allerySolve $\dfrac{dy}{dx} = \dfrac{1 + y^2}{1 + x^2}$.
A$\tan^{-1}y = \tan^{-1}x + c$
B$\tan^{-1}y - \tan^{-1}x = 0$
C$y = x + c$
D$y^2 = x^2 + c$
Answer & Solution
Correct answer: A. $\tan^{-1}y = \tan^{-1}x + c$
**Variables separable:** $dy/(1+y^2) = dx/(1+x^2)$.
**Integrating:** $\tan^{-1}y = \tan^{-1}x + c$.
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