<s>[INST] Justify the solution of $x^2-x+\arctan{x}=0$ It is obvious that a solution of  $x^2-x+\arctan{x}=0$ is $x=0$, but I would like you to show me how this can be derived more conceptually than by plugging in $0$, and how can one prove that such solution is unique.
 [/INST] like you conjectured, $x = 0$ is the only solution.  i will first show that there are no positive solutions.
$$\arctan x = \int_0^x {dt \over 1 + t^2} = \int_0^x \left( 1 - t^2 + {t^4 \over 1 + t^2} \right) dt = x -{1 \over 3}x^3 + \int_0^x {t^4 \over 1 + t^4}dt$$
for $x > 0, \arctan x > x - {1 \over 3}x^3,$ therefore
$$x^2 - x + \arctan x > x^2 + {1 \over 3}x^3 > 0$$ the last inequality establishes the claim for $x > 0$
now for the case $x < 0$
$$\arctan x = \int_0^x {dt \over 1 + t^2} = \int_0^x \left( 1  - {t^2 \over 1 + t^2} \right) dt = x - \int_0^x {t^2 \over 1 + t^4}dt = x + \int_x^0 {t^2 \over 1 + t^4}dt$$
$$x^2 - x + \arctan x = x^2 + \int_x^0 {t^2 \over 1 + t^4}dt > x^2 > 0$$ the last inequality establishes the claim for $x < 0$
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