly increasing on $[1;+\infty)$, we can see that this solution corresponds to a global minimum of $h$.
Hence, we can show the existence and uniqueness of the solution and that it is equal to:
$$a=argmin_{x\in I}h(x)$$
