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**Hibachi** So $\displaystyle \int\dfrac{x}{x^2 + 1}dx = \int\dfrac{y}{y^2+1} dy$

$\displaystyle \therefore \;\;\; \dfrac{1}{2}\log(x^2+1) = \dfrac{1}{2}\log(y^2+1)+C$

when $\displaystyle y = 0$, $\displaystyle x = 2$

$\displaystyle \therefore \;\;\; \dfrac{1}{2}\log(3) = C $

$\displaystyle \therefore \;\;\; \dfrac{1}{2}\log(x^2+1) = \dfrac{1}{2}\log(y^2+1)+\log\sqrt{3}.

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The question said in the begening express teh solution as an implicit equation relating $\displaystyle x$ and $\displaystyle y$. That's the way it already is, right?