5/x²+2x+1=2/1-x²+1/x-1 помогите решить

5/x²+2x+1=2/1-x²+1/x-1
$\frac{5}{x^2+2x+1} = \frac{2}{1-x^2} + \frac{1}{x-1} \\ \\ \frac{5}{(x+1)^2} - \frac{2}{1-x^2} - \frac{1}{x-1} =0\\ \\ \frac{5}{(x+1)^2} + \frac{2}{x^2-1} - \frac{1}{x-1} =0\\ \\ \frac{5}{(x+1)^2} + \frac{2}{(x-1)(x+1)} - \frac{1}{x-1} =0\\ \\ \frac{5(x-1)+2(x+1)-(x+1)^2}{(x+1)^2(x-1)} =0\\ \\ \frac{5x-5+2x+2-x^2-2x-1}{(x+1)^2(x-1)} =0\\ \\ \frac{-x^2+5x-4}{(x+1)^2(x-1)} =0\\ \\ \frac{x^2-5x+4}{(x+1)^2(x-1)} =0$
$\frac{(x-4)(x-1)}{(x+1)^2(x-1)} =0 \\ \\ \frac{x-4}{(x+1)^2} =0 \\ \\ x=4$