Question

Solve the following inequality:
$\frac{x+1}{(x-1{)}^2}<1$

Answer 1

$\frac{x+1}{(x-1{)}^2}<1$

$\frac{x+1}{(x-1{)}^2}-1<0$

$\frac{(x+1)-(x^2-2x+1)}{(x-1{)}^2}<0$

$\frac{-x^2+3x}{(x-1{)}^2}<0$

$\frac{x(x-3)}{(x-1{)}^2}>0$

$xϵ(-\infty ,0)\cup (3,\infty )$