jmc

Rewriting the right-hand side under a common denominator, we have $$\frac{1}{x^2+1}>\frac{30+17x}{10x}.$$Then we can write $$\frac{1}{x^2+1}-\frac{30+17x}{10x}>0,$$or $$\frac{-17x^3-30x^2-7x-30}{10x(x^2+1)}>0.$$Multiplying both sides by $-10$ and flipping the inequality sign, we get $$\frac{17x^3+30x^2+7x+30}{x(x^2+1)}<0.$$Looking for rational roots of the numerator, we see that $x=-2$ makes the numerator zero, so $x+2$ is a factor, by the factor theorem. Doing the polynomial division, we have $$17x^3+30x^2+7x+30=(x+2)(17x^2-4x+15),$$so $$\frac{(x+2)(17x^2-4x+15)}{x(x^2+1)}<0.$$Since $x^2+1$ is positive for all real numbers $x$, it does not affect the sign on the left side. Similarly, since $y=17x^2-4x+15$ is the graph of a parabola that opens upward, and its disciminant is $4^2-4\cdot 17\cdot 15,$ which is negative, we see that $17x^2-4x+15>0$ for all $x.$ Therefore, the given inequality is equivalent to $$\frac{x+2}{x}<0.$$Letting $f(x)=\frac{x+2}{x},$ we construct a sign table: \begin{array}{c|cc|c} &$x+2$ &$x$ &$f(x)$ \\ \hline$x<-2$ &$-%%DISP_0%%amp;$-%%DISP_0%%amp;$+$\\ [.1cm]$-2<x<0$ &$+%%DISP_0%%amp;$-%%DISP_0%%amp;$-$\\ [.1cm]$x>0$ &$+%%DISP_0%%amp;$+%%DISP_0%%amp;$+$\\ [.1cm]\end{array}Therefore, $f(x)<0$ when $x\in \boxed{(-2,0)}.$