Answer
$x=\dfrac{-1+ i\sqrt {3}}{2},\dfrac{-1 - i\sqrt {3}}{2}$
Work Step by Step
Given: $\dfrac{x}{x+1}=1+x$
The equation is not an identity.
$\dfrac{x}{x+1}=1+x$
This gives: $x=(1+x)^2$
$\implies x^2-x+1=0$
This implies that
$x=\dfrac{-1\pm \sqrt {1-4}}{2}=\dfrac{-1\pm i\sqrt {3}}{2}$
Thus, $x=\dfrac{-1+ i\sqrt {3}}{2},\dfrac{-1 - i\sqrt {3}}{2}$
