jmc

Combining the terms on the right-hand side, we have $$\frac{x^2}{x+1}\ge \frac{5x+3}{4(x-1)}.$$Then, moving all the terms to the left-hand side and combining denominators again, we get $$\begin{array}{rl}\frac{x^2}{x+1}-\frac{5x+3}{4(x-1)}& \ge 0\\ \frac{4x^2(x-1)-(x+1)(5x+3)}{(x+1)(x-1)}& \ge 0\\ \frac{4x^3-9x^2-8x-3}{(x+1)(x-1)}& \ge 0.\end{array}$$We try to factor the numerator. Using the rational root theorem to test for rational roots, we see that $x=3$ is a root of $4x^3-9x^2-8x-3.$ Then, doing the polynomial division gives $$4x^3-9x^2-8x-3=(x-3)(4x^2+3x+1),$$so we have $$\frac{(x-3)(4x^2+3x+1)}{(x+1)(x-1)}\ge 0.$$Since $4x^2+3x+1$ has a positive $x^2$ coefficient, and its discriminant is $3^2-4\cdot 4=-7,$ which is negative, it follows that $4x^2+3x+1>0$ for all $x.$ Thus, the above inequality is equivalent to $$f(x)=\frac{x-3}{(x+1)(x-1)}\ge 0.$$We make a sign table for $f(x)$: \begin{array}{c|ccc|c} &$x-3$ &$x+1$ &$x-1$ &$f(x)$ \\ \hline$x<-1$ &$-%%DISP_0%%amp;$-%%DISP_0%%amp;$-%%DISP_0%%amp;$-$\\ [.1cm]$-1<x<1$ &$-%%DISP_0%%amp;$+%%DISP_0%%amp;$-%%DISP_0%%amp;$+$\\ [.1cm]$1<x<3$ &$-%%DISP_0%%amp;$+%%DISP_0%%amp;$+%%DISP_0%%amp;$-$\\ [.1cm]$x>3$ &$+%%DISP_0%%amp;$+%%DISP_0%%amp;$+%%DISP_0%%amp;$+$\\ [.1cm]\end{array}We see that $f(x)>0$ when $-1<x<1$ or $x>3.$ Since the inequality is nonstrict, we also include the values of $x$ such that $f(x)=0,$ that is, only $x=3.$ Therefore, the solution to the inequality is $$x\in \boxed{(-1,1)\cup [3,\infty )}.$$